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University Physics Volume 1

6.1 Solving Problems with Newton’s Laws

University Physics Volume 16.1 Solving Problems with Newton’s Laws
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  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

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

  1. Identify the physical principles involved by listing the givens and the quantities to be calculated.
  2. Sketch the situation, using arrows to represent all forces.
  3. Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
  4. Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
  5. 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 6.2(a). Then, as in Figure 6.2(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.
Figure 6.2 (a) A grand piano is being lifted to a second-story apartment. (b) Arrows are used to represent all forces: TT is the tension in the rope above the piano, FTFT is the force that the piano exerts on the rope, and ww is the weight of the piano. All other forces, such as the nudge of a breeze, are assumed to be negligible. (c) Suppose we are given the piano’s mass and asked to find the tension in the rope. We then define the system of interest as shown and draw a free-body diagram. Now FTFT is no longer shown, because it is not a force acting on the system of interest; rather, FTFT acts on the outside world. (d) Showing only the arrows, the head-to-tail method of addition is used. It is apparent that if the piano is stationary, T=wT=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 6.2(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 6.2(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 6.2(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:

Fx=max,Fy=may.Fx=max,Fy=may.

(If, for example, the system is accelerating horizontally, then you can then set ay=0.ay=0.) 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.

Example 6.1

Different Tensions at Different Angles Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in Figure 6.3. 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.
Figure 6.3 A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The free-body diagram for the traffic light is also shown. (d) The forces projected onto vertical (y) and horizontal (x) axes. The horizontal components of the tensions must cancel, and the sum of the vertical components of the tensions must equal the weight of the traffic light. (e) The free-body diagram shows the vertical and horizontal forces acting on the traffic light.

Strategy The system of interest is the traffic light, and its free-body diagram is shown in Figure 6.3(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 6.3(d). There are two unknowns in this problem (T1T1 and T2T2), 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.

Solution First consider the horizontal or x-axis:

Fnetx=T2x+T1x=0.Fnetx=T2x+T1x=0.

Thus, as you might expect,

|T1x|=|T2x|.|T1x|=|T2x|.

This gives us the following relationship:

T1cos30°=T2cos45°.T1cos30°=T2cos45°.

Thus,

T2=1.225T1.T2=1.225T1.

Note that T1T1 and T2T2 are not equal in this case because the angles on either side are not equal. It is reasonable that T2T2 ends up being greater than T1T1 because it is exerted more vertically than T1.T1.

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

Fnety=T1y+T2yw=0.Fnety=T1y+T2yw=0.

This implies

T1y+T2y=w.T1y+T2y=w.

Substituting the expressions for the vertical components gives

T1sin30°+T2sin45°=w.T1sin30°+T2sin45°=w.

There are two unknowns in this equation, but substituting the expression for T2T2 in terms of T1T1 reduces this to one equation with one unknown:

T1(0.500)+(1.225T1)(0.707)=w=mg,T1(0.500)+(1.225T1)(0.707)=w=mg,

which yields

1.366T1=(15.0kg)(9.80m/s2).1.366T1=(15.0kg)(9.80m/s2).

Solving this last equation gives the magnitude of T1T1 to be

T1=108N.T1=108N.

Finally, we find the magnitude of T2T2 by using the relationship between them, T2=1.225T1T2=1.225T1, found above. Thus we obtain

T2=132N.T2=132N.

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.

Example 6.2

Drag Force on a Barge Two tugboats push on a barge at different angles (Figure 6.4). The first tugboat exerts a force of 2.7×105N2.7×105N in the x-direction, and the second tugboat exerts a force of 3.6×105N3.6×105N in the y-direction. The mass of the barge is 5.0×106kg5.0×106kg and its acceleration is observed to be 7.5×10−2m/s27.5×10−2m/s2 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.
Figure 6.4 (a) A view from above of two tugboats pushing on a barge. (b) The free-body diagram for the ship contains only forces acting in the plane of the water. It omits the two vertical forces—the weight of the barge and the buoyant force of the water supporting it cancel and are not shown. Note that FappFapp is the total applied force of the tugboats.

Strategy The directions and magnitudes of acceleration and the applied forces are given in Figure 6.4(a). We define the total force of the tugboats on the barge as FappFapp so that

Fapp=F1+F2.Fapp=F1+F2.

The drag of the water FDFD is in the direction opposite to the direction of motion of the boat; this force thus works against Fapp,Fapp, as shown in the free-body diagram in Figure 6.4(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 F1F1 and F2.F2. The problem quickly becomes a one-dimensional problem along the direction of FappFapp, since friction is in the direction opposite to Fapp.Fapp. Our strategy is to find the magnitude and direction of the net applied force FappFapp and then apply Newton’s second law to solve for the drag force FD.FD.

Solution Since FxFx and FyFy are perpendicular, we can find the magnitude and direction of FappFapp directly. First, the resultant magnitude is given by the Pythagorean theorem:

Fapp=F12+F22=(2.7×105N)2+(3.6×105N)2=4.5×105N.Fapp=F12+F22=(2.7×105N)2+(3.6×105N)2=4.5×105N.

The angle is given by

θ=tan−1(F2F1)=tan−1(3.6×105N2.7×105N)=53.1°.θ=tan−1(F2F1)=tan−1(3.6×105N2.7×105N)=53.1°.

From Newton’s first law, we know this is the same direction as the acceleration. We also know that FDFD is in the opposite direction of Fapp,Fapp, since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as Fapp,Fapp, but its magnitude is slightly less than Fapp.Fapp. The problem is now one-dimensional. From the free-body diagram, we can see that

Fnet=FappFD.Fnet=FappFD.

However, Newton’s second law states that

Fnet=ma.Fnet=ma.

Thus,

FappFD=ma.FappFD=ma.

This can be solved for the magnitude of the drag force of the water FDFD in terms of known quantities:

FD=Fappma.FD=Fappma.

Substituting known values gives

FD=(4.5×105N)(5.0×106kg)(7.5×102m/s2)=7.5×104N.FD=(4.5×105N)(5.0×106kg)(7.5×102m/s2)=7.5×104N.

The direction of FDFD has already been determined to be in the direction opposite to Fapp,Fapp, or at an angle of 53°53° south of west.

Significance 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 FDFD 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.

Example 6.3

What Does the Bathroom Scale Read in an Elevator? Figure 6.5 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 1.20m/s2,1.20m/s2, 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.
Figure 6.5 (a) The various forces acting when a person stands on a bathroom scale in an elevator. The arrows are approximately correct for when the elevator is accelerating upward—broken arrows represent forces too large to be drawn to scale. TT is the tension in the supporting cable, ww is the weight of the person, wsws is the weight of the scale, wewe is the weight of the elevator, FsFs is the force of the scale on the person, FpFp is the force of the person on the scale, FtFt is the force of the scale on the floor of the elevator, and NN is the force of the floor upward on the scale. (b) The free-body diagram shows only the external forces acting on the designated system of interest—the person—and is the diagram we use for the solution of the problem.

Strategy If the scale at rest is accurate, its reading equals FpFp, the magnitude of the force the person exerts downward on it. Figure 6.5(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 6.5(b). Analysis of the free-body diagram using Newton’s laws can produce answers to both Figure 6.5(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 ww and the upward force of the scale Fs.Fs. According to Newton’s third law, FpFp and FsFs are equal in magnitude and opposite in direction, so that we need to find FsFs in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,

Fnet=ma.Fnet=ma.

From the free-body diagram, we see that Fnet=Fsw,Fnet=Fsw, so we have

Fsw=ma.Fsw=ma.

Solving for FsFs gives us an equation with only one unknown:

Fs=ma+w,Fs=ma+w,

or, because w=mg,w=mg, simply

Fs=ma+mg.Fs=ma+mg.

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 Fsw=ma.Fsw=ma.)

Solution

  1. We have a=1.20m/s2,a=1.20m/s2, so that
    Fs=(75.0kg)(9.80m/s2)+(75.0kg)(1.20m/s2)Fs=(75.0kg)(9.80m/s2)+(75.0kg)(1.20m/s2)

    yielding
    Fs=825N.Fs=825N.
  2. Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because a=ΔvΔta=ΔvΔt and Δv=0.Δv=0. Thus,
    Fs=ma+mg=0+mgFs=ma+mg=0+mg

    or
    Fs=(75.0kg)(9.80m/s2),Fs=(75.0kg)(9.80m/s2),

    which gives
    Fs=735N.Fs=735N.

Significance The scale reading in Figure 6.5(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:

Fnet=ma=0=FswFnet=ma=0=Fsw
Fs=w=mgFs=w=mg
Fs=(75.0kg)(9.80m/s2)=735N.Fs=(75.0kg)(9.80m/s2)=735N.

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 6.5(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 6.1

Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20m/s2.1.20m/s2.

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.

Example 6.4

Two Attached Blocks Figure 6.6 shows a block of mass m1m1 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 m2.m2. Find the acceleration of the blocks and the tension in the string in terms of m1,m2,andg.m1,m2,andg.

(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.
Figure 6.6 (a) Block 1 is connected by a light string to block 2. (b) The free-body diagrams of the blocks.

Strategy We draw a free-body diagram for each mass separately, as shown in Figure 6.6. 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: T+w1+N=m1a1T+w1+N=m1a1

For block 2: T+w2=m2a2.T+w2=m2a2.

Notice that TT 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

Solution 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:

Block 1Block 2Fx=maxFy=mayTx=m1a1xTym2g=m2a2y.Block 1Block 2Fx=maxFy=mayTx=m1a1xTym2g=m2a2y.

When block 1 moves to the right, block 2 travels an equal distance downward; thus, a1x=a2y.a1x=a2y. Writing the common acceleration of the blocks as a=a1x=a2y,a=a1x=a2y, we now have

T=m1aT=m1a

and

Tm2g=m2a.Tm2g=m2a.

From these two equations, we can express a and T in terms of the masses m1andm2,andg:m1andm2,andg:

a=m2m1+m2ga=m2m1+m2g

and

T=m1m2m1+m2g.T=m1m2m1+m2g.

Significance 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 T=m2g.T=m2g. You can see from the free-body diagram of block 2 that cannot be correct if the block is accelerating.

Check Your Understanding 6.2

Calculate the acceleration of the system, and the tension in the string, when the masses are m1=5.00kgm1=5.00kg and m2=3.00kg.m2=3.00kg.

Example 6.5

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 6.7, m1=2.00kgm1=2.00kg and m2=4.00kg.m2=4.00kg. Consider the pulley to be frictionless. (a) If m2m2 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.
Figure 6.7 An Atwood machine and free-body diagrams for each of the two blocks.

Strategy 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 a2a2 in the downward direction, block 1 accelerates upward with acceleration a1a1. Thus, a=a1=a2.a=a1=a2.

Solution

  1. We have
    Form1,Fy=Tm1g=m1a.Form2,Fy=Tm2g=m2a.Form1,Fy=Tm1g=m1a.Form2,Fy=Tm2g=m2a.

    (The negative sign in front of m2am2a indicates that m2m2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is
    (m2m1)g=(m1+m2)a.(m2m1)g=(m1+m2)a.

    Solving for a:
    a=m2m1m1+m2g=4kg2kg4kg+2kg(9.8m/s2)=3.27m/s2.a=m2m1m1+m2g=4kg2kg4kg+2kg(9.8m/s2)=3.27m/s2.
  2. Observing the first block, we see that
    Tm1g=m1aT=m1(g+a)=(2kg)(9.8m/s2+3.27m/s2)=26.1N.Tm1g=m1aT=m1(g+a)=(2kg)(9.8m/s2+3.27m/s2)=26.1N.

Significance The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system, (m2m1)g(m2m1)g, to the total mass of the system, m1+m2m1+m2. We can also use the Atwood machine to measure local gravitational field strength.

Check Your Understanding 6.3

Determine a general formula in terms of m1,m2m1,m2 and g for calculating the tension in the string for the Atwood machine shown above.

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.

Example 6.6

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.

Strategy 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.

Solution

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

    Substituting the known values yields
    a=8.00m/s2.50s=3.20m/s2.a=8.00m/s2.50s=3.20m/s2.
  2. Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is,
    Fnet=ma.Fnet=ma.

    Substituting the known values of m and a gives
    Fnet=(70.0kg)(3.20m/s2)=224N.Fnet=(70.0kg)(3.20m/s2)=224N.

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.

Significance 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.

Check Your Understanding 6.4

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?

Example 6.7

What Force Acts on a Model Helicopter? A 1.50-kg model helicopter has a velocity of 5.00j^m/s5.00j^m/s at t=0.t=0. It is accelerated at a constant rate for two seconds (2.00 s) after which it has a velocity of (6.00i^+12.00j^)m/s.(6.00i^+12.00j^)m/s. What is the magnitude of the resultant force acting on the helicopter during this time interval?

Strategy We can easily set up a coordinate system in which the x-axis (i^(i^ direction) is horizontal, and the y-axis (j^(j^ direction) is vertical. We know that Δt=2.00sΔt=2.00s and (6.00i^+12.00j^m/s)(5.00j^m/s).(6.00i^+12.00j^m/s)(5.00j^m/s). From this, we can calculate the acceleration by the definition; we can then apply Newton’s second law.

Solution We have

a=ΔvΔt=(6.00i^+12.00j^m/s)(5.00j^m/s)2.00s=3.00i^+3.50j^m/s2a=ΔvΔt=(6.00i^+12.00j^m/s)(5.00j^m/s)2.00s=3.00i^+3.50j^m/s2
F=ma=(1.50kg)(3.00i^+3.50j^m/s2)=4.50i^+5.25j^N.F=ma=(1.50kg)(3.00i^+3.50j^m/s2)=4.50i^+5.25j^N.

The magnitude of the force is now easily found:

F=(4.50N)2+(5.25N)2=6.91N.F=(4.50N)2+(5.25N)2=6.91N.

Significance The original problem was stated in terms of i^j^i^j^ vector components, so we used vector methods. Compare this example with the previous example.

Check Your Understanding 6.5

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

Example 6.8

Baggage Tractor Figure 6.8(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 F=(820.0t)N,F=(820.0t)N, 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.
Figure 6.8 (a) A free-body diagram is shown, which indicates all the external forces on the system consisting of the tractor and baggage carts for carrying airline luggage. (b) A free-body diagram of the tractor only is shown isolated in order to calculate the tension in the cable to the carts.

Strategy 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 T,T, which is our objective.

Solution

  1. Fx=msystemaxFx=msystemax and Fx=820.0t,Fx=820.0t, so
    820.0t=(650.0+250.0+150.0)aa=0.7809t.820.0t=(650.0+250.0+150.0)aa=0.7809t.

    Since acceleration is a function of time, we can determine the velocity of the tractor by using a=dvdta=dvdt with the initial condition that v0=0v0=0 at t=0.t=0. We integrate from t=0t=0 to t=3:t=3:
    dv=adt,03dv=03.00adt=03.000.7809tdt,v=0.3905t2]03.00=3.51m/s.dv=adt,03dv=03.00adt=03.000.7809tdt,v=0.3905t2]03.00=3.51m/s.
  2. Refer to the free-body diagram in Figure 6.8(b).
    Fx=mtractorax820.0tT=mtractor(0.7805)t(820.0)(3.00)T=(650.0)(0.7805)(3.00)T=938N.Fx=mtractorax820.0tT=mtractor(0.7805)t(820.0)(3.00)T=(650.0)(0.7805)(3.00)T=938N.

Significance 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 6.8(a), whereas only the mass of the truck (since it supplied the force) was of use in Figure 6.8(b).

Recall that v=dsdtv=dsdt and a=dvdta=dvdt. 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 dt=dsvdt=dsv and dt=dva.dt=dva. Now, equating these expressions, we have dsv=dva.dsv=dva. We can rearrange this to obtain ads=vdv.ads=vdv.

Example 6.9

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 6.9). Determine the maximum height it will travel if atmospheric resistance is measured as FD=(0.0100v2)N,FD=(0.0100v2)N, 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.
Figure 6.9 (a) The mortar fires a shell straight up; we consider the friction force provided by the air. (b) A free-body diagram is shown which indicates all the forces on the mortar shell. (credit a: modification of work by OS541/DoD; The appearance of U.S. Department of Defense (DoD) visual information does not imply or constitute DoD endorsement.)

Strategy 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.

Solution Initially, y0=0y0=0 and v0=50.0m/s.v0=50.0m/s. At the maximum height y=h,v=0.y=h,v=0. The free-body diagram shows FDFD to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

Fy=may Fy=may
FDw=may0.0100v298.0=10.0aa=−0.00100v29.80.FDw=may0.0100v298.0=10.0aa=−0.00100v29.80.

The acceleration depends on v and is therefore variable. Since a=f(v),a=f(v), we can relate a to v using the rearrangement described above,

ads=vdv.ads=vdv.

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

ady=vdv,(−0.00100v29.80)dy=vdv.ady=vdv,(−0.00100v29.80)dy=vdv.

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

0hdy=50.00vdv(−0.00100v29.80)0hdy=50.00vdv(0.00100v2+9.80)=(−5×103)ln(0.00100v2+9.80)|50.00.0hdy=50.00vdv(−0.00100v29.80)0hdy=50.00vdv(0.00100v2+9.80)=(−5×103)ln(0.00100v2+9.80)|50.00.

Thus, h=114m.h=114m.

Significance 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.

Check Your Understanding 6.6

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

Interactive

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).

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