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Solving problems which involve forces, friction, and Newton's Laws: A step-by-step guide

This step-by-step guide is meant to show you how to approach problems where you have to deal with moving objects subject to friction and other forces, and you need to apply Newton's Laws. We will go through many problems, so you can have a clear idea of the process involved in solving them.

The problems we will examine include objects that

  • are pushed/pulled horizontally with an angle
  • move up or down an incline
  • hang from ropes attached to the ceiling
  • hang from ropes that run over pulleys
  • move connected by a string
  • are pushed in contact with each other (Coming soon!)
  • Box pulled at an angle over a horizontal surface
  • Block pushed over the floor with a downward and forward force
  • Object moving at constant velocity over a horizontal surface
  • Block pushed up a frictionless ramp
  • Mass pulled up an incline with friction
  • A mass hanging from two ropes
  • Two hanging objects connected by a rope
  • Two masses on a pulley
  • Two blocks connected by a string are pulled horizontally

Dynamics: Force and Newton’s Laws of Motion

Problem-solving strategies, learning objective.

By the end of this section, you will be able to:

  • Understand and apply a problem-solving procedure to solve problems using Newton’s laws of motion.

Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy for Newton’s Laws of Motion

Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation . Such a sketch is shown in Figure 1(a). Then, as in Figure 1(b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent (whenever sufficient information exists).

Figure 1. (a) A sketch of Tarzan hanging from a vine. (b) Arrows are used to represent all forces. T is the tension in the vine above Tarzan, F T is the force he exerts on the vine, and w is his weight. All other forces, such as the nudge of a breeze, are assumed negligible. (c) Suppose we are given the ape man’s mass and asked to find the tension in the vine. We then define the system of interest as shown and draw a free-body diagram. F T is no longer shown, because it is not a force acting on the system of interest; rather, F T  acts on the outside world. (d) Showing only the arrows, the head-to-tail method of addition is used. It is apparent that T = –w , if Tarzan is stationary.

Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and unknowns. Then carefully determine the system of interest . This decision is a crucial step, since Newton’s second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 1(c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in clearly defining systems will be beneficial in later chapters as well.

A diagram showing the system of interest and all of the external forces is called a free-body diagram . Only forces are shown on free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure 1(c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem . This is done in Figure 1(d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known.

Applying Newton’s Second Law

image

F net x  = ma ,

F net y = 0.

You will need this information in order to determine unknown forces acting in a system.

Step 4. As always, check the solution to see whether it is reasonable . In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake.

Section Summary

To solve problems involving Newton’s laws of motion, follow the procedure described:

  • Draw a sketch of the problem.
  • Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions from the dot. If vectors act in directions that are not horizontal or vertical, resolve the vectors into horizontal and vertical components and draw them on the free-body diagram.
  • Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the x-direction) then  F net x  = 0 . If the object does accelerate in that direction,  F net x  = ma .
  • Check your answer. Is the answer reasonable? Are the units correct?

Problems & Exercises

1. A 5.00 × 10 5 -kg rocket is accelerating straight up. Its engines produce 1.250 × 10 7  of thrust, and air resistance is 4.50 × 10 6 N. What is the rocket’s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

2. The wheels of a midsize car exert a force of 2100 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 250 N and the acceleration of the car is 1.80 m/s 2 , what is the mass of the car plus its occupants? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. For this situation, draw a free-body diagram and write the net force equation.

3. Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

4. When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

5. A freight train consists of two 8.00 × 10 4   engines and 45 cars with average masses of 5.50 × 10 4 kg . (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00 × 10 -2  if the force of friction is 7.50 × 10 5 , assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

6. Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. (a) An 1800-kg tractor exerts a force of 1.75 × 10 5  backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is 0.150 m/s 2 , what is the mass of the airplane? (b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane. (c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each.

7. A 1100-kg car pulls a boat on a trailer. (a) What total force resists the motion of the car, boat, and trailer, if the car exerts a 1900-N force on the road and produces an acceleration of 0.550 m/s 2 ? The mass of the boat plus trailer is 700 kg. (b) What is the force in the hitch between the car and the trailer if 80% of the resisting forces are experienced by the boat and trailer?

8. (a) Find the magnitudes of the forces F 1 and F 2  that add to give the total force F tot  shown in Figure 4. This may be done either graphically or by using trigonometry. (b) Show graphically that the same total force is obtained independent of the order of addition of   F 1 and F 2 . (c) Find the direction and magnitude of some other pair of vectors that add to give F tot . Draw these to scale on the same drawing used in part (b) or a similar picture.

A right triangle is shown made up of three vectors. The first vector, F sub one, is along the triangle’s base toward the right; the second vector, F sub two, is along the perpendicular side pointing upward; and the third vector, F sub tot, is along the hypotenuse pointing up the incline. The magnitude of F sub tot is twenty newtons. In a free-body diagram, F sub one is shown by an arrow pointing right and F sub two is shown by an arrow acting vertically upward.

9. Two children pull a third child on a snow saucer sled exerting forces  F 1 and F 2 as shown from above in Figure 4 . Find the acceleration of the 49.00-kg sled and child system. Note that the direction of the frictional force is unspecified; it will be in the opposite direction of the sum of  F 1 and F 2 .

An overhead view of a child sitting on a snow saucer sled. Two forces, F sub one equal to ten newtons and F sub two equal to eight newtons, are acting toward the right. F sub one makes an angle of forty-five degrees from the x axis and F sub two makes an angle of thirty degrees from the x axis in a clockwise direction. A friction force f is equal to seven point five newtons, shown by a vector pointing in negative x direction. In the free-body diagram, F sub one and F sub two are shown by arrows toward the right, making a forty-five degree angle above the horizontal and a thirty-degree angle below the horizontal respectively. The friction force f is shown by an arrow along the negative x axis.

10. Suppose your car was mired deeply in the mud and you wanted to use the method illustrated in Figure 6 to pull it out. (a) What force would you have to exert perpendicular to the center of the rope to produce a force of 12,000 N on the car if the angle is 2.00°? In this part, explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. (b) Real ropes stretch under such forces. What force would be exerted on the car if the angle increases to 7.00° and you still apply the force found in part (a) to its center?

Figure of car stuck in the mud and a rope connected to a tree trunk in an attempt to pull out the car.

11. What force is exerted on the tooth in Figure 7 if the tension in the wire is 25.0 N? Note that the force applied to the tooth is smaller than the tension in the wire, but this is necessitated by practical considerations of how force can be applied in the mouth. Explicitly show how you follow steps in the Problem-Solving Strategy for Newton’s laws of motion.

Cross-section of jaw with sixteen teeth is shown. Braces are along the outside of the teeth. Three forces are acting on the protruding tooth. The applied force, F sub app, is shown by an arrow vertically downward; a second force, T, is shown by an arrow making an angle of fifteen degrees below the positive x axis; and a third force, T, is shown by an arrow making an angle of fifteen degrees below the negative x axis.

Figure 7. Braces are used to apply forces to teeth to realign them. Shown in this figure are the tensions applied by the wire to the protruding tooth. The total force applied to the tooth by the wire, F app , points straight toward the back of the mouth.

12. Figure 9 shows Superhero and Trusty Sidekick hanging motionless from a rope. Superhero’s mass is 90.0 kg, while Trusty Sidekick’s is 55.0 kg, and the mass of the rope is negligible. (a) Draw a free-body diagram of the situation showing all forces acting on Superhero, Trusty Sidekick, and the rope. (b) Find the tension in the rope above Superhero. (c) Find the tension in the rope between Superhero and Trusty Sidekick. Indicate on your free-body diagram the system of interest used to solve each part.

Two caped superheroes hang on a rope suspended vertically from a bar.

Figure 9. Superhero and Trusty Sidekick hang motionless on a rope as they try to figure out what to do next. Will the tension be the same everywhere in the rope?

13. A nurse pushes a cart by exerting a force on the handle at a downward angle 35.0º below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is 60.0 N. (a) Draw a free-body diagram for the system of interest. (b) What force must the nurse exert to move at a constant velocity?

14. Construct Your Own Problem Consider the tension in an elevator cable during the time the elevator starts from rest and accelerates its load upward to some cruising velocity. Taking the elevator and its load to be the system of interest, draw a free-body diagram. Then calculate the tension in the cable. Among the things to consider are the mass of the elevator and its load, the final velocity, and the time taken to reach that velocity.

15. Construct Your Own Problem Consider two people pushing a toboggan with four children on it up a snow-covered slope. Construct a problem in which you calculate the acceleration of the toboggan and its load. Include a free-body diagram of the appropriate system of interest as the basis for your analysis. Show vector forces and their components and explain the choice of coordinates. Among the things to be considered are the forces exerted by those pushing, the angle of the slope, and the masses of the toboggan and children.

16. Unreasonable Results (a) Repeat Exercise 7, but assume an acceleration of 1.20 m/s 2  is produced. (b) What is unreasonable about the result? (c) Which premise is unreasonable, and why is it unreasonable?

17. Unreasonable Results (a) What is the initial acceleration of a rocket that has a mass of 1.50 × 10 6  at takeoff, the engines of which produce a thrust of 2.00 × 10 6 ? Do not neglect gravity. (b) What is unreasonable about the result? (This result has been unintentionally achieved by several real rockets.) (c) Which premise is unreasonable, or which premises are inconsistent? (You may find it useful to compare this problem to the rocket problem earlier in this section.)

Selected Solutions to Problems & Exercises

1. Using the free-body diagram:

An object of mass m is shown. Three forces acting on it are tension T, shown by an arrow acting vertically upward, and friction f and gravity m g, shown by two arrows acting vertically downward.

  • [latex]{F}_{\text{net}}=T-f-mg=\text{ma}\\[/latex] ,

[latex]a=\frac{T-f-\text{mg}}{m}=\frac{1\text{.}\text{250}\times {\text{10}}^{7}\text{N}-4.50\times {\text{10}}^{\text{6}}N-\left(5.00\times {\text{10}}^{5}\text{kg}\right)\left(9.{\text{80 m/s}}^{2}\right)}{5.00\times {\text{10}}^{5}\text{kg}}=\text{6.20}{\text{m/s}}^{2}\\[/latex]

3. Use Newton’s laws of motion.

Two forces are acting on an object of mass m: F, shown by an arrow pointing upward, and its weight w, shown by an arrow pointing downward. Acceleration a is represented by a vector arrow pointing upward. The figure depicts the forces acting on a high jumper.

[latex]F=\left(\text{70.0 kg}\right)\left[\left(\text{39}\text{.}{\text{2 m/s}}^{2}\right)+\left(9\text{.}{\text{80 m/s}}^{2}\right)\right]\\[/latex] [latex]=3.\text{43}\times {\text{10}}^{3}\text{N}\\[/latex].  The force exerted by the high-jumper is actually down on the ground, but F is up from the ground and makes him jump.

  • This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of 10 3 N.

5. (a) 4.41 × 10 5 N (b) 1.50 × 10 5 N

7. (a) 910 N (b) 1.11 × 10 3

9. (a) a = 0.139 m/s, θ = 12.4º

11. Use Newton’s laws since we are looking for forces.

  • Draw a free-body diagram:

A horizontal dotted line with two vectors extending downward from the mid-point of the dotted line, both at angles of fifteen degrees. A third vector points straight downward from the intersection of the first two angles, bisecting them; it is perpendicular to the dotted line.

  • The tension is given as T = 25.0 N. Find F app . Using Newton’s laws gives:[latex]\sigma{F}_{y}=0\\[/latex], so that applied force is due to the y -components of the two tensions: F app = 2 T  sin  θ  = 2(25.0 N) sin(15º) = 12.9 N The x -components of the tension cancel. [latex]\sum{F}_{x}=0\\[/latex].
  • This seems reasonable, since the applied tensions should be greater than the force applied to the tooth.
  • College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License
  • 1.7 Solving Problems in Physics
  • Introduction
  • 1.1 The Scope and Scale of Physics
  • 1.2 Units and Standards
  • 1.3 Unit Conversion
  • 1.4 Dimensional Analysis
  • 1.5 Estimates and Fermi Calculations
  • 1.6 Significant Figures
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Scalars and Vectors
  • 2.2 Coordinate Systems and Components of a Vector
  • 2.3 Algebra of Vectors
  • 2.4 Products of Vectors
  • 3.1 Position, Displacement, and Average Velocity
  • 3.2 Instantaneous Velocity and Speed
  • 3.3 Average and Instantaneous Acceleration
  • 3.4 Motion with Constant Acceleration
  • 3.5 Free Fall
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.3 Projectile Motion
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.1 Solving Problems with Newton’s Laws
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.1 Conditions for Static Equilibrium
  • 12.2 Examples of Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.6 Bernoulli’s Equation
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
  • 16.2 Mathematics of Waves
  • 16.3 Wave Speed on a Stretched String
  • 16.4 Energy and Power of a Wave
  • 16.5 Interference of Waves
  • 16.6 Standing Waves and Resonance
  • 17.1 Sound Waves
  • 17.2 Speed of Sound
  • 17.3 Sound Intensity
  • 17.4 Normal Modes of a Standing Sound Wave
  • 17.5 Sources of Musical Sound
  • 17.7 The Doppler Effect
  • 17.8 Shock Waves
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

  • Describe the process for developing a problem-solving strategy.
  • Explain how to find the numerical solution to a problem.
  • Summarize the process for assessing the significance of the numerical solution to a problem.

Problem-solving skills are clearly essential to success in a quantitative course in physics. More important, the ability to apply broad physical principles—usually represented by equations—to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in this text and for applying physics in everyday life.

As you are probably well aware, a certain amount of creativity and insight is required to solve problems. No rigid procedure works every time. Creativity and insight grow with experience. With practice, the basics of problem solving become almost automatic. One way to get practice is to work out the text’s examples for yourself as you read. Another is to work as many end-of-section problems as possible, starting with the easiest to build confidence and then progressing to the more difficult. After you become involved in physics, you will see it all around you, and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text.

Although there is no simple step-by-step method that works for every problem, the following three-stage process facilitates problem solving and makes it more meaningful. The three stages are strategy, solution, and significance. This process is used in examples throughout the book. Here, we look at each stage of the process in turn.

Strategy is the beginning stage of solving a problem. The idea is to figure out exactly what the problem is and then develop a strategy for solving it. Some general advice for this stage is as follows:

  • Examine the situation to determine which physical principles are involved . It often helps to draw a simple sketch at the outset. You often need to decide which direction is positive and note that on your sketch. When you have identified the physical principles, it is much easier to find and apply the equations representing those principles. Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws of nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical solution is meaningless.
  • Make a list of what is given or can be inferred from the problem as stated (identify the “knowns”) . Many problems are stated very succinctly and require some inspection to determine what is known. Drawing a sketch can be very useful at this point as well. Formally identifying the knowns is of particular importance in applying physics to real-world situations. For example, the word stopped means the velocity is zero at that instant. Also, we can often take initial time and position as zero by the appropriate choice of coordinate system.
  • Identify exactly what needs to be determined in the problem (identify the unknowns) . In complex problems, especially, it is not always obvious what needs to be found or in what sequence. Making a list can help identify the unknowns.
  • Determine which physical principles can help you solve the problem . Since physical principles tend to be expressed in the form of mathematical equations, a list of knowns and unknowns can help here. It is easiest if you can find equations that contain only one unknown—that is, all the other variables are known—so you can solve for the unknown easily. If the equation contains more than one unknown, then additional equations are needed to solve the problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems it is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may have to use two (or more) different equations to get the final answer.

The solution stage is when you do the math. Substitute the knowns (along with their units) into the appropriate equation and obtain numerical solutions complete with units . That is, do the algebra, calculus, geometry, or arithmetic necessary to find the unknown from the knowns, being sure to carry the units through the calculations. This step is clearly important because it produces the numerical answer, along with its units. Notice, however, that this stage is only one-third of the overall problem-solving process.

Significance

After having done the math in the solution stage of problem solving, it is tempting to think you are done. But, always remember that physics is not math. Rather, in doing physics, we use mathematics as a tool to help us understand nature. So, after you obtain a numerical answer, you should always assess its significance:

  • Check your units. If the units of the answer are incorrect, then an error has been made and you should go back over your previous steps to find it. One way to find the mistake is to check all the equations you derived for dimensional consistency. However, be warned that correct units do not guarantee the numerical part of the answer is also correct.
  • Check the answer to see whether it is reasonable. Does it make sense? This step is extremely important: –the goal of physics is to describe nature accurately. To determine whether the answer is reasonable, check both its magnitude and its sign, in addition to its units. The magnitude should be consistent with a rough estimate of what it should be. It should also compare reasonably with magnitudes of other quantities of the same type. The sign usually tells you about direction and should be consistent with your prior expectations. Your judgment will improve as you solve more physics problems, and it will become possible for you to make finer judgments regarding whether nature is described adequately by the answer to a problem. This step brings the problem back to its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able to solve a problem mechanically.
  • Check to see whether the answer tells you something interesting. What does it mean? This is the flip side of the question: Does it make sense? Ultimately, physics is about understanding nature, and we solve physics problems to learn a little something about how nature operates. Therefore, assuming the answer does make sense, you should always take a moment to see if it tells you something about the world that you find interesting. Even if the answer to this particular problem is not very interesting to you, what about the method you used to solve it? Could the method be adapted to answer a question that you do find interesting? In many ways, it is in answering questions such as these that science progresses.

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  • Authors: William Moebs, Samuel J. Ling, Jeff Sanny
  • Publisher/website: OpenStax
  • Book title: University Physics Volume 1
  • Publication date: Sep 19, 2016
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
  • Section URL: https://openstax.org/books/university-physics-volume-1/pages/1-7-solving-problems-in-physics

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High School Physics : Calculating Force

Study concepts, example questions & explanations for high school physics, all high school physics resources, example questions, example question #1 : net force.

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Plug these into the equation to solve for acceleration.

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Example Question #2 : Calculating Force

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Example Question #3 : Calculating Force

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(Assume the only two forces acting on the object are friction and Derek).

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Plug in the information we've been given so far to find the force of friction.

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Friction will be negative because it acts in the direction opposite to the force of Derek.

Example Question #5 : Calculating Force

problem solving in force physics

Newton's third law states that when one object exerts a force on a second object, the second object exerts a force equal in size, but opposite in direction to the first. That means that the force of the hammer on the nail and the nail on the hammer will be equal in size, but opposite in direction.

problem solving in force physics

Example Question #6 : Calculating Force

problem solving in force physics

We can find the net force by adding the individual force together.

problem solving in force physics

Example Question #7 : Calculating Force

problem solving in force physics

If the object has a constant velocity, that means that the net acceleration must be zero.

problem solving in force physics

In conjunction with Newton's second law, we can see that the net force is also zero. If there is no net acceleration, then there is no net force.

problem solving in force physics

Since Franklin is lifting the weight vertically, that means there will be two force acting upon the weight: his lifting force and gravity. The net force will be equal to the sum of the forces acting on the weight.

problem solving in force physics

We know the mass of the weight and we know the acceleration, so we can solve for the lifting force.

problem solving in force physics

Example Question #8 : Calculating Force

problem solving in force physics

We are given the mass, but we will need to calculate the acceleration to use in the formula.

problem solving in force physics

Plug in our given values and solve for acceleration.

problem solving in force physics

Now we know both the acceleration and the mass, allowing us to solve for the force.

problem solving in force physics

Example Question #9 : Calculating Force

problem solving in force physics

We can calculate the gravitational force using the mass.

problem solving in force physics

Example Question #10 : Calculating Force

problem solving in force physics

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6 Applications of Newton’s Laws

6.1 Solving Problems with Newton’s Laws

Learning objectives.

By the end of the section, you will be able to:

  • Apply problem-solving techniques to solve for quantities in more complex systems of forces
  • Use concepts from kinematics to solve problems using Newton’s laws of motion
  • Solve more complex equilibrium problems
  • Solve more complex acceleration problems
  • Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy: Applying Newton’s Laws of Motion

  • Identify the physical principles involved by listing the givens and the quantities to be calculated.
  • Sketch the situation, using arrows to represent all forces.
  • Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
  • Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
  • Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure (a). Then, as in Figure (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

This figure shows the development of the free body diagram of a piano being lifted and passed through a window. Figure a is a sketch showing the piano hanging from a crane and part way through a window. Figure b identifies the forces. It shows the same sketch with the addition of the forces, represented as labeled vector arrows. Vector T points up, vector F sub T points down, vector w points down. Figure c defines the system of interest. The sketch is shown again with the piano circled and identified as the system of interest. Only vectors T up and w down are included in this diagram. The downward force F sub T is not a force on the system of interest since it is exerted on the outside world. It must be omitted from the free body diagram. The free body diagram is shown as well. It consists of a dot, representing the system of interest, and the vectors T pointing up and w pointing down, with their tails at the dot. Figure d shows the addition of the forces. Vectors T and w are shown. We are told that these forces must be equal and opposite since the net external force is zero. Thus T is equal to minus w.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set [latex]{a}_{y}=0.[/latex]) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Different Tensions at Different Angles

Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in Figure . Find the tension in each wire, neglecting the masses of the wires.

A sketch of a traffic light suspended from two wires supported by two poles is shown. (b) Some forces are shown in this system. Tension T sub one pulling the top of the left-hand pole is shown by the vector arrow along the left wire from the top of the pole, and an equal but opposite tension T sub one is shown by the arrow pointing up along the left-hand wire where it is attached to the light; the left-hand wire makes a thirty-degree angle with the horizontal. Tension T sub two is shown by a vector arrow pointing downward from the top of the right-hand pole along the right-hand wire, and an equal but opposite tension T sub two is shown by the arrow pointing up along the right-hand wire, which makes a forty-five degree angle with the horizontal. The traffic light is suspended at the lower end of the wires, and its weight W is shown by a vector arrow acting downward. (c) The traffic light is the system of interest, indicated by circling the traffic light. Tension T sub one starting from the traffic light is shown by an arrow along the wire making an angle of thirty degrees with the horizontal. Tension T sub two starting from the traffic light is shown by an arrow along the wire making an angle of forty-five degrees with the horizontal. The weight W is shown by a vector arrow pointing downward from the traffic light. A free-body diagram is shown with three forces acting on a point. Weight W acts downward; T sub one and T sub two act at an angle with the vertical. A coordinate system is shown, with positive x to the right and positive y upward. (d) Forces are shown with their components. T sub one is decomposed into T sub one y pointing vertically upward and T sub one x pointing along the negative x direction. The angle between T sub one and T sub one x is thirty degrees. T sub two is decomposed into T sub two y pointing vertically upward and T sub two x pointing along the positive x direction. The angle between T sub two and T sub two x is forty five degrees. Weight W is shown by a vector arrow acting downward. (e) The net vertical force is zero, so the vector equation is T sub one y plus T sub two y equals W. T sub one y and T sub two y are shown on a free body diagram as equal length arrows pointing up. W is shown as a downward pointing arrow whose length is twice as long as each of the T sub one y and T sub two y arrows. The net horizontal force is zero, so vector T sub one x is equal to minus vector T sub two x. T sub two x is shown by an arrow pointing toward the right, and T sub one x is shown by an arrow pointing toward the left.

The system of interest is the traffic light, and its free-body diagram is shown in Figure (c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in Figure (d). There are two unknowns in this problem ([latex]{T}_{1}[/latex] and [latex]{T}_{2}[/latex]), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.

First consider the horizontal or x -axis:

Thus, as you might expect,

This gives us the following relationship:

Note that [latex]{T}_{1}[/latex] and [latex]{T}_{2}[/latex] are not equal in this case because the angles on either side are not equal. It is reasonable that [latex]{T}_{2}[/latex] ends up being greater than [latex]{T}_{1}[/latex] because it is exerted more vertically than [latex]{T}_{1}.[/latex]

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for [latex]{T}_{2}[/latex] in terms of [latex]{T}_{1}[/latex] reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of [latex]{T}_{1}[/latex] to be

Finally, we find the magnitude of [latex]{T}_{2}[/latex] by using the relationship between them, [latex]{T}_{2}=1.225{T}_{1}[/latex], found above. Thus we obtain

Significance

Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion .

Particle Acceleration

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Drag Force on a Barge

Two tugboats push on a barge at different angles ( Figure ). The first tugboat exerts a force of [latex]2.7\times {10}^{5}\,\text{N}[/latex] in the x -direction, and the second tugboat exerts a force of [latex]3.6\times {10}^{5}\,\text{N}[/latex] in the y -direction. The mass of the barge is [latex]5.0\times {10}^{6}\,\text{kg}[/latex] and its acceleration is observed to be [latex]7.5\times {10}^{-2}\,{\text{m/s}}^{2}[/latex] in the direction shown. What is the drag force of the water on the barge resisting the motion? ( Note: Drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object. Since the barge is flat bottomed, we can assume that the drag force is in the direction opposite of motion of the barge.)

(a) A view from above of two tugboats pushing on a barge. One tugboat is pushing with the force F sub 1 equal to two point seven times by ten to the five newtons, shown by a vector arrow acting toward the right in the x direction. Another tugboat is pushing with a force F sub 2 equal to three point six times by ten to the five newtons acting upward in the positive y direction. Acceleration of the barge, a, is shown by a vector arrow directed fifty-three point one degree angle above the x axis. In the free-body diagram, the mass is represented by a point, F sub 2 is acting upward on the point, F sub 1 is acting toward the right, and F sub D is acting approximately southwest. (b) The vectors F sub 1 and F sub 2 are the sides of a right triangle. The resultant is the hypotenuse of this triangle, vector F sub app, making a fifty-three point one degree angle from the base vector F sub 1. The vector F sub app plus the vector force F sub D, pointing down the incline, is equal to the force vector F sub net, which points up the incline.

The directions and magnitudes of acceleration and the applied forces are given in Figure (a). We define the total force of the tugboats on the barge as [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] so that

The drag of the water [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] is in the direction opposite to the direction of motion of the boat; this force thus works against [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] as shown in the free-body diagram in Figure (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x – and y -axes are in the same direction as [latex]{\mathbf{\overset{\to }{F}}}_{1}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{2}.[/latex] The problem quickly becomes a one-dimensional problem along the direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex], since friction is in the direction opposite to [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}.[/latex] Our strategy is to find the magnitude and direction of the net applied force [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] and then apply Newton’s second law to solve for the drag force [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}.[/latex]

Since [latex]{F}_{x}[/latex] and [latex]{F}_{y}[/latex] are perpendicular, we can find the magnitude and direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] directly. First, the resultant magnitude is given by the Pythagorean theorem:

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] is in the opposite direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] but its magnitude is slightly less than [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}.[/latex] The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water [latex]{F}_{\text{D}}[/latex] in terms of known quantities:

Substituting known values gives

The direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] has already been determined to be in the direction opposite to [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] or at an angle of [latex]53^\circ[/latex] south of west.

The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where [latex]{F}_{\text{D}}[/latex] is less than 1/600th of the weight of the ship.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

What Does the Bathroom Scale Read in an Elevator?

Figure shows a 75.0-kg man (weight of about 165 lb.) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of [latex]1.20\,{\text{m/s}}^{2},[/latex] and (b) if the elevator moves upward at a constant speed of 1 m/s.

A person is standing on a bathroom scale in an elevator. His weight w is shown by an arrow near his chest, pointing downward. F sub s is the force of the scale on the person, shown by a vector starting from his feet pointing vertically upward. W sub s is the weight of the scale, shown by a vector starting at the scale pointing pointing vertically downward. W sub e is the weight of the elevator, shown by a broken arrow starting at the bottom of the elevator pointing vertically downward. F sub p is the force of the person on the scale, drawn starting at the scale and pointing vertically downward. F sub t is the force of the scale on the floor of the elevator, pointing vertically downward, and N is the normal force of the floor on the scale, starting on the elevator near the scale pointing upward. (b) The same person is shown on the scale in the elevator, but only a few forces are shown acting on the person, which is our system of interest. W is shown by an arrow acting downward, and F sub s is the force of the scale on the person, shown by a vector starting from his feet pointing vertically upward. The free-body diagram is also shown, with two forces acting on a point. F sub s acts vertically upward, and w acts vertically downward. An x y coordinate system is shown, with positive x to the right and positive y upward.

If the scale at rest is accurate, its reading equals [latex]{\mathbf{\overset{\to }{F}}}_{\text{p}}[/latex], the magnitude of the force the person exerts downward on it. Figure (a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn, as in Figure (b). Analysis of the free-body diagram using Newton’s laws can produce answers to both Figure (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weight [latex]\mathbf{\overset{\to }{w}}[/latex] and the upward force of the scale [latex]{\mathbf{\overset{\to }{F}}}_{\text{s}}.[/latex] According to Newton’s third law, [latex]{\mathbf{\overset{\to }{F}}}_{\text{p}}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{\text{s}}[/latex] are equal in magnitude and opposite in direction, so that we need to find [latex]{F}_{\text{s}}[/latex] in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,

From the free-body diagram, we see that [latex]{\mathbf{\overset{\to }{F}}}_{\text{net}}={\mathbf{\overset{\to }{F}}}_{s}-\mathbf{\overset{\to }{w}},[/latex] so we have

Solving for [latex]{F}_{s}[/latex] gives us an equation with only one unknown:

or, because [latex]w=mg,[/latex] simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes [latex]{F}_{s}-w=\text{−}ma.[/latex])

which gives

The scale reading in Figure (a) is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding

Now calculate the scale reading when the elevator accelerates downward at a rate of [latex]1.20\,{\text{m/s}}^{2}.[/latex]

[latex]{F}_{\text{s}}=645\,\text{N}[/latex]

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Two Attached Blocks

Figure shows a block of mass [latex]{m}_{1}[/latex] on a frictionless, horizontal surface. It is pulled by a light string that passes over a frictionless and massless pulley. The other end of the string is connected to a block of mass [latex]{m}_{2}.[/latex] Find the acceleration of the blocks and the tension in the string in terms of [latex]{m}_{1},{m}_{2},\,\text{and}\,g.[/latex]

(a) Block m sub 1 is on a horizontal surface. It is connected to a string that passes over a pulley then hangs straight down and connects to block m sub 2. Block m sub 1 has acceleration a sub 1 directed to the right. Block m sub 2 has acceleration a sub 2 directed downward. (b) Free body diagrams of each block. Block m sub 1 has force w sub 1 directed vertically down, N directed vertically up, and T directed horizontally to the right. Block m sub 2 has force w sub 2 directed vertically down, and T directed vertically up. The x y coordinate system has positive x to the right and positive y up.

We draw a free-body diagram for each mass separately, as shown in Figure . Then we analyze each one to find the required unknowns. The forces on block 1 are the gravitational force, the contact force of the surface, and the tension in the string. Block 2 is subjected to the gravitational force and the string tension. Newton’s second law applies to each, so we write two vector equations:

For block 1: [latex]\mathbf{\overset{\to }{T}}+{\mathbf{\overset{\to }{w}}}_{1}+\mathbf{\overset{\to }{N}}={m}_{1}{\mathbf{\overset{\to }{a}}}_{1}[/latex]

For block 2: [latex]\mathbf{\overset{\to }{T}}+{\mathbf{\overset{\to }{w}}}_{2}={m}_{2}{\mathbf{\overset{\to }{a}}}_{2}.[/latex]

Notice that [latex]\mathbf{\overset{\to }{T}}[/latex] is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

The component equations follow from the vector equations above. We see that block 1 has the vertical forces balanced, so we ignore them and write an equation relating the x -components. There are no horizontal forces on block 2, so only the y -equation is written. We obtain these results:

When block 1 moves to the right, block 2 travels an equal distance downward; thus, [latex]{a}_{1x}=\text{−}{a}_{2y}.[/latex] Writing the common acceleration of the blocks as [latex]a={a}_{1x}=\text{−}{a}_{2y},[/latex] we now have

From these two equations, we can express a and T in terms of the masses [latex]{m}_{1}\,\text{and}\,{m}_{2},\,\text{and}\,g:[/latex]

Notice that the tension in the string is less than the weight of the block hanging from the end of it. A common error in problems like this is to set [latex]T={m}_{2}g.[/latex] You can see from the free-body diagram of block 2 that cannot be correct if the block is accelerating.

Calculate the acceleration of the system, and the tension in the string, when the masses are [latex]{m}_{1}=5.00\,\text{kg}[/latex] and [latex]{m}_{2}=3.00\,\text{kg}.[/latex]

[latex]a=3.68\,{\text{m/s}}^{2},[/latex] [latex]T=18.4\,\text{N}[/latex]

Atwood Machine

A classic problem in physics, similar to the one we just solved, is that of the Atwood machine , which consists of a rope running over a pulley, with two objects of different mass attached. It is particularly useful in understanding the connection between force and motion. In Figure , [latex]{m}_{1}=2.00\,\text{kg}[/latex] and [latex]{m}_{2}=4.00\,\text{kg}\text{.}[/latex] Consider the pulley to be frictionless. (a) If [latex]{m}_{2}[/latex] is released, what will its acceleration be? (b) What is the tension in the string?

An Atwood machine consists of masses suspended on either side of a pulley by a string passing over the pulley. In the figure, mass m sub 1 is on the left and mass m sub 2 is on the right. The free body diagram of block one shows mass one with force vector T pointing vertically up and force vector w sub one pointing vertically down. The free body diagram of block two shows mass two with force vector T pointing vertically up and force vector w sub two pointing vertically down.

We draw a free-body diagram for each mass separately, as shown in the figure. Then we analyze each diagram to find the required unknowns. This may involve the solution of simultaneous equations. It is also important to note the similarity with the previous example. As block 2 accelerates with acceleration [latex]{a}_{2}[/latex] in the downward direction, block 1 accelerates upward with acceleration [latex]{a}_{1}[/latex]. Thus, [latex]a={a}_{1}=\text{−}{a}_{2}.[/latex]

(The negative sign in front of [latex]{m}_{2}a[/latex] indicates that [latex]{m}_{2}[/latex] accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is

Solving for a :

  • Observing the first block, we see that [latex]\begin{array}{c}T-{m}_{1}g={m}_{1}a\hfill \\ T={m}_{1}(g+a)=(2\,\text{kg})(9.8\,{\text{m/s}}^{2}+3.27\,{\text{m/s}}^{2})=26.1\,\text{N}\text{.}\hfill \end{array}[/latex]

The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system, [latex]({m}_{2}-{m}_{1})g[/latex], to the total mass of the system, [latex]{m}_{1}+{m}_{2}[/latex]. We can also use the Atwood machine to measure local gravitational field strength.

Determine a general formula in terms of [latex]{m}_{1},{m}_{2}[/latex] and g for calculating the tension in the string for the Atwood machine shown above.

[latex]T=\frac{2{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}g[/latex] (This is found by substituting the equation for acceleration in Figure (a), into the equation for tension in Figure (b).)

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

What Force Must a Soccer Player Exert to Reach Top Speed?

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

Substituting the known values yields

Substituting the known values of m and a gives

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

What Force Acts on a Model Helicopter?

A 1.50-kg model helicopter has a velocity of [latex]5.00\mathbf{\hat{j}}\,\text{m/s}[/latex] at [latex]t=0.[/latex] It is accelerated at a constant rate for two seconds (2.00 s) after which it has a velocity of [latex](6.00\mathbf{\hat{i}}+12.00\mathbf{\hat{j}})\text{m/s}\text{.}[/latex] What is the magnitude of the resultant force acting on the helicopter during this time interval?

We can easily set up a coordinate system in which the x -axis [latex](\mathbf{\hat{i}}[/latex] direction) is horizontal, and the y -axis [latex](\mathbf{\hat{j}}[/latex] direction) is vertical. We know that [latex]\Delta t=2.00s[/latex] and [latex](6.00\mathbf{\hat{i}}+12.00\mathbf{\hat{j}}\,\text{m/s})-(5.00\mathbf{\hat{j}}\,\text{m/s}).[/latex] From this, we can calculate the acceleration by the definition; we can then apply Newton’s second law.

The magnitude of the force is now easily found:

The original problem was stated in terms of [latex]\mathbf{\hat{i}}-\mathbf{\hat{j}}[/latex] vector components, so we used vector methods. Compare this example with the previous example.

Find the direction of the resultant for the 1.50-kg model helicopter.

49.4 degrees

Baggage Tractor

Figure (a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by [latex]F=(820.0t)\,\text{N,}[/latex] find the speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?

Figure (a) shows a baggage tractor driving to the left and pulling two luggage carts. The external forces on the system are shown. The forces on the tractor are F sub tractor, horizontally to the left, N sub tractor vertically up, and w sub tractor vertically down. The forces on the cart immediately behind the tractor, cart A, are N sub A vertically up, and w sub A vertically down. The forces on cart B, the one behind cart A, are N sub B vertically up, and w sub B vertically down. Figure (b) shows the free body diagram of the tractor, consisting of F sub tractor, horizontally to the left, N sub tractor vertically up, w sub tractor vertically down, and T horizontally to the right.

A free-body diagram shows the driving force of the tractor, which gives the system its acceleration. We only need to consider motion in the horizontal direction. The vertical forces balance each other and it is not necessary to consider them. For part b, we make use of a free-body diagram of the tractor alone to determine the force between it and cart A. This exposes the coupling force [latex]\mathbf{\overset{\to }{T}},[/latex] which is our objective.

Since acceleration is a function of time, we can determine the velocity of the tractor by using [latex]a=\frac{dv}{dt}[/latex] with the initial condition that [latex]{v}_{0}=0[/latex] at [latex]t=0.[/latex] We integrate from [latex]t=0[/latex] to [latex]t=3\text{:}[/latex]

  • Refer to the free-body diagram in Figure (b). [latex]\begin{array}{ccc}\hfill \sum {F}_{x}& =\hfill & {m}_{\text{tractor}}{a}_{x}\hfill \\ \hfill 820.0t-T& =\hfill & {m}_{\text{tractor}}(0.7805)t\hfill \\ \hfill (820.0)(3.00)-T& =\hfill & (650.0)(0.7805)(3.00)\hfill \\ \hfill T& =\hfill & 938\,\text{N}.\hfill \end{array}[/latex]

Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving Figure (a), whereas only the mass of the truck (since it supplied the force) was of use in Figure (b).

Recall that [latex]v=\frac{ds}{dt}[/latex] and [latex]a=\frac{dv}{dt}[/latex]. If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have [latex]dt=\frac{ds}{v}[/latex] and [latex]dt=\frac{dv}{a}.[/latex] Now, equating these expressions, we have [latex]\frac{ds}{v}=\frac{dv}{a}.[/latex] We can rearrange this to obtain [latex]{a}^{}ds={v}^{}dv.[/latex]

Motion of a Projectile Fired Vertically

A 10.0-kg mortar shell is fired vertically upward from the ground, with an initial velocity of 50.0 m/s (see Figure ). Determine the maximum height it will travel if atmospheric resistance is measured as [latex]{F}_{\text{D}}=(0.0100{v}^{2})\,\text{N,}[/latex] where v is the speed at any instant.

(a) A photograph of a soldier firing a mortar shell straight up. (b) A free body diagram of the mortar shell shows forces F sub D and w, both pointing vertically down. Force w is larger than force F sub D.

The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

Initially, [latex]{y}_{0}=0[/latex] and [latex]{v}_{0}=50.0\,\text{m/s}\text{.}[/latex] At the maximum height [latex]y=h,v=0.[/latex] The free-body diagram shows [latex]{F}_{\text{D}}[/latex] to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

The acceleration depends on v and is therefore variable. Since [latex]a=f(v)\text{,}[/latex] we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, [latex]h=114\,\text{m}\text{.}[/latex]

Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t ), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

  • Newton’s laws of motion can be applied in numerous situations to solve motion problems.
  • Some problems contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whether [latex]{F}_{\text{net}}=ma[/latex] or [latex]{F}_{\text{net}}=0.[/latex]
  • The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating vertically, the normal force is less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force is always less than the full weight of the object.
  • Some problems contain several physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics to solve these problems.

Conceptual Questions

To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at g . Why do they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

The scale is in free fall along with the astronauts, so the reading on the scale would be 0. There is no difference in the apparent weightlessness; in the aircraft and in orbit, free fall is occurring.

A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force [latex]\mathbf{\overset{\to }{F}}[/latex] so that the swing ropes are [latex]30.0^\circ[/latex] with respect to the vertical. (a) Calculate the tension in each of the two ropes supporting the swing under these conditions. (b) Calculate the magnitude of [latex]\mathbf{\overset{\to }{F}}.[/latex]

a. 170 N; b. 170 N

Find the tension in each of the three cables supporting the traffic light if it weighs 2.00 × 10 2 N.

A sketch of a traffic light suspended by a cable that is in turn suspended from two other cables is shown. Tension T sub 3 is the tension in the cable connecting the traffic light to the upper cables. Tension T sub one is the tension in the upper cable pulling up and to the left, making a 41 degree angle with the horizontal. Tension T sub two is the tension pulling up and to the right, making a 63 degree angle with the horizontal. Force vector w equal to 200 Newtons pulls vertically downward on the traffic light.

Three forces act on an object, considered to be a particle, which moves with constant velocity [latex]v=(3\mathbf{\hat{i}}-2\mathbf{\hat{j}})\,\text{m/s}\text{.}[/latex] Two of the forces are [latex]{\mathbf{\overset{\to }{F}}}_{1}=(3\mathbf{\hat{i}}+5\mathbf{\hat{j}}-6\mathbf{\hat{k}})\,\text{N}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{2}=(4\mathbf{\hat{i}}-7\mathbf{\hat{j}}+2\mathbf{\hat{k}})\,\text{N}\text{.}[/latex] Find the third force.

[latex]{\mathbf{\overset{\to }{F}}}_{3}=(-7\mathbf{\hat{i}}+2\mathbf{\hat{j}}+4\mathbf{\hat{k}})\,\text{N}[/latex]

A flea jumps by exerting a force of [latex]1.20\times {10}^{-5}\,\text{N}[/latex] straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of [latex]0.500\times {10}^{-6}\,\text{N}[/latex] on the flea while the flea is still in contact with the ground. Find the direction and magnitude of the acceleration of the flea if its mass is [latex]6.00\times {10}^{-7}\,\text{kg}[/latex]. Do not neglect the gravitational force.

Two muscles in the back of the leg pull upward on the Achilles tendon, as shown below. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?

An Achilles tendon is shown in the figure with two forces exerted on it by the lateral and medial heads of the gastrocnemius muscle. F sub one, equal to two hundred Newtons, is shown as a vector making an angle twenty degrees to the right of vertical, and F sub two, equal to two hundred Newtons, is shown making an angle of twenty degrees left of vertical.

After a mishap, a 76.0-kg circus performer clings to a trapeze, which is being pulled to the side by another circus artist, as shown here. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.

A circus performer hanging from a trapeze is being pulled to the right by another performer using a rope. Her weight is shown by a vector w acting vertically downward. The trapeze rope exerts a tension, T sub one, up and to the left, making an angle of fifteen degrees with the vertical. The second performer pulls with tension T sub two, making an angle of ten degrees above the positive x direction.

A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow the first dolphin if it was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)

When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?

A large rocket has a mass of [latex]2.00\times {10}^{6}\,\text{kg}[/latex] at takeoff, and its engines produce a thrust of [latex]3.50\times {10}^{7}\,\text{N}.[/latex] (a) Find its initial acceleration if it takes off vertically. (b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust?

a. [latex]7.70\,{\text{m/s}}^{2}[/latex]; b. 4.33 s

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110.0 kg.

A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110.0 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

a. 46.4 m/s; b. [latex]2.40\times {10}^{3}\,{\text{m/s}}^{2}\text{;}[/latex] c. 5.99 × 10 3 N; ratio of 245

A 0.500-kg potato is fired at an angle of [latex]80.0^\circ[/latex] above the horizontal from a PVC pipe used as a “potato gun” and reaches a height of 110.0 m. (a) Neglecting air resistance, calculate the potato’s velocity when it leaves the gun. (b) The gun itself is a tube 0.450 m long. Calculate the average acceleration of the potato in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the potato in the gun? Express your answer in newtons and as a ratio to the weight of the potato.

An elevator filled with passengers has a mass of [latex]1.70\times {10}^{3}\,\text{kg}[/latex]. (a) The elevator accelerates upward from rest at a rate of [latex]1.20\,{\text{m/s}}^{2}[/latex] for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of [latex]0.600\,{\text{m/s}}^{2}[/latex] for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?

a. [latex]1.87\times {10}^{4}\,\text{N;}[/latex] b. [latex]1.67\times {10}^{4}\,\text{N;}[/latex] c. [latex]1.56\times {10}^{4}\,\text{N;}[/latex] d. 19.4 m, 0 m/s

A 20.0-g ball hangs from the roof of a freight car by a string. When the freight car begins to move, the string makes an angle of [latex]35.0^\circ[/latex] with the vertical. (a) What is the acceleration of the freight car? (b) What is the tension in the string?

A student’s backpack, full of textbooks, is hung from a spring scale attached to the ceiling of an elevator. When the elevator is accelerating downward at [latex]3.8\,{\text{m/s}}^{2}[/latex], the scale reads 60 N. (a) What is the mass of the backpack? (b) What does the scale read if the elevator moves upward while slowing down at a rate [latex]3.8\,{\text{m/s}}^{2}[/latex]? (c) What does the scale read if the elevator moves upward at constant velocity? (d) If the elevator had no brakes and the cable supporting it were to break loose so that the elevator could fall freely, what would the spring scale read?

a. 10 kg; b. 90 N; c. 98 N; d. 0

A service elevator takes a load of garbage, mass 10.0 kg, from a floor of a skyscraper under construction, down to ground level, accelerating downward at a rate of [latex]1.2\,{\text{m/s}}^{2}[/latex]. Find the magnitude of the force the garbage exerts on the floor of the service elevator?

A roller coaster car starts from rest at the top of a track 30.0 m long and inclined at [latex]20.0^\circ[/latex] to the horizontal. Assume that friction can be ignored. (a) What is the acceleration of the car? (b) How much time elapses before it reaches the bottom of the track?

a. [latex]3.35\,{\text{m/s}}^{2}[/latex]; b. 4.2 s

The device shown below is the Atwood’s machine considered in Figure . Assuming that the masses of the string and the frictionless pulley are negligible, (a) find an equation for the acceleration of the two blocks; (b) find an equation for the tension in the string; and (c) find both the acceleration and tension when block 1 has mass 2.00 kg and block 2 has mass 4.00 kg.

An Atwood machine consisting of masses suspended on either side of a pulley by a string passing over the pulley is shown. Mass m sub 1 is on the left and mass m sub 2 is on the right.

Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. The table and the pulley are frictionless. (a) Find the acceleration of the system. (b) Find the tension in the rope. (c) Find the speed with which the hanging mass hits the floor if it starts from rest and is initially located 1.0 m from the floor.

Block m sub 1 is on a horizontal table. It is connected to a string that passes over a pulley at the edge of the table. The string then hangs straight down and connects to block m sub 2, which is not in contact with the table. Block m sub 1 has acceleration a sub 1 directed to the right. Block m sub 2 has acceleration a sub 2 directed downward.

Shown below are two carts connected by a cord that passes over a small frictionless pulley. Each cart rolls freely with negligible friction. Calculate the acceleration of the carts and the tension in the cord.

Two carts connected by a string passing over a pulley are on either side of a double inclined plane. The string passes over a pulley attached to the top of the double incline. On the left, the incline makes an angle of 37 degrees with the horizontal and the cart on that side has mass 10 kilograms. On the right, the incline makes an angle of 53 degrees with the horizontal and the cart on that side has mass 15 kilograms.

A 2.00 kg block (mass 1) and a 4.00 kg block (mass 2) are connected by a light string as shown; the inclination of the ramp is [latex]40.0^\circ[/latex]. Friction is negligible. What is (a) the acceleration of each block and (b) the tension in the string?

Block 1 is on a ramp inclined up and to the right at an angle of 40 degrees above the horizontal. It is connected to a string that passes over a pulley at the top of the ramp, then hangs straight down and connects to block 2. Block 2 is not in contact with the ramp.

6.1 Solving Problems with Newton’s Laws Copyright © 2016 by OpenStax. All Rights Reserved.

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Video transcript

StickMan Physics

StickMan Physics

Animated Physics Lessons

F=ma Practice Problems

F=ma problem set.

Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations.

In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems.

F=ma Equations

1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right?

1

a = 33.33 m/s 2 Right

2. What is the acceleration of the 25 kg box that has 50 N of force applied to the right?

2

                a=2.0 m/s 2 Right

3. What is the acceleration of the 3 kg box that has 25 N of force applied to the right and 55 N left?

3

               a = 10.0 m/s 2 Left

4. What is the acceleration of the 5 kg box that has a 25 N force and 50 N force applied both right?

4

            a = 15.0 m/s 2 Right

5. What is the acceleration of the 25 kg box that has a 100 N force north and 50 N force east applied?

5

             a= 4.47 m/s 2

5b. What direction would this box accelerate?

            63.43° North of East

6. Does a 795 kg Lorinser speedy, 6300 kg elephant, or 8.6 kg wagon have more inertia and why?

            6300 kg Elephant

            The more mass the more inertia

different masses

7. How much force is required to accelerate a 795 kg Lorinser Speedy by 15 m/s 2 ?

            F = 11925 N

8. How much force is required to accelerate an 8.6 kg wagon by 15 m/s 2 ?

            F = 129 N

9. How much does a 6300 kg elephant accelerate when you apply 500 N of force?

            a = 0.0794 m/s 2

10. What is the mass of an object if it takes a net force of 40 N to accelerate at a rate of 0.88 m/s 2 ?

            m = 45.45 kg

11. How much force is required to accelerate a 0.142 kg baseball to 44.7 m/s during a pitchers 1.5 meter delivery?

            F = 94.58 N

12. A 0.050 kg golf ball leaves the tee at a speed of 75.0 m/s. The club is in contact with the ball for 0.020 s. What is the net force of the club on the ball?

                F = 187.5 N

13. A 90.0 kg astronaut receives a 30.0 N force from her jetpack. How much faster is she be moving after 2.00 seconds?

            0.667 m/s faster

14. A 795 kg car starts from rest and travels 41 m in 3.0 s. How much force did the car engine provide?

            F = 7242 N

15. Joe and his sailboat have a combined weight of 450 kg. How far has Joe sailed when he started at 5 m/s and a gust of wind provided 600 Newtons of force for 4 seconds?

            x = 30.64 m

16. Tom pulls a 45 kilogram wagon with a force of 200 Newtons at a 15° angle to the horizontal from rest. How much faster will the wagon be moving after 2 seconds?

force at an angle

            v f = 8.58 m/s

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problem solving in force physics

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problem solving in force physics

Simplifying a Difficult Problem

Consider the situation below in which a force is directed at an angle to the horizontal. In such a situation, the applied force could be resolved into two components. These two components can be considered to replace the applied force at an angle. By doing so, the situation simplifies into a familiar situation in which all the forces are directed horizontally and vertically.

Once the situation has been simplified, the problem can be solved like any other problem. The task of determining the acceleration involves first determining the net force by adding up all the forces as vectors and then dividing the net force by the mass to determine the acceleration. In the above situation, the vertical forces are balanced (i.e., F grav , F y , and F norm add up to 0 N), and the horizontal forces add up to 29.3 N, right (i.e., 69.3 N, right + 40 N, left = 29.3 N, right). The net force is 29.3 N, right and the mass is 10 kg (m = F grav /g); therefore, the acceleration is 2.93 m/s/s, right.

Your Turn to Practice

To test your understanding, analyze the two situations below to determine the net force and the acceleration. When finished, click the button to view the answers.

The net force is 69.9 N, right and the acceleration is 3.5 m/s/s, right .

Note that the vertical forces balance but the horizontal forces do not. The net force is

F net = 129.9 N, right - 60 N, left = 69.9 N, right

The mass is

m = (F grav / g) = 20 kg

So the acceleration is

a = (69.9 N) / (20 kg) =3.50 m/s/s.

The net force is 30.7 N, right and the acceleration is 1.23 m/s/s, right .

F net = 70.7 N, right - 40 N, left = 30.7 N, right
m = (F grav / g) = 25 kg
a = (30.7 N) / (25 kg) =1.23 m/s/s.

What's Up with the Normal Force?

There is one peculiarity about these types of problems that you need to be aware of. The normal force (F norm ) is not necessarily equal to the gravitational force (F grav ) as it has been in problems that we have previously seen. The principle is that the vertical forces must balance if there is no vertical acceleration. If an object is being dragged across a horizontal surface, then there is no vertical acceleration. For this reason, the normal force (F norm ) plus the vertical component (F y ) of the applied force must balance the gravitational force (F grav ). A quick review of these problems shows that this is the case. If there is an acceleration for an object being pulled across a floor, then it is a horizontal acceleration; and thus the only imbalance of force would be in the horizontal direction .

Now consider the following situation in which a force analysis must be conducted to fill in all the blanks and to determine the net force and acceleration. In a case such as this, a thorough understanding of the relationships between the various quantities must be fully understood. Make an effort to solve this problem. When finished, click the button to view the answers. (When you run into difficulties, consult the help from a previous unit .)

 The F grav is

F grav = m • g = (10 kg) • (9.8 m/s/s) = 98 N

Using the sine function,

F y = (60 N) • sine (30 degrees) = 30 N

Since vertical forces are balanced, F norm = 68 N.

Now F frict can be found

F frict = mu • F norm = ( 0.3) • (68 N) = 20.4 N

Using the cosine function,

F x = (60 N) • cosine (30 degrees) = 52.0 N

Now since all the individual force values are known, the F net can be found:

F net = 52.0 N,right + 20.4 N, left = 31.6 N, right.

The acceleration is

a = 3.16 m/s/ s, right.

In conclusion, a situation involving a force at an angle can be simplified by using trigonometric relations to resolve that force into two components. Such a situation can be analyzed like any other situation involving individual forces. The net force can be determined by adding all the forces as vectors and the acceleration can be determined as the ratio of Fnet/mass.

   

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Check your understanding.

The following problems provide plenty of practice with F net = m • a problems involving forces at angles. Try each problem and then click the button to view the answers.

Glen Brook and Warren Peace are incorrect. Warren Peace perhaps believes that the Fnorm = Fgrav; but this is only the case when there are only two vertical forces and no vertical acceleration; sorry Warren - there is a second vertical force in this problem (F app ).

Glen Brook perhaps thinks that the F app force is 50 N upwards and thus the F norm must be 50 N upwards to balance the Fgrav. Sorry Glen - the F app is only 25 N upwards (50 N) • sine 30 degrees).

"Datagal Olive!"

2. A box is pulled at a constant speed of 0.40 m/s across a frictional surface. Perform an extensive analysis of the diagram below to determine the values for the blanks.

First use the mass to determine the force of gravity.

F grav = m • g = (20 kg) • (9.8 m/s/s) = 196 N

Now find the vertical component of the applied force using a trigonometric function.

F y = (80 N) • sine (45 degrees) = 56.7 N

Thus, F norm =139.3 N in order for the vertical forces to balance.

The horizontal component of the applied force can be found as

F x = (80 N) • cosine (45 degrees) = 56.7 N

Since the speed is constant, the horizontal forces must also balance; and so F frict = 56.7 N.

The value of "mu" can be found using the equation

3. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

Answer: "mu" = 0.25

 The F grav can be calculated from the mass of the object.

The vertical component of the applied force can be calculated using a trigonometric function:

F y = (80 N) • sine (30 degrees) = 40 N

In order for the vertical forces to balance, F norm + F y = F grav . Thus,

F norm = F grav - F y = = 196 N - 40 N = 156 N

The horizontal component of the applied force can be calculated using a trigonometric function:

F x = (80 N) • cosine (30 degrees) = 69.2 N

The net force is the sum of all the forces when added as vectors. Thus,

F net = (69.2 N, right) + (40 N,left) = 29.2 N, right
a = F net / m = (29.2 N, right) / (20 kg) = 1.46 m/s/s, right.

The value of "mu" can be found using the equation "mu" = F frict / F norm .

4. The 5-kg mass below is moving with a constant speed of 4 m/s to the right. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

The F grav can be calculated from the mass of the object.

F grav = m • g = (5 kg) • (9.8 m/s/s) = 49 N

The vertical component of the applied force can be calculated using a trigonometric function.

F y = (15 N) • sine (45 degrees) =10.6 N
F norm = F grav - F y = = 49 N - 10.6 N = 38.4 N
F x = (15 N) • cosine (45 degrees) = 10.6 N

Since the speed is constant, the horizontal forces must balance. Therefore, F frict = 10.6 N.

The value of "mu" can be found using the equation "mu"= F frict / F norm :

"mu" = 0.276

5. The following object is being pulled at a constant speed of 2.5 m/s. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

Since the velocity is constant, the acceleration and the net force are 0 m/s/s and 0 N respectively. The F grav can be calculated from the mass of the object.

The object moves at constant speed; thus, the horizontal forces must balance. For this reason, F x = 10 N.

The applied force can now be found using a trigonometric function and the horizontal component:

cosine (60 degrees) = (10 N) / (F app )

Proper algebra yields

F app = (10 N) / [cosine (60 degrees) ] = 20 N
F y = (20 N) • sine (60 degrees) = 17.3 N
F norm = F grav - F y = 49 N - 17.3 N = 31.7 N

6. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

The slope of a velocity-time graph is the acceleration of the object. In this case, a = +2 m/s/s.

The net force can be calculated as:

F net = m • a = (10 kg) • (2 m/s/s) = 20 N, right.
F x = (70 N) • cosine (45 degrees) = 49.5 N

The net force is the vector sum of all the forces. Thus, F net = F x + F frict . That is,

20 N, right = 49.5 N, right + F frict

Therefore, F frict must be 29.5 N, left.

F y = (70 N) • sine (45 degrees) = 49.5 N
F norm = F grav - F y = = 98 N - 49.5 N = 48.5 N

The value of "mu" can be found using the equation "mu" = F frict / F norm :

7. Study the diagram below and determine the acceleration of the box and its velocity after being pulled by the applied force for 2.0 seconds.

F y = (50 N) • sine (45 degrees) = 35.4 N

In order for the vertical forces to balance, F norm + F y = F grav . Algebraic rearrangement leads to:

F norm = F grav - F y = 196 N - 35.4 N = 160.6 N
F x = (50 N) • cosine (45 degrees) = 35.4 N

The F net is 35.4 N, right since the only force which is not balanced is F x .

The acceleration is:

a = F net / m = (35.4 N) / (20 kg) = 1.77 m/s/s

The velocity after 2.0 seconds can be calculated using a kinematic equation:

v f = v i + a • t = 0 m/s + (1.77 m/s/s) • (2.0 s) v f = 3.54 m/s

8. A student pulls a 2-kg backpack across the ice (assume friction-free) by pulling at a 30-degree angle to the horizontal. The velocity-time graph for the motion is shown. Perform a careful analysis of the situation and determine the applied force.

The slope of a velocity-time graph is the acceleration of the object. In this case, a = +0.125 m/s/s.

F net = m • a = (2 kg) • (0.125 m/s/s) = 0.250 N, right

Since the acceleration is horizontal, the vertical forces balance each other. The horizontal component of the applied force (F x ) supplies the horizontal force required for the acceleration. Thus, the horizontal component of the applied force is 0.250 N.

Using trigonometry, the applied force (F app ) can be calculated:

cosine (30 degrees) = ( 0.250 N) / (F app )

Algebraic rearrangement of this equation leads to:

F app = 0.289 N

9. The following object is moving to the right and encountering the following forces. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F grav = m • g = (10 kg) • (10 m/s/s) = 100 N
F norm = F grav - F y = 100 N - 35.4 N = 64.6 N

This x-component of the applied force (F x ) is directed leftward. This horizontal component of force is not counteracted by a rightward force. For this reason, the net force is 35.4 N.  Knowing F net , allows us to determine the acceleration of the object:

a = F net / m = (35.4 N) / (10 kg) = ~3.5 m/s/s

The acceleration of an object is the velocity change per time. For an acceleration of 3.5 m/s/s, the velocity change should be 3.5 m/s for each second of time change. In the velocity-time table, the velocity is decreasing by 3.5 m/s each second. Thus, the values should read 17.5 m/s, 14.0 m/s, 10.5 m/s, 7.0 m/s, 3.5 m/s, 0 m/s.

10. The 10-kg object is being pulled to the left at a constant speed of 2.5 m/s. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F y = F grav - F norm = 98 N - 80 N = 18 N

The applied force can be determined using a trigonometric function:

sine 30 (degrees) = (18 N) / F app

Algebraic rearrangement leads to:

F app = (18 N) / [ sine (30 degrees) ] = 36 N

Similar trigonometry allows one to determine the x-component of the applied force:

F x = (36 N) • cosine(30 degrees) = 31.2 N

Since the speed is constant, the horizontal forces must also balance. Thus the force of friction is equal to the F x value. F frict = 31.2 N

The value of "mu" can be found using the equation "mu" = F frict / F norm

11. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F y = (100 N) • sine (45 degrees) = 70.7 N
F norm = F grav - F y = 98 N - 70.7 N = 27.3 N
F x = (100 N) • cosine (45 degrees) = 70. N

The net force is the vector sum of all the forces.

F net = 50 N, right + 70.7 N, left = 20.7 N, left

The acceleration is can be found from a = F net / m :

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Title: physics simulation capabilities of llms.

Abstract: [Abridged abstract] Large Language Models (LLMs) can solve some undergraduate-level to graduate-level physics textbook problems and are proficient at coding. Combining these two capabilities could one day enable AI systems to simulate and predict the physical world. We present an evaluation of state-of-the-art (SOTA) LLMs on PhD-level to research-level computational physics problems. We condition LLM generation on the use of well-documented and widely-used packages to elicit coding capabilities in the physics and astrophysics domains. We contribute $\sim 50$ original and challenging problems in celestial mechanics (with REBOUND), stellar physics (with MESA), 1D fluid dynamics (with Dedalus) and non-linear dynamics (with SciPy). Since our problems do not admit unique solutions, we evaluate LLM performance on several soft metrics: counts of lines that contain different types of errors (coding, physics, necessity and sufficiency) as well as a more "educational" Pass-Fail metric focused on capturing the salient physical ingredients of the problem at hand. As expected, today's SOTA LLM (GPT4) zero-shot fails most of our problems, although about 40\% of the solutions could plausibly get a passing grade. About $70-90 \%$ of the code lines produced are necessary, sufficient and correct (coding \& physics). Physics and coding errors are the most common, with some unnecessary or insufficient lines. We observe significant variations across problem class and difficulty. We identify several failure modes of GPT4 in the computational physics domain. Our reconnaissance work provides a snapshot of current computational capabilities in classical physics and points to obvious improvement targets if AI systems are ever to reach a basic level of autonomy in physics simulation capabilities.

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CBS News

How quantum computers may change the world as we know it

Posted: December 4, 2023 | Last updated: December 7, 2023

Advances in quantum computing are bringing us closer to a world where new types of computers may solve problems in minutes that would take today's supercomputers millions of years.

Today's transistor-based computers have their limitations, but quantum computers could give us answers to problems in physics, chemistry, engineering and medicine that currently seem impossible. 

"There are many, many problems that are so complex that we can make that statement that, actually, classical computers will never be able to solve that problem, not now, not 100 years from now, not 1,000 years from now," IBM Director of Research Dario Gil said. "You actually require a different way to represent information and process information. That's what quantum gives you."

What is quantum computing?

Computers have processed information on transistors for decades, with advancements being made as companies squeeze more transistors onto chips. Faster, more powerful computing requires more transistors because each transistor holds information in only two states: zero or one.

Quantum computing ditches transistors and instead encodes information using qubits, which act like artificial atoms. Qubits aren't binary—they can be zero or one or anything in between.

A reliable, general purpose, quantum computer is a tough climb. Charina Chou, chief operating officer at Google Quantum AI, showed 60 Minutes correspondent Scott Pelley the processor that holds the qubits at Google's quantum computer lab.

The sealed quantum computer at Google's lab is one of the coldest places in the universe, according to Chou. The deep freeze eliminates electrical resistance and isolates qubits from outside vibrations.

Companies and countries around the world are racing to be first to develop quantum computing technology. China named quantum a top national priority and the U.S. government is spending nearly a billion dollars a year on research. 

What's so special about quantum computing?

Physicist Michio Kaku, of the City University of New York, already calls today's computers "classical." He uses a maze to explain quantum's difference.

"Let's look at a classical computer calculating how a mouse navigates a maze. It is painful. One by one, it has to map every single left turn, right turn, left turn, right turn before it finds the goal. Now a quantum computer scans all possible routes simultaneously. This is amazing," Kaku said. "How many turns are there? Hundreds of possible turns, right? Quantum computers do it all at once."

Kaku's book, "Quantum Supremacy," explains the stakes. 

"We're looking at a race, a race between China, between IBM, Google, Microsoft, Honeywell," Kaku said. "All the big boys are in this race to create a workable, operationally efficient quantum computer. Because the nation or company that does this will rule the world economy."

It's not just the economy quantum computing could impact. A quantum computer is set up at Cleveland Clinic, where Chief Research Officer Dr. Serpil Erzurum believes the technology could revolutionize the world of health care. 

Quantum computers can potentially model the behavior of proteins, the molecules that regulate all life, Erzurum said. Proteins change their shape to change their function in ways that are too complex to follow, but quantum computing could change that understanding.  

"I need to understand the shape it's in when it's doing an interaction or a function that I don't want it to do for that patient. Cancer, autoimmunity — it's a problem," Dr. Erzurum said. "We are limited completely by the computational ability to look at the structure in real time for any, even one, molecule."

Quantum computers could also potentially break some of the encryption codes for today's online security. Next year, the federal government plans to publish a new standard for all encryption to resist quantum computers. 

How close are we to a future with quantum computing?

Quantum researchers are still working on tough problems, including a trick called coherence. To achieve coherence, qubits must vibrate in unison, but maintaining this has been a challenge.

"We're making about one error in every hundred or so steps," Chou said. "Ultimately, we think we're going to need about one error in every million or so steps. That would probably be identified as one of the biggest barriers."

Still, the founder of Google's quantum lab, Hartmut Neven, feels optimistic the company can mitigate those errors, extend coherence time and scale up to larger machines.  

"We don't need any more fundamental breakthroughs. We need little improvements here and there," Neven said. "We have all the pieces together. We just need to integrate them well to build larger and larger systems."

Neven believes this can be accomplished by the end of the decade. IBM's Gil also predicts around the same timeline. 

IBM is set to unveil its latest quantum computer, Quantum System Two, on Monday. It has three times the qubits as the quantum computer at Cleveland Clinic. System Two has room to expand to thousands of qubits, Gil said. 

"It's a machine unlike anything we have ever built," Gil said. 

He sees a future with even more potential. 

"We don't see an obstacle right now that would prevent us from building systems that will have tens of thousands and even a hundred thousand qubits working with each other," Gil said. "We are highly confident that we will get there."

Charina Chou and Scott Pelley

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These 10 scientists are on the cusp of changing the world

It's the Brilliant 10 class of 2023.

By Molly Glick , Hannah Seo | Published Dec 5, 2023 10:00 AM EST

These 10 scientists are on the cusp of changing the world

Leaving a tangible mark on your scientific field is a staggering achievement at any stage of your career. Each year,  Popular Science  honors 10 early-career researchers who’ve gotten a head start: The Brilliant 10. These researchers already stand out as innovators and change makers in their fields. They are asking the unasked questions, adopting novel methods, and pursuing remedies where none exist. Whether they are driven by the desire to fill a need, the pursuit of justice, or sheer fascination, there’s little doubt that each of these awardees will change the world for the better. From more equitable AI to self-assembling lab organs to potential new laws of physics, the groundbreaking work of these up-and-coming researchers offers us a sneak peek at the cutting-edge science of tomorrow. While they’ve already turned heads and earned some of the most prestigious academic awards out there, these experts are just getting started. What on earth will they think of next?

MyDzung Chu: Addressing environmental health disparities within Asian-American communities Woojin Han: Engineering microenvironments to grow muscle stem cells Mary Caswell Stoddard: Investigating the wonders of bird eggs to build a better world Kara McKinley: Understanding how the uterus regenerates Tina Lasisi: Understanding human skin and hair diversity Brendan Keith: Breathing new life into old math Carlos Argüelles-Delgado: Solving space mysteries with Antarctic ice Quinton Smith: Growing organs to solve health disparities Robin Brewer: Empowering marginalized communities with AI Ronald Garcia Ruiz: Aiming lasers at the universe’s origins

Addressing environmental health disparities within Asian-American communities

MyDzung Chu: an environmental epidemiologist at Tufts Medical Center

There’s a big problem with how the United States collects health data from its fastest-growing immigrant population. Asians come to the US from dozens of countries, each with its own languages and cultures. Some come as students, others on worker visas, and many others arrive as refugees. Asian immigrants run the socioeconomic gamut, but public health researchers often lump their data together into one cohort. 

This aggregation makes it seem as if, on average, Asians in America are pretty healthy. But that buries the experiences of vulnerable minority groups within the community, says MyDzung Chu, an environmental epidemiologist at Tufts Medical Center. 

Chu’s mission is to tease apart the nuances of health issues among Asians in the US, especially on the local level. She studies communities in Boston’s Chinatown neighborhood and investigates environmental health disparities that may get hidden in bigger trends. Her research incorporates input and feedback from local groups and individuals. One ongoing project, called Chinatown HEROS and done in collaboration with John Durant, Ponnapa Prakkamakul, and the Asian Community Development Corporation, involves mapping high temperatures and pollution—two climate-change-related variables known to directly harm health—in parks and other open-air spaces throughout Chinatown, with the ultimate aim of educating the community and advocating for changes to make these spaces healthier. 

Chu says Chinatown is notorious for being Boston’s hottest neighborhood. Looking at her research group’s initial map of open-air spaces, it’s not hard to see why. Many of the area’s public spaces are fully paved and bake in the sun, while shade-filled green spaces tend to be at the outer edges of Chinatown.

To get a deeper look at the health implications of these disparities, Chu and her team spent the summer of 2023 installing sensors at each open-air site to measure particulate matter, heat, and humidity. They drove around the area in a van that acted as a mobile monitoring station, measuring pollutants like carbon dioxide, soot, and nitrogen dioxide. They also assessed the setup of these spaces, noting how much of each area was paved, shaded, or provided with greenery, among other factors. They’ll begin synthesizing the data later this fall. 

The ultimate goal is “to share all this data back, in a very digestible educational way, to the residents of Chinatown,” says Chu—and to use that data to spur change. Chu’s team will work with community partners to create forums and workshops to educate residents of Chinatown on climate and heat hazards, as well as strategies to stay healthy. They’ll also ask locals for feedback, including what kinds of modifications they’d like to see in their parks and green spaces. “We’re going to document that and share it with the city,” says Chu. 

This kind of work feels natural for Chu, who grew up in a close-knit Vietnamese community in western Massachusetts. Her father was a public health worker who did tuberculosis outreach. “He was always out in the community seeing patients, seeing new immigrants and refugees in the Vietnamese community who had TB or were at risk for TB,” she recalls. After a Bachelor’s in neuroscience from Smith College, a master’s in public health from Emory University, and a Ph.D. at Harvard T.H. Chan School of Public Health, during which she learned how to layer rigorous research methods on top of community outreach, “It made so much sense” to follow in his footsteps. 

Chu has learned that doing this work involves an immense amount of trust and cooperation between researchers and the people they hope to study. That means attending community events, listening to people’s concerns, and understanding local and cultural context. This path has led her, in collaboration with the local organizations Boston Chinatown Neighborhood Center and Asian Women for Health, to research how to improve cultural responsiveness training for community members, often the first responders during a mental health crisis. Chu and her collaborators conducted focus groups to assess the mental health needs of Chinatown’s residents and learn which barriers—like stigma or lack of health insurance—prevented them from seeking care. They evaluated existing training curricula for first responders and found gaps that left trainees ill-equipped to respond to certain problems, such as Asian youth suicidality or the mental health challenges facing transgender Asian individuals. Chu and her colleagues reported their findings to organizations that run mental health first responder training. There’s still more work to do, especially as these trainings get rolled out in different languages, says Chu, “But I’m really proud of this work because I think we were able to do something actually impactful, even with just a small bit of funding.”

It’s validating for people to see their experiences reflected in data, says Chu, and the insights can empower them to make evidence-based arguments to city and county officials to improve their communities. But she stresses that none of that can happen if researchers don’t listen to—and learn from—the people they want to study. “That’s how we’re really able to create something innovative, relevant, and hopefully impactful,” she says.  —H.S

Engineering microenvironments to grow muscle stem cells

Woojin Han: Assistant Professor in the Leni and Peter W. May Department of Orthopaedics at the Icahn School of Medicine at Mount Sinai

If you entered Woojin Han’s New York City lab, you might see rows of tiny Jell-O-like samples, each about the size of a nickel. While these colorless blobs may look unassuming, Han explains that each one is essentially a 3D petri dish. An assistant professor of orthopedics at the Icahn School of Medicine at Mount Sinai, Han is trying to bioengineer hydrogels—those colorless blobs—to resemble the natural microscopic environments of skeletal muscle stem cells.

The goal is to craft a medium where stem cells can “proliferate without losing their self-renewal potency, or ‘stemness,’ as we call it,” says Han. Stem cells are special precisely because they are not yet specialized. They are the raw materials of the body, holding the potential to develop into various different kinds of cells. Besides the totally flexible stem cells that embryos make to start building a body, there are also stem cells in most adult tissues, including muscle. Muscle stem cells can generate new skeletal muscle—the most common kind of muscle in the body—on demand. But this variability makes stem cells difficult to culture in the lab. When they’re placed in a petri dish, they immediately start differentiating—turning into fully fledged cells of one type or another. If Han’s team can mimic the specific microscopic environments where these stem cells grow in our bodies, that could open up new avenues for stem cell therapies and transplantation. In 2022, he received a $2.2 million grant from the National Institutes of Health to pursue this work. The unassuming hydrogel blobs are the key.

To create hydrogel that will allow stem cells to thrive in their ambiguous state, Han and his team have to consider a plethora of variables such as stiffness, shape, chemical composition, and more. He’s found that softer gels better recreate the relative firmness of muscle tissue and help stem cells maintain stemness, and that hydrogels seem to have more success when extruded in the shape of a muscle fiber.  

Looking at how skeletal muscle stem cells thrive in the body is key. Normally they are wedged between muscle fibers on one side and the basement membrane, which runs between the body’s tissues, on the other. “Evidence … suggests that this asymmetrically partitioned microenvironment plays a very important role in controlling how the cells establish their polarity and begin to guide their cell division processes,” he says. It creates a delicate balance that keeps the stem cells proliferating, but not differentiating. To recreate this asymmetry, Han and his team have landed on a “sandwich hydrogel system,” whereby the cells are embedded between two different hydrogels. “We’ve made a lot of progress on the material engineering end, and now we are starting to get into the nitty-gritty of the biology.”

Han was already interested in regenerative medicine when he started as a postdoctoral fellow at the Georgia Institute of Technology in 2015. “I always was intrigued by the fact that even though our muscles have this very surprising capacity to regenerate in minor injury contexts, they don’t regenerate in more severe injury or disease,” he says. Han began working with biomaterials, like hydrogels, to study how to improve cell survival after engraftment, which is when donor skeletal muscle stem cells begin making new muscle fibers in their host. He saw a world of potential. A better medium for culturing skeletal muscle stem cells could help make stem cell therapy a viable treatment for people suffering from severe muscle disease.

Research on such treatments is very much still in progress. But for now, “there’s no way to get large quantities of these cells, because they’re quite scarce in the body,” says Han. In the long term, Han hopes to be able to extract skeletal muscle stem cells from patient biopsies. With Han’s hydrogels as a medium, labs would be able to proliferate those stem cells, collect them, and inject them back into patients. This could aid muscle healing and regrowth in people with traumatic volumetric muscle loss and rotator cuff injuries. In theory, Han says, this could even reverse damage seen in diseases like Duchenne muscular dystrophy, a genetic condition that leads muscle fibers to grow progressively weaker. But that’s all in the far future, and “there are still a lot of unknowns.” 

Han acknowledges that adequately remaking the microenvironment needed to keep muscle stem cells potent in perpetuity is a huge task. But Han and his colleagues will keep honing their hydrogels, working to stretch that window of “stemness” as long as they can. —H.S

Investigating the wonders of bird eggs to build a better world

Mary Caswell Stoddard: Associate Professor in the Princeton Department of Ecology and Evolutionary Biology

Birds have always been a part of Mary Caswell “Cassie” Stoddard’s life. Growing up with a mother and grandmother who were both avid bird-watchers meant she absorbed plenty of admiration for the winged creatures. 

After an undergraduate degree in biology from Yale University, that appreciation led Stoddard to her Ph.D. in zoology at the University of Cambridge. She began studying the common cuckoo—a bird that stealthily lays its eggs in the nests of other bird species. “In order to get away with this, cuckoos have evolved excellent egg color and pattern mimicry,” she says. “Cuckoos were the entry point that led me to all these questions about eggs.” She took that interest with her in 2016 when she joined Princeton University, where she is now an associate professor of ecology and evolutionary biology.

Egg shape in particular became a major subject of her curiosity. Stoddard can wax poetic about the many possible variations. “You have your typical chicken-egg shape, but many seabirds lay very pointy eggs,” she says. “Hummingbird eggs look like a Tic Tac, and owls lay golf-ball-like round eggs.” But scientists have long lacked a solid understanding of how and why this variation occurs. 

One theory suggests egg shape is related to nutrition. Perhaps the spherical eggs of owls, with their smaller ratio of surface area to volume, arose as a solution to the lack of calcium in the birds’ diets. Another school of thought holds that the number of eggs a bird tends to lay in each clutch might affect shape, to ensure they all fit optimally in a nest. Yet another theory suggests that pointier shapes, as seen in eggs laid by seabirds nesting on cliff edges, prevent incubating chicks from rolling off such precipices. But none of these theories have been tested with large enough sample sizes or studied comprehensively enough for significant conclusions. 

To get at the why and how of egg shape diversity, Stoddard led a multidisciplinary team of biologists, computer scientists, mathematicians, and more in analyzing more than 49,000 eggs from about 1,400 bird species. This research opened Stoddard’s lab to a variety of interdisciplinary methods. Her lab is also studying other facets of bird biology, including with an ongoing project in the Colorado Rockies on hummingbird vision.

The team translated the shapes of the eggs into mathematical models and incorporated data such as habitat, diet, and how many eggs each species laid at a time. They found that one of the best predictors of egg shape was flight ability. “Birds that are strong fliers—birds that tend to fly a lot or migrate long distances—tend to lay eggs that are more elliptical, or more pointy,” says Stoddard. The team published its findings in the journal Science in 2017. 

Its theory is that good fliers have more streamlined body plans, which favor pelvises that can push out only longer, thinner eggs. This is probably not the only evolutionary force behind egg size and shape, Stoddard concedes, and different species could have come to lay outwardly similar eggs due to different evolutionary pressures. But when you look at the amalgam of data from 1,400 bird species, she says, flight is a strong predictor. 

Now Stoddard’s lab has turned its attention to the shells themselves. “Right now, one main focus in our group is understanding [the shell’s] biomechanical properties,” says Stoddard. “And that’s taken us down this whole new path with a whole new suite of questions and tools.”

Eggshell is a fascinating and unique material, she says: It’s quite lightweight and comes together very quickly. Crucially, it’s both strong and breakable. The same egg that protects a growing chick must also allow its young resident to break free when the time comes. “My collaborators and I think that what we are learning about eggshell could be applied to the design of novel synthetic materials with special mechanical properties,” Stoddard says. “For example, it is sometimes desirable to have glass windows that are hard to break from the outside but easy to break from the inside.”

The study of eggshells has fascinating scientific implications too. Unpacking the details of how and why eggs form can teach us how birds thrived as their fellow dinosaurs went extinct, as well as how they might be affected by climate change. 

Stoddard says that in her lab, questions are always rooted in avian evolution. “Our questions begin with birds … and are rooted in evolutionary biology,” she says. But she and her colleagues use whatever tools and disciplines they can to get at those answers—and they follow the research wherever it leads. —H.S

Understanding how the uterus regenerates

Credit: Courtesy Kara McKinley

As a postdoctoral fellow , Kara McKinley came across studies that piqued her interest and altered her scientific trajectory: Researchers had just developed new methods for culturing mini uteruses in petri dishes. At the time, McKinley was studying the regenerative healing properties of the intestine at the University of California at San Francisco. But the more she learned about the biology of the uterus, the more she wanted to know. 

The uterus is a bit of an anomaly. As we get older, our bodies experience damage due to aging, disease, and trauma. A body can often repair or replace parts of itself, but not perfectly. Scars form and accumulate, making most tissues less functional over time. 

But things are a little different for the uterus. For every month of menstruation, the uterus sheds all or most of its lining (the endometrium) and then perfectly rebuilds what was lost, without any scarring. It’s not impervious to injury; certain medical procedures and infections can leave scar tissue behind. But the fact that the uterus can usually bounce back from the physical trauma of pregnancy and birth is something of a marvel. “It’s a uniquely powerful system to understand regeneration in humans,” says McKinley, now an assistant professor of stem cell and regenerative biology at Harvard University and a Freeman Hrabowski scholar at the Howard Hughes Medical Institute. 

Research on menstruation and uterine regeneration is still scant, she says. But studying these processes can open up vast possibilities for future medical treatments. For example, understanding the mechanisms behind this regular, scarless regeneration could help scientists encourage similar healing in other tissues and organs. More comprehensive knowledge of how menstruation works could also lead to treatments for heavy or painful periods. 

Many scientists in the field study these possibilities by culturing small uterus analogues in petri dishes. “There’s a lot of beautiful work” that has come from these methods, says McKinley. But her lab is more interested in studying the uterus in context—inside a living body. Scientists studying human organ functions tend to use mice or rats as models. In some ways, their biology is quite similar to ours. But the species generally used in labs do not menstruate. In fact, most mammals don’t have menstrual cycles. Among rodents, just one—the Cairo spiny mouse ( Acomys cahirinus )—is known to experience them. That’s the animal McKinley’s lab uses alongside regular lab mice, who can be induced to experience a process analogous to menstruation. 

Looking at menstruation in the context of an animal’s whole body is key to understanding our own bodily systems, McKinley says. “Menstruation is a very complicated process, and we don’t yet know all of the components that affect it.” Studying spiny mouse menstruation will help her tease out which variables contribute to the endometrium’s regenerative abilities. McKinley plans to investigate which cells are responsible for rebuilding the uterine lining. Her latest work on the subject is a paper in Annual Reviews that summarizes what scientists currently know about how uterine repairs happen without scarring, and why scarring does occur after certain medical procedures.

Reproductive health is a natural target for McKinley’s academic passions. She has long been interested in issues surrounding gender equity in science. In 2019, she founded a program called Leading Edge that is dedicated to increasing the presence of women and other gender minorities in the biomedical research field. 

According to the World Health Organization, there are almost 2 billion women of reproductive age around the world. But “studies of the uterus, particularly the nonpregnant uterus, have received very little attention from the scientific community,” says McKinley. Some of that is due to prevailing taboos and stigma around menstruation. There is also a long-standing tendency to focus on fetal development over all other aspects of female reproductive health. But there are also technical constraints: The uterus is hardly the most accessible body part, and menstruation isn’t always regular or reliable in its timing. And until the 2016 confirmation of the Cairo spiny mouse’s menstruation, researchers also lacked a good animal model for studying the process. Those are factors scientists have had to take into account when designing studies. 

McKinley doesn’t think those barriers should get in the way of good science. She believes that the approximately 2 billion people worldwide whose uteruses are capable of shedding and regenerating every 28 days or so deserve solid research on reproductive health. Exploring a field with so many unknowns—and with a potential impact on so many people—is incredibly exciting. But McKinley sees more than just an opportunity for discovery. “We also see it as an obligation,” she says.  —H.S

Understanding human skin and hair diversity

Tina Lasisi: Assistant Professor of Anthropology; University of Michigan

Despite growing up in a multicultural household , Tina Lasisi hadn’t given much thought to the biological science of diversity. Diversity, yes, but the biology of it hadn’t much crossed her mind. That changed in 2011, during an undergraduate class in biological anthropology at the University of Cambridge. A professor showed Lasisi and her classmates how the map of skin pigmentation variation almost perfectly overlapped with differences in ultraviolet radiation exposure across the world. A lightbulb went off. 

Lasisi was fascinated to see skin color and melanation explained as a protective force against harmful UV rays. “Cool, that explains skin color and why I’m brown,” she remembers thinking. But she asked an immediate second question: “What about my hair?” She explains, “That’s a very Black-woman thing to do.” Like skin, hair is a highly racialized physical feature that Black people, especially women, often feel pressured to defend or alter. 

If skin tone can be measured by melanin and mapped against the intensity of UV ray exposure, Lasisi wondered, was there a similar method for quantifying hair and trying to find links with geography? She started looking for an evolutionary explanation for hair diversity and specifically for hair morphology and curliness. No one could give her a good answer. The scientific literature was sparse. The few studies she found that focused on hair diversity didn’t inquire about the evolutionary function of different hair textures—and none went about trying to quantify the different textures. 

So Lasisi began honing her own methodology. She developed a method to measure variation in human hair shape by taking pieces from individual strands of hair, measuring their curves, and using a National Institutes of Health program called ImageJ to analyze and quantify hair from different people. She published her results in 2016 in the American Journal of Physical Anthropology (now known as the American Journal of Biological Anthropology ) and a follow-up study in Nature Scientific Reports in 2021.

Many of the existing ways of describing curly hair are laden with bias, Lasisi says: as frizzy, kinky, hard to comb, fragile. “A lot of these metrics have prejudice and bias baked into them,” she says, “because they have a value system baked into them.”

In the 12 years since that formative undergrad class, Lasisi worked to answer unasked questions about hair shape and diversity as a Ph.D. candidate at Pennsylvania State University before completing a science-communication-focused postdoctoral research position at the same institution and a second postdoc focusing on forensic genetic genealogy at the University of Southern California. Now she’s running the Lasisi Lab as an assistant professor of anthropology at the University of Michigan. “I really want to focus on setting up my lab so that we can be a hub for people who want hair samples measured quantitatively and accurately, for all kinds of purposes,” she says.

Now that Lasisi has become a leading expert on hair form and function, the questions she hopes to explore reach far beyond variations in curl pattern. For example, Lasisi is excited to study the complicated interplay between hormones and hair growth. “You can have lots of testosterone coursing through you and still not be able to grow a full beard,” she says, “or still experience hair loss.” These relationships are not well understood but could change treatments for hair loss and options for gender-affirming care. Lasisi is also setting up research into the diversity of facial hair and skin to help reduce bias in artificial intelligence. You can train AI on huge datasets, “but if you don’t understand the structure of a given variation, you might have unwittingly created a biased training dataset,” she says.

Lasisi’s current focus in this area is gathering as much data as possible about hair, including genetic information. “There are a lot of fantastic resources and databases with genetic data, all kinds of scientific data,” Lasisi says, “but the data they include on hair, if they include it at all, is categorical, subjective, and usually self-described,” meaning people provided their own nonstandardized descriptions of their own hair. The Lasisi Lab must continue her work of collecting hair samples to get measurements precise and accurate enough to answer the questions she’s asking. It will also swab saliva to gather genetic data on its hair donors. 

Beyond potential applications in medicine and AI, an increased understanding of the evolution behind human diversity can be empowering. Lasisi’s most recent paper presents evidence that human scalp hair evolved to protect the head from the sun, and that tightly curled hair provides the most protection against solar rays while allowing for airy ventilation to cool the head down. Knowing how your traits came to be can change the way you think about yourself, says Lasisi. Learning of that research linking UV rays and skin melanation gave her a newfound appreciation for why she looks the way she does. She hopes to do the same for many others in the years to come. —H.S.

Breathing new life into old math

Brendan Keith, assistant professor of applied mathematics at Brown University

Computer simulations touch everything in our lives. Every branch of manufacturing—shipbuilding, engineering, aerospace, and more—runs on a family of algorithms developed over the last six or so decades, says Brendan Keith, an assistant professor of applied mathematics at Brown University. These types of algorithms, known as finite element methods, help researchers like Keith predict how objects might behave in different environments. You can see why this would be particularly useful for, say, engineers working on massive skyscrapers or manufacturers constructing giant cargo freighters.

As a kid, Brendan Keith never expected to turn into a self-described “computational science nerd.” But he fell in love with high school physics, eager to model how far various careening objects would fly through the air and estimate where they’d land. Eventually, he shifted his focus from physics to geometry and finally to applied math, which he studied for his master’s degree at Canada’s McGill University in 2013. Five years later, he earned his Ph.D. in computational science, engineering and mathematics at the University of Texas at Austin. Now he wields theory in an attempt to explain the world around us.

As an assistant professor at Brown University, Keith is breathing new life into old math. He’s currently trying to update vintage techniques with a method he calls proximal Galerkin (a recent addition to a group of methods named after Soviet mathematician Boris Galerkin). 

The problems in question involve keeping simulations in line with how the world works. This includes preventing one object from penetrating another during computer simulations, an error that makes it harder to predict how things will play out off-screen. To take example, when plane manufacturers model how air moves past a jet wing during flight, they need to make sure the air pressure conditions reflect reality. So far, it’s been tricky to simulate such a scenario accurately without violating the laws of real-life physics and thus blowing up a computer model. So Keith and his collaborator Thomas Surowiec found a way to “cut up the problem into little pieces” on a computer, do mathematical transformations on each piece, run their simulations, and then recombine them.

Their new solution ends up reducing the time and costs involved with this type of work, and, most importantly, prevents computer models from going haywire; for example, the proximal Galerkin method can help a car crash simulation run correctly by preventing steel from entering, rather than wrapping around, a telephone pole. This year, Keith received a grant of more than $800,000 from the Department of Energy to fine-tune this research, along with other new numerical methods he’s cooking up.

Keith has also received attention for much loftier calculations. In 2021, he helped develop a new machine learning technique for peering into black holes. Specifically, he found a way to model the motion of a type of black hole based on data from the gravitational waves such black holes produce. Researchers have long accomplished this by solving Einstein’s equations of general relativity. (He suggested that certain chaotic events, like two black holes smashing together, mess with what he called space-time to create the ripples known as gravitational waves.) But that process is time-consuming and costly and can require supercomputers. With Keith’s technique, the math can be done on any laptop—including the nearly decade-old Mac he used for the project. And while someone might need weeks to get an accurate result using traditional models, the new machine learning tool can crank out the answer in less than an hour.

With his new funding, Keith wants to apply his proximal Galerkin method to a whole host of problems. He has been working with researchers from a wide range of fields and is now thinking up ways to better understand bone fractures, design large bridges, and simulate car crashes. In fact, there are so many possibilities that he says it’s hard to decide what his team should tackle first. He does know one thing for sure: He hopes his algorithms continue to prove useful for future generations of researchers.

“My dream would be to actually develop an algorithm that people are still using in 50 years,” Keith says. “Methods come and go, people are always studying the newest-fangled thing on the same old problems, but it’s very rare that somebody comes across something that has that kind of staying power—that’s what I’d really like to be someday.” —M.G.

Solving space mysteries with Antarctic ice

Carlos Argüelles-Delgado, Assistant Professor of Physics, Harvard University

For decades, physicists have widely accepted a theory that supposedly explains our universe. But some say the Standard Model of physics, which was developed in the early 1970s, is beginning to crumble under pressure. The Standard Model holds that nature is composed of a few types of particles and governed by a handful of forces. But this theory doesn’t explain certain astronomical anomalies. For example, it can’t explain how gravity exists, nor account for the mysterious dark matter and energy that take up most of our universe.

Another hole in the Standard Model: It doesn’t explain a key fact about neutrinos , a teeny type of electrically neutral particle. Each second, about 65 billion of them surge through each square centimeter of your body. While the rules posited by the Standard Model suggest that neutrinos should lack mass, they’re actually the most abundant particles with mass in the universe.

“Particle physicists are like detectives,” says Carlos Argüelles-Delgado, an assistant professor of physics at Harvard University. “You look at the places where things seem to be misbehaving. Neutrinos misbehave.”

This glaring discrepancy intrigued Argüelles-Delgado while they completed a physics master’s program in their native country of Peru just over a decade ago. Neutrino research was starting to really pick up steam. Like many Latin American physicists at the time, Argüelles-Delgado says, their academic adviser lacked the funds for experimental physics equipment. That gave Argüelles-Delgado the opportunity to build up extensive experience on the theory side of things. They credit this focus on theory for spurring them to dig into calculations and attempt to get to know the puzzling particles.

This is a particularly difficult task because neutrinos rarely interact with other types of matter, earning them the nickname “ghost particles.” This means they’re difficult to detect and study. But there’s a major upside here. Since neutrinos hardly touch other forms of matter, they allow researchers to study scenes that light-based telescopes can’t capture—such as the inner workings of the sun or even other galaxies—without stuff like electromagnetic fields and gas getting in the way.

Neutrinos may be hard to spot, but the task is not impossible: You just have to look in unusual places. Argüelles-Delgado’s work at the University of Wisconsin at Madison, where they earned their Ph.D. in physics in 2015, led them to the world’s largest neutrino detector. Fittingly called IceCube , the massive tool incorporates thousands of optical sensors buried up to 8,000 feet within Antarctic ice. When neutrinos interact with the ice, they create electrically charged particles that emit a blue glow called Cherenkov radiation. IceCube’s sensors record the light pattern to gauge the direction and energy of an arriving neutrino. Ice aids in the endeavor because it’s clear enough to provide glimpses of the blue glow, and the light can easily travel within its depths.

Since it launched in 2011, IceCube has detected neutrinos packed with energy from some of the “most violent” locations in the universe, Argüelles-Delgado says. Since they travel in a straight line—a product of their inability to be interrupted or redirected by most other forms of matter—these chaotic neutrinos readily offer clues about their origins. 

While many observed neutrinos have zipped over from the sun or within our atmosphere, last year IceCube caught some from a galaxy 47 million light-years away (still relatively close to us in cosmic terms) that’s powered by a supermassive black hole. The hole pulls in particles at high speeds. When these collide, they spawn neutrinos. Argüelles-Delgado also studies neutrinos from our own galactic neighborhood. Back in Cambridge, Argüelles-Delgado sifts through this info for new insights. For instance, they pay close attention to the various types or “flavors” of neutrinos IceCube measures, such as the rarely observed tau neutrino, and what secrets they hold. They have even pioneered a new quantum computing method to model how neutrinos morph into these different varieties.

Such IceCube data, which is being studied by Argüelles-Delgado and their colleagues around the globe, could help in the hunt for elusive dark matter, offer a closer look at black holes, and even identify new laws of physics, because neutrinos point straight toward the phenomena under investigation, like a perp’s footprints at the scene of the crime.

Over the next decade, IceCube will receive fancier sensors and calibration devices and grow around eight times larger. A sharper resolution could lead to new out-of-this-world insights. “We could find a new force, we could find new matter,” Argüelles-Delgado says. “It could just totally change physics.” —M.G.

Growing organs to solve health disparities

Quinton Smith, Assistant Professor of Chemical and Biomolecular Engineering at the University of California, Irvine

Up to 90 percent of all new drugs fail in human trials, often because substances that proved safe for animal subjects end up being toxic for people. But a high-tech solution could bring us more effective medications with fewer hiccups: growing tissue from stem cells to mimic our organs and how they interact with drugs. 

This futuristic field appealed to Quinton Smith, who initially studied to be a chemical engineer at the University of New Mexico starting in 2007 and had considered becoming a doctor. But after realizing that patient interaction didn’t appeal to him as much as research, he decided he wanted to build things to help improve human health and began working in biology labs. He applied his engineering expertise to the human body when he earned his Ph.D. in chemical and biomolecular engineering from Johns Hopkins University in 2017. Smith studied under Sharon Gerecht, an early leader in stem cell engineering. Now he harnesses groups of lab-grown cells called organoids to research, and potentially treat, some of the deadliest and trickiest conditions. 

“Having that stem cell perspective and an engineering background is a really powerful tool,” Smith says. “We have this idea: Can we actually replace animal studies and create a body on a chip to really study how tissues interact?”

Researchers have spent decades laying the foundations for today’s organoid experiments. Today, labs can begin with an adult’s own blood or skin cells and transform them into any adult cell type thanks to cocktails of chemicals that direct their growth. 

Smith’s lab at the University of California at Irvine, which was founded in 2021, is one of many around the globe tinkering with these organ models. But his team stands apart with a game-changing new technique. Scientists have long struggled to add structures similar to vascular tissue to cells in lab dish experiments. It’s tricky to match the specific types of cells found in blood vessels in the body. But blood vessels are a crucial feature, “the holy grail” of tissue engineering according to Smith: They supply oxygen and nutrients to cells and could enable increasingly large groups of lab-grown structures to communicate and thrive. 

Thanks to his engineering know-how, Smith joined a community of scientists who take inspiration from the intricate circuits built into the silicon chips that run our computers. His team first prints tiny channels onto its “organ” chips, which are made of a compound that contains silicon. They fill those divots with a gel containing protein found in blood clots. Inside this gel, they’ve also added lab-grown blood vessel cells and a type of cell called a fibroblast that helps form connective tissue, which work together to form tubes that mimic the vasculature inside our bodies. The final product looks nothing like the slimy tissues inside us, but is rather a see-through gadget with branching wires inside, ranging from the size of a quarter to just a few micrometers across.

Smith’s team can send liquids through the faux blood vessels to learn, for example, how changes in blood flow contribute to disease development, or how a specific drug might affect circulation. Smith says this type of work on liver models, which he first delved into during his postdoctoral research, is particularly important. No FDA-approved drugs exist for end-stage liver diseases like cirrhosis, which disproportionately affects marginalized groups such as Latino and Black people.  

By using patients’ own stem cells, he could even create bespoke models to test their unique reactions to drugs and develop personalized stem cell therapies. Smith envisions a future where doctors can restore the function of a diseased organ by implanting stem cells in easy-to-reach places, like under the arm. Such supplementation probably wouldn’t make organ transplants a thing of the past, Smith says, but it could help lower demand and shorten wait times for those in need.

Smith is also harnessing his faux organs to tackle another glaring healthcare gap. He’s interested in investigating the causes of preeclampsia, a potentially deadly complication of pregnancy that causes high blood pressure. It is a risk for any birthing person but is especially prominent among Black women in the United States. In fact, Black women born in the US experience higher rates of preeclampsia than those who immigrate here, perhaps due to stress and discrepancies in hospital experiences tied to systemic racism. But the exact mechanisms behind this phenomenon remain unclear. “There seems to be not only an ancestral contribution, but social, economic, or other factors that can really impact maternal health,” Smith says.

To solve this urgent puzzle, Smith is now using the techniques he’s developed to create placentalike cells. He’s using them to figure out how a pregnant person’s environment might spur inflammation in the placenta and interfere with blood vessels. This is particularly critical work because there are logistical and ethical obstacles to doing studies on pregnant people. Smith’s insights could help scientists around the world illuminate what he calls “the black box” of human development and help ensure safer pregnancies for marginalized communities. —M.G.

Empowering marginalized communities with AI

Robin Brewer Assistant Professor of Information, School of Information, U Michigan

Growing up as an only child, Robin Brewer says she had lots of time to tinker on her computer. She devoured math programs along with roller coaster simulators and typing games. Her parents worked in IT for the government, and her household was one of the first in her Maryland neighborhood to own a PC. When she attended a science and tech high school, her self-described “bossy” only-child nature compelled her to learn how to order computers around with the help of languages like Java and C++. This curiosity led her to study computer science and human-centered computing for her bachelor’s and master’s degrees at the University of Maryland.

But by the time she was working on her Ph.D. at Northwestern University in 2013, Brewer had decided to take a step back and learn how other people interact with technology. She noticed that devices tend to cater to young, nondisabled people, with accessibility features often added as an afterthought.

“I’m really an advocate for more aging-first and disability-first approaches to research and design,” says Brewer, now an assistant professor of information at the University of Michigan.

To help remedy these disparities, Brewer develops accessibility tools based on feedback from older adults and people with disabilities. In recent years, she has focused on the role of AI and voice assistants like Siri in engaging these communities and helping connect people with shared experiences. In June 2023, she co-authored a paper exploring ways to improve cultural sensitivity in voice technologies for older Black users. Some of the participants, for instance, expressed a desire to hear voices that sounded more like their own without relying on stereotypes.

Her work could prove particularly useful as the US contends with a significant shortage of home healthcare workers. While Brewer doesn’t think we should replace human professionals with automatons in this context, she thinks tech can help ease the stress of supporting seniors at home—in 2020, more than 50 million people in the US reportedly cared for someone with particular health needs or disabilities. 

Right now, Brewer is working on a program that listens to caregivers and care recipients describe their daily challenges. It then provides a human- or AI-generated summary of what they’ve shared to help them navigate challenging conversations. For example, a care recipient may desire more independence during tasks like grocery shopping, so a voice assistant can work as an intermediary and help communicate this to their caregiver.

While Brewer believes that the artificial intelligence behind much of her work can benefit marginalized populations, she’s also dedicated to confronting the biases baked into the models AI uses. In a study published in June 2023, she collaborated with Google research scientists to highlight the biases that large language models—such as those that power chatbots like ChatGPT—hold toward people with disabilities. The team found that these models tend to view disability in narrow terms, as mostly pertaining to physical limitations, and can make harmful remarks about people who rely on others for assistance. 

While researchers have conducted work on the prejudices AI presents toward communities of color and older people, Brewer says that harms against the disabled community have been harder to quantify and act on. She says it’s likely possible to retrain models to be more useful and display more sensitivity. 

Last year, Brewer was awarded more than $500,000 from the National Science Foundation’s prestigious early-career prize to study how voice technologies could help older adults avoid lies online. Given the potential shortcomings of tech, along with the risks of invasive surveillance, she wants to help users build digital boundaries and learn to spot misinformation—particularly as chatbots have been found to “ hallucinate ” and make things up or peddle conspiracy theories.

“It’s not all rainbows and sunshine,” she says. “Every technology has some type of risk to it.” —M.G.

Aiming lasers at the universe’s origins

Ronald Garcia Ruiz, Assistant Professor of Physics, MIT

Although our universe is billions of years old , there are still a lot of unanswered questions about how the big bang played out. Matter composes everything around us, but how did it evolve? To tackle this existential question, researchers have devised some mind-boggling experiments. Take the Large Hadron Collider at the European Organization for Nuclear Research (CERN) spanning the border between France and Switzerland, where scientists smash together high-energy particles at nearly the speed of light. The team there studies the resulting interactions to test predictions from the Standard Model theory that explains our universe.

But the Large Hadron Collider can’t provide all the answers scientists are looking for, according to MIT assistant professor of physics Ronald Garcia Ruiz. After working at CERN while earning his Ph.D. at KU Leuven in Belgium, Garcia Ruiz decided to approach the same question with the help of lasers. 

He compares digging into the beginnings of the universe to peering inside a watermelon. There are two distinct approaches to glimpsing its innards: one rather violent and one more meticulous. You can see inside it by blowing it up with a high-energy particle and then reconstructing the interior, which is essentially the protocol at the Large Hadron Collider (sort of like shooting the fruit and then gluing it back together). Instead, Garcia Ruiz realized he could probe the inside of this “watermelon” with an electron, offering a more clear-cut and precise view.

He and his colleagues create radioactive atoms and molecules to get the job done, since they have an imbalance of neutrons and protons. This asymmetry makes the atoms and molecules highly sensitive to funky physics happenings in the lab and could help explain the uneven ratio of matter and antimatter in the universe, a reality that violates the Standard Model.

It’s hard to concoct these radioactive substances, but Garcia Ruiz and his team work at specialized nuclear physics labs, including CERN and the Facility for Rare Isotope Beams, a U.S. Department of Energy Office of Science user facility operated by Michigan State University. They then wield their unique, highly sensitive laser system to measure the interactions among protons, neutrons, and electrons within these radioactive substances. They can use the laser to measure the minuscule structures of these particles and catch any shifts in energy among different isotopes of the same element. So-called isotope shifts matter because they may reveal phenomena that defy the known fundamental forces of nature.

It all has to happen in a flash, since these radioactive substances can disappear in less than a fraction of a second. This gives Garcia Ruiz an unprecedented look at hard-to-capture properties of the particles and forces that serve as the foundation of the universe’s visible matter. These studies could even identify (or rule out the existence of) never-before-seen particles.

This laser-precise technique nicely complements the Large Hadron Collider method, because it offers scientists a close-up view of the building blocks of nature. “I can do this in a very elegant way,” Garcia Ruiz says.

Garcia Ruiz’s work has made waves outside the world of particle physics too. His team is collaborating with astronomers to help hunt for radioactive molecules in space. By measuring the radiation emitted by these molecules at the lab, Garcia Ruiz offers “fingerprints” scientists can use to seek out similar measurements in the cosmos. This work can also help astronomers pinpoint when astrophysical events went down due to the time-telling properties of radioactive decay. Garcia Ruiz has even tailored his experiments to create radioactive nuclei with tons of neutrons to help teams studying the dynamics of neutron-rich matter such as neutron stars—which form when a giant star runs out of fuel and collapses.

Ultimately, he aims to help plug some of the gaps in the Standard Model and delve deeper into the origins of life, the universe, and everything. Garcia Ruiz even suspects that he could analyze how dark matter, a little-understood substance that constitutes 85 percent of the universe’s matter, engages with the protons, neutrons, and electrons in his lab.

“I think we are really at the age of a revolution in our understanding of fundamental physics,” he says. —M.G.

Hannah Seo

Hannah Seo is a science contributor at Popular Science. She started as an intern in 2020 and has since regularly contributed to both Popular Science’s website and quarterly magazine. Hannah’s reporting has covered everything from COVID-19 to rare archeological finds, and they’re always down to talk about quirky marine creatures or the mysteries of neuropsychology.

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IMAGES

  1. Learn How to Solve Force Problems

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  2. 😎 Problem solving in physics. Science Problems Help. 2019-02-15

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  3. Dynamics: Force, Friction, FBD, & Problem Solving

    problem solving in force physics

  4. Force Physics Problem Solving, IIT-JEE physics classes

    problem solving in force physics

  5. gravity

    problem solving in force physics

  6. Spice of Lyfe: Acceleration Formula Physics Examples

    problem solving in force physics

VIDEO

  1. FORCE AND MOTION problem solving #physics #easy

  2. Chapter 9

  3. Work done calculation by Force-displacement graph. #class11 #jeemain #neet #engineeringmechanics

  4. Work done calculation by force-displacement graph. #jeemain #neet #engineeringmechanics #class11

  5. Work done by a variable force. #class11 #jeemain #neet #physics #engineeringmechanics

  6. 24.Physics

COMMENTS

  1. 6.1 Solving Problems with Newton's Laws

    Problem-Solving Strategy Applying Newton's Laws of Motion Identify the physical principles involved by listing the givens and the quantities to be calculated. Sketch the situation, using arrows to represent all forces. Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.

  2. 6.2: Solving Problems with Newton's Laws (Part 1)

    Apply problem-solving techniques to solve for quantities in more complex systems of forces Use concepts from kinematics to solve problems using Newton's laws of motion Solve more complex equilibrium problems

  3. Solving problems which involve forces, friction, and Newton's ...

    Solving problems which involve forces, friction, and Newton's Laws: A step-by-step guide | Phyley Support Ukraine 🇺🇦 Help Ukrainian Army Humanitarian Assistance to Ukrainians In this tutorial you will learn how to examine and solve problems which involve forces and applications of Newton's Laws.

  4. Forces and Newton's laws of motion

    Science Physics library Unit 3: Forces and Newton's laws of motion 300 possible mastery points Mastered Proficient Familiar Attempted Not started Quiz Unit test About this unit This unit is part of the Physics library. Browse videos, articles, and exercises by topic. Newton's laws of motion Learn Newton's first law of motion introduction

  5. 4.6 Problem-Solving Strategies

    Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 4.20 (a).

  6. 6.3: Solving Problems with Newton's Laws (Part 2)

    Solution. We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ Δ v = 8.00 m/s . We are given the elapsed time, so Δ Δ t = 2.50 s. The unknown is acceleration, which can be found from its definition: a = Δv Δt. (6.3.1) (6.3.1) a = Δ v Δ t.

  7. Problem-Solving Strategies

    Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 1 (a).

  8. Ch. 1 Problems & Exercises

    4.1 Development of Force Concept; 4.2 Newton's First Law of Motion: Inertia; 4.3 Newton's Second Law of Motion: Concept of a System; 4.4 Newton's Third Law of Motion: Symmetry in Forces; 4.5 Normal, Tension, and Other Examples of Forces; 4.6 Problem-Solving Strategies; 4.7 Further Applications of Newton's Laws of Motion

  9. 8.1 Linear Momentum, Force, and Impulse

    We can solve for Δ p by rearranging the equation. F net = Δ p Δ t. to be. Δ p = F net Δ t . F net Δ t is known as impulse and this equation is known as the impulse-momentum theorem. From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts.

  10. 1.7 Solving Problems in Physics

    Problem-solving skills are clearly essential to success in a quantitative course in physics. More important, the ability to apply broad physical principles—usually represented by equations—to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts.

  11. 4.9: Problem-Solving Strategies

    Problem-Solving Strategy for Newton's Laws of Motion. Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure(a).

  12. Calculating Force

    What is the resultant acceleration? ball rolls forward with a net acceleration of . What is the net force on the ball? Newton's second law states that Plug in the values given to us and solve for the force. Louisa rolls a ball with of force. She observes that it has a constant linear acceleration of . What is the mass of the ball?

  13. 6.1 Solving Problems with Newton's Laws

    If we are solving for force and end up with units of millimeters per second, then we have made a mistake. There are many interesting applications of Newton's laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills.

  14. Force Problems

    Force Problems On this page I put together a collection of force problems to help you understand forces better. The required equations and background reading to solve these problems are given on the friction page, the equilibrium page, and Newton's second law page . Problem # 1 A ball of mass m is hanging on a wall with a string, as shown.

  15. Centripetal force problem solving (video)

    And so, now's a good time for me to let you in on a little secret. The secret to solving centripetal force problems is that you solve them the same way you solve any force problem. In other words, first, you draw a quality force diagram. And then you use Newton's second law for one of the directions at a time.

  16. Vectors and Forces Problem Sets

    Problem 1: For each collection of listed forces, determine the vector sum or the net force. Audio Guided Solution Show Answer Problem 2: Hector is walking his dog (Fido) around the neighborhood. Upon arriving at Fidella's house (a friend of Fido's), Fido turns part mule and refuses to continue on the walk.

  17. F=ma Practice Problems

    Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations. In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems. 1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right? See Answer. 2.

  18. Forces in Physics, tutorials and Problems with Solutions

    The concepts of forces, friction forces, action and reaction forces, free body diagrams, tension of string, inclined planes, etc. are discussed and through examples, questions with solutions and clear and self explanatory diagrams. Questions to practice for the SAT Physics test on forces are also included with their detailed solutions.

  19. Solving Force Problems in Physics by Using Free-Body Diagrams

    Follow this seven-step method to solve force problems: Identify the forces acting on each object. For each force acting on one of the objects from Step 1, draw an arrow that indicates the direction of the force, as shown in the following figure. Note that the tail of the arrow indicates which part of the object the force is acting on.

  20. Newton's Law Problem Sets

    Problem 1: An African elephant can reach heights of 13 feet and possess a mass of as much as 6000 kg. Determine the weight of an African elephant in Newtons and in pounds. (Given: 1.00 N = .225 pounds) Audio Guided Solution Show Answer Problem 2: About twenty percent of the National Football League weighs more than 300 pounds.

  21. Net Force Problems Revisited

    Resolution of Forces Equilibrium and Statics Net Force Problems Revisited Inclined Planes Two-Body Problems This part of Lesson 3 focuses on net force-acceleration problems in which an applied force is directed at an angle to the horizontal.

  22. 1.4: Solving Physics Problems

    Trigonometry and Solving Physics Problems. In physics, most problems are solved much more easily when a free body diagram is used. Free body diagrams use geometry and vectors to visually represent the problem. Trigonometry is also used in determining the horizontal and vertical components of forces and objects.

  23. Tension, String, Forces Problems with Solutions

    Problem 1. A block of mass 5 Kg is suspended by a string to a ceiling and is at rest. Find the force F c exerted by the ceiling on the string. Assume the mass of the string to be negligible. Solution. a) The free body diagram below shows the weight W and the tension T 1 acting on the block. Tension T 2 acting on the ceiling and F c the reaction ...

  24. [2312.02091] Physics simulation capabilities of LLMs

    Physics simulation capabilities of LLMs. Mohamad Ali-Dib, Kristen Menou. [Abridged abstract] Large Language Models (LLMs) can solve some undergraduate-level to graduate-level physics textbook problems and are proficient at coding. Combining these two capabilities could one day enable AI systems to simulate and predict the physical world.

  25. Tech giants compete to build 1st reliable, general-purpose ...

    The machines could revolutionize problem solving in medicine, physics, chemistry and engineering. ... new engineering-- all leading to this processor that computes with the atomic forces that ...

  26. How quantum computers may change the world as we know it

    Quantum computers could give us answers to impossible problems in physics, chemistry, engineering and medicine. ... classical computers will never be able to solve that problem, not now, not 100 ...

  27. These 10 scientists are on the cusp of changing the world

    Each year, Popular Science honors 10 early-career researchers who've gotten a head start: The Brilliant 10. These researchers already stand out as innovators and change makers in their fields ...