4.1. Solve: A force is basically a push or a pull on an object. There are five basic characteristics of forces. (i) Aforce has an agent that is the direct and immediate source of the push or pull. (ii) Most forces are contact forcesthat occur at a point of contact between the object and its environment. (iii) A very few forces, such as gravity andmagnetism, are long-range forces that require no contact. (iv) A force is a vector quantity, having both a magnitude(or size) and a direction. (v) When multiple forces act on an object, the forces combine through vector addition to

give a net force r rF Finet = ∑ .

4.2. Visualize:

Assess: Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse walls.

4.3. Visualize:

4.4. Model: Assume friction is negligible compared to other forces.Visualize:

4.5. Visualize: Please refer to Figure Ex4.5.Solve: Mass is defined to be

m = 1

slope of the acceleration-versus-force graph

A larger slope implies a smaller mass. We know m2 = 0.20 kg, and we can find the other masses relative to m2 bycomparing their slopes. Thus

m

m

m m

1

2

1 2

1

1

1

5 2

2

50 40

0 40 0 40 0 20 0 08

= = = = =

⇒ = = × =

/

/.

. . . .

slope 1

slope 2

slope 2

slope 1

kg kg

Similarly,

m

m

m m

3

2

3 2

1

1

1

2 5

5

22 50

2 50 2 50 0 20 0 50

= = = = =

⇒ = = × =

/

/.

. . . .

slope 3

slope 2

slope 2

slope 3

kg kg

Assess: From the initial analysis of the slopes we had expected m3 > m2 and m1 < m2 . This is consistent with ournumerical answers.

4.6. Model: An object’s acceleration is linearly proportional to the net force.Solve: (a) One rubber band produces a force F, two rubber bands produce a force 2F, and so on. Because F a∝and two rubber bands (force 2F) produce an acceleration of 1.2 m/s2 , four rubber bands will produce anacceleration of 2.4 m/s2.(b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). Using F ma= ,

2F = (2m)a ⇒ a = F/m = 0.6 m/s2

4.7. Force is not necessary for motion. Constant velocity motion occurs in the absence of forces, that is, when thenet force on an object is zero. Thus, it is incorrect to say that “force causes motion.” Instead, force causesacceleration. That is, force causes a change in the motion of an object, and acceleration is the kinematic quantity

that measures a change of motion. Newton’s second law quantifies this idea by stating that the net forcerFnet on an

object of mass m causes the object to undergo an acceleration:

rr

aF

m= net

The acceleration vector and the net force vector must point in the same direction.

4.8. Visualize:

Solve: Newton’s second law is F ma= . When F = 2 N, we have 2 N = (0.5 kg)a, hence a = 4 m/s2. Afterrepeating this procedure at various points, the above graph is obtained.

4.9. Visualize: Please refer to Figure Ex4.9.Solve: Newton’s second law is F ma= . We can read a force and an acceleration from the graph, and hence findthe mass. Choosing the force F = 1 N gives us a = 4 m/s2. Newton’s second law yields m = 0.25 kg.

4.10. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 4.3gives us no information on textbooks, but does give the weight of a one-pound object. Place a pound weight in onehand and the textbook on the other. The sensation on your hand is the weight of the object. The sensation from thetextbook is about five times the sensation from the pound weight. So we conclude the weight of the textbook isabout five times the weight of the one-pound object or about 25 N.(b) According to Table 4.3, the propulsion force on a car is 5000 N. Now a bicycle (including the rider) is about100 kg. This is about one-tenth of the mass of a car which is about 1000 kg for a compact model. The accelerationof a bicycle is somewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law asfollows:

F bicycle mass of car acceleration of car N

50 N( ) = ( ) × ( ) = =1

10

1

5

5000100

So we would roughly estimate the propulsion force of a bicycle to be 100 N.

4.11. Solve: (a) This problem calls for an estimate so we are looking for an approximate answer. Table 4.3gives us no information on pencils, but does give us the weight of the U.S. quarter. Put the quarter on one hand anda pencil on the other hand. The sensation on your hand is the weight of the object. The sensation from the quarter isabout the same as the sensation from the pencil. So they both have about the same weight. We can estimate theweight of the pencil to be 0.05 N.(b) According to Table 4.3, the propulsion force on a car is 5000 N. Now the mass of a sprinter is about 100 kg. Thisis about one-tenth of the mass of a car which is about 1000 kg for a compact model. The acceleration of a sprinter issomewhat less than that of a car, let’s guess about one-fifth. We can write Newton’s second law as follows:

F sprinter mass of car acceleration of car N

50 N( ) = ( ) × ( ) = =1

10

1

5

5000100

So, we would roughly estimate the propulsion force of a sprinter to be 100 N.Assess: This is the same estimated number as we obtained in problem 4.10. This is reasonable since in both thecases the propulsion force comes from a human and it probably does not matter how the human is providing that force.

4.12. Visualize:

Solve: The object will be in equilibrium if rF3 has the same magnitude as

r rF F1 2+ but is in the opposite direction

so that the sum of all the three forces is zero.

4.13. Visualize:

Solve: The object will be in equilibrium if rF3 has the same magnitude as

r rF F1 2+ but is in the opposite direction

so that the sum of all three forces is zero.

4.14. Visualize:

Solve: The object will be in equilibrium if rF3 has the same magnitude as

r rF F1 2+ but is in the opposite direction

so that the sum of all the three forces is zero.

4.15. Visualize:

Solve: The free-body diagram shows two equal and opposite forces such that the net force is zero. The forcedirected down is labeled as a weight, and the force directed up is labeled as a tension. With zero net force theacceleration is zero. So, a possible description is: “An object hangs from a rope and is at rest.” Or, “An object hangingfrom a rope is moving up or down with a constant speed.”

4.16. Visualize:

Solve: The free-body diagram shows three forces with a net force (and therefore net acceleration) upward. There is a

force labeled rw directed down, a force

rFthrust directed up, and a force

rD directed down. So a possible description is:

“A rocket accelerates upward.”

4.17. Visualize:

Solve: The free-body diagram shows three forces. There is a weight force rw which is down. There is a normal

force labeled rn which is up. The forces

rw and

rn are shown with vectors of the same length so they are equal in

magnitude and the net vertical force is zero. So we have an object on the ground which is not moving vertically.

There is also a force rfk to the left. This must be a frictional force and we need to decide whether it is static or

kinetic friction. The frictional force is the only horizontal force so the net horizontal force must be rfk . This means

there is a net force to the left producing an acceleration to the left. This all implies motion and therefore thefrictional force is kinetic. A possible description is: “A baseball player is sliding into second base.”

4.18. Visualize:

4.19. Visualize:

Assess: The problem says that there is no friction and it tells you nothing about any drag; so we do not includeeither of these forces. The only remaining forces are the weight and the normal force.

4.20. Visualize:

Assess: There are only two forces on the elevator. The weight rw is directed down and the tension in the cable is

directed up. Since the elevator is descending at a steady speed there is no acceleration and the two forces must beequal in magnitude.

4.21. Visualize:

Assess: The problem uses the word “sliding.” Any real situation involves friction with the surface. Since we arenot told to neglect it, we show that force.

4.22. Visualize:

Figure (a) shows velocity as downward, so the object is moving down. The length of the vector increases with each

step showing that the speed is increasing (like a dropped ball). Thus, the acceleration is directed down. Since r rF ma=

the force is in the same direction as the acceleration and must be directed down.Figure (b), however, shows the velocity as upward, so the object is moving upward. But the length of the vectordecreases with each step showing that the speed is decreasing (like a ball thrown up). Thus, the acceleration is alsodirected down. As in part (a) the net force must be directed down.

4.23. Visualize:

The velocity vector in figure (a) is shown downward and to the left. So movement is downward and to the left. Thevelocity vectors get successively longer which means the speed is increasing. Therefore the acceleration is

downward and to the left. By Newton’s second law r rF ma= , the net force must be in the same direction as the

acceleration. Thus, the net force is downward and to the left.The velocity vector in (b) is shown to be upward and to the right. So movement is upward and to the right. Thevelocity vector gets successively shorter which means the speed is decreasing. Therefore the acceleration isdownward and to the left. From Newton’s second law, the net force must be in the direction of the acceleration andso it is directed downward and to the left.

4.24. Visualize:

Solve: According to Newton’s second law F ma= , the force at any time is found simply by multiplying thevalue of the acceleration by the mass of the object.

4.25. Visualize:

Solve: According to Newton’s second law F ma= , the force at any time is found simply by multiplying thevalue of the acceleration by the mass of the object.

4.26. Visualize:

Solve: According to Newton’s second law F ma= , the acceleration at any time is found simply by dividing thevalue of the force by the mass of the object.

4.27. Visualize:

Solve: According to Newton’s second law F ma= , the acceleration at any time is found simply by dividing thevalue of the force by the mass of the object.

4.28. Model: Use the particle model for the object.Solve: (a) We are told that for an unknown force (call it F0) acting on an unknown mass (call it m0) the accelerationof the mass is 10 m/s2. According to Newton’s second law, F0 = m0(10 m/s2). The force then becomes 1

2 0F . Newton’ssecond law gives

1

2

1

2100 0 0F m a m= = ( )[ ] m s2

This means a is 5 m/s2.(b) The force is F0 and the mass is now 1

2 0m . Newton’s second law gives

F m a m0 0 0

1

210= = ( ) m s2

This means a = 20 m/s2.(c) A similar procedure gives a = 10 m/s2.(d) A similar procedure gives a = 2.5 m/s2.

4.29. Model: Use the particle model for the object.Solve: (a) We are told that for an unknown force (call it F0) acting on an unknown mass (call it m0) theacceleration of the mass is 8 m/s2. According to Newton’s second law, F0 = m0(8 m/s2). The force then becomes2 0F . Newton’s second law gives

2 2 80 0 0F m a m= = ( )[ ] m s2

This means a is 16 m/s2.(b) The force is F0 and the mass is now 2 0m . Newton’s second law gives

F m a m0 0 02 8= = ( ) m s2

This means a = 4 m/s2.(c) A similar procedure gives a = 8 m/s2.(d) A similar procedure gives a = 32 m/s2.

4.30. Visualize:

Solve: (d) There are a normal force and a weight which are equal and opposite, so this is an object on a horizontalsurface. The description could be: “A tow truck pulls a stuck car out of the mud.”

4.31. Visualize:

Solve: (d) There is a normal force and a weight which are equal and opposite, so this is an object on a horizontalsurface. The description of this free-body diagram could be: “A jet plane is flying at constant speed.”

4.32. Visualize:

Solve: (d) This is an object on a surface because w n= . It must be moving to the left because the kinetic friction is tothe right. The description of the free-body diagram could be: “A compressed spring is shooting a plastic block to the left.”

4.33. Visualize:

Solve: (d) There is only a single force of weight. We are unable to tell the direction of motion. The description is:“Galileo has dropped a ball from the Leaning Tower of Pisa.”

4.34. Visualize:

Solve: (d) There is an object on an inclined surface. The net force is down the plane so the acceleration is downthe plane. The net force includes both the frictional force and the component of the weight. The direction of theforce of kinetic friction implies that the object is moving upward. The description could be: “A car is skidding upan embankment.”

4.35. Visualize:

Solve: (d) There is an object on an inclined surface with a tension force down the surface. There is a smallfrictional force up the surface implying that the object is sliding down the slope. A description could be: “A sled isbeing pulled down a slope with a rope which is parallel to the slope.”

4.36. Visualize:

Solve: (d) There is a thrust at an angle to the horizontal and a weight. There is no normal force so the object is noton a surface. The description could be: “A rocket is fired at an angle to the horizontal and there is no drag force.”

4.37. Visualize:

Tension is the only contact force. The downward acceleration implies that w T> .

4.38. Visualize:

4.39. Visualize:

The normal force is perpendicular to the ground. The thrust force is parallel to the ground and in the direction ofacceleration. The drag force is opposite to the direction of motion.

4.40. Visualize:

The normal force is perpendicular to the hill. The frictional force is parallel to the hill.

4.41. Visualize:

The normal force is perpendicular to the hill. The kinetic frictional force is parallel to the hill and directed upwardopposite to the direction of motion. The wind force is given as horizontal. Since the skier stays on the slope (that is,there is no acceleration away from the slope) the net force must be parallel to the slope.

4.42. Visualize:

The ball is rolling, so there is no indication of a speed change (no acceleration, no net force) and no indication offriction. Thus the weight and the normal force are the only forces.

4.43. Visualize:

The drag force due to air is opposite the motion.

4.44. Visualize:

The ball rests on the floor of the barrel because the weight is equal to the normal force. There is a force of thespring to the right which causes an acceleration.

4.45. Visualize:

There are no contact forces on the rock. Weight is the only force acting on the rock.

4.46. Visualize:

The gymnast experiences the long range force of weight. There is also a contact force from the trampoline which isthe normal force of the trampoline on the gymnast. The gymnast is moving downward and the trampoline isdecreasing her speed, so the acceleration is upward and there is a net force upward. Thus the normal force must belarger than the weight. The actual behavior of the normal force will be complicated as it involves the stretching ofthe trampoline and therefore tensions.

4.47. Visualize:

You can see from the motion diagram that the box accelerates to the right along with the truck. According toNewton’s second law,

r rF ma= , there must be a force to the right acting on the box. This is friction, but not kinetic

friction. The box is not sliding against the truck. Instead, it is static friction, the force that prevents slipping. Were itnot for static friction, the box would slip off the back of the truck. Static friction acts in the direction needed toprevent slipping. In this case, friction must act in the forward (toward the right) direction.

4.48. Visualize:

You can see from the motion diagram that the bag accelerates to the left along with the car as the car slows down.According to Newton’s second law,

r rF ma= , there must be a force to the left acting on the bag. This is friction, but

not kinetic friction. The bag is not sliding against across the seat. Instead, it is static friction, the force that preventsslipping. Were it not for static friction, the bag would slide off the seat as the car stops. Static friction acts in thedirection needed to prevent slipping. In this case, friction must act in the backward (toward the left) direction.

4.49. Visualize: (a)

(b) (c)

(d) The ball accelerates downward until the instant when it makes contact with the ground. Once it makes contact,it begins to compress and to slow down. The compression takes a short but nonzero distance, as shown in themotion diagram. The point of maximum compression is the turning point, where the ball has an instantaneousspeed of v = 0 m/s and reverses direction. The ball then expands and speeds up until it loses contact with theground. The motion diagram shows that the acceleration vector

ra points upward the entire time that the ball is in

contact with the ground. An upward acceleration implies that there is a net upward force rFnet on the ball. The only

two forces on the ball are its weight downward and the normal force of the ground upward. To have a net forceupward requires n w> . So the ball bounces because the normal force of the ground exceeds the weight, causing anet upward force during the entire time that the ball is in contact with the ground. This net upward force slows theball, turns it, and accelerates it upward until it loses contact with the ground. Once contact with the ground is lost,the normal force vanishes and the ball is simply in free fall.

4.50. Visualize: (a)

You are sitting on a seat driving along to the right. Both you and the seat are moving with a constant speed. Thereis a force on you due to your weight which is directed down. There is a contact force between you and the seatwhich is directed up. Since you are not accelerating up or down the net vertical force on you is zero, which meansthe two vertical forces are equal in magnitude. The statement of the problem gives no indication of any othercontact forces. Specifically, we are told that the seat is very slippery. We can take this to mean there is no frictionalforce. So our force diagram includes only the normal force up, the weight down, and no horizontal force.(b) The above considerations lead to the free-body diagram that is shown.(c) The car (and therefore the seat) slows down. Does this create any new force on you? No. The forces remain thesame. This means the pictorial representation and the free-body diagram are unchanged.(d) The car slows down because of some new contact force on the car (maybe the brakes lock the wheels and theroad exerts a force on the tires). But there is no new contact force on you. So the force diagram for you remainsunchanged. There are no horizontal forces on you. You do not slow down and you continue at an unchangedvelocity until something in the picture changes for you (for example, you fall off the seat or hit the windshield).(e) The net force on you has remained zero because the net vertical force is zero and there are no horizontal forcesat all. According to Newton’s first law if the net force on you is zero, then you continue to move in a straight linewith a constant velocity. That is what happens to you when the car slows down. You continue to move forwardwith a constant velocity. The statement that you are “thrown forward” is misleading and incorrect. To be “thrown”there would need to be a net force on you and there is none. It might be correct to say that the car has been “thrownbackward” leaving you to continue onward (until you part company with the seat).(f) We are now asked to consider what happens if the bench is NOT slippery. That implies there is a frictional forcebetween the seat and you. This force is certainly horizontal (parallel to the surface of the seat). Is the frictionalforce directed forward (in the direction of motion) or backward? The car is slowing down and you are staying onthe bench. That means you are slowing down with the bench. Your velocity to the right is decreasing (you aremoving right and slowing down) so you are accelerating to the left. By Newton’s second law that means the forceproducing the acceleration must be to the left. That force is the force of static friction and it is shown on the free-body diagram below. Of course, when the car accelerates (increases in speed to the right) and you accelerate withit, then your acceleration is to the right and the frictional force must be to the right.