PH201  Chapter  4  Homework  –  Solutions    4.1. Set Up and Solve: The horizontal component of the force is to the right, and the vertical component is downward.  

4.8. Set Up: Take to be the direction in which the skater is moving initially. (comes to rest).

Solve: so

The only horizontal force on the skater is the friction force, so

The force is 46.7 N, directed opposite to the motion of the skater.

*4.9. Set Up: Take to be in the direction in which the cheetah moves.

Solve: (a) so

(b) The force is exerted on the cheetah by the ground. Reflect: The net force on the cheetah is in the same direction as the acceleration of the cheetah.    

4.10. Set Up: Let be the direction of the force. Use to calculate the acceleration of the puck. Then use constant acceleration equations to find the speed and displacement of the puck for

Solve:

*4.11. Set Up: Let be the direction of the force. Use the information about the motion to find the acceleration and then use to calculate m.

Solve: gives

Reflect: The mass determines the amount of acceleration produced by a given force.  

4.13. Set Up: We must use to find the mass of the boulder.

Solve:

Then Reflect: We must use mass in Newton’s second law. Mass and weight are proportional.

4.14. Set Up: An object of mass 1 kg weighs 2.205 lb.

Solve: (a)

(b) Yes, an adult collie should weigh much more than this, perhaps 40 lbs.

(c) Yes, this is over twice the weight of an average person.

 

4.18. Set Up:

Solve: (a) too heavy for a normal adult.

(b) this is reasonable. (c) this is reasonable.  

4.33. Set Up: The suitcase is in contact with the floor and this surface exerts an upward force and a horizontal force that is directed opposite to the motion of the suitcase. Solve: The free-body diagram for the suitcase is shown in the figure below.

 

4.37. Set Up: Use coordinates with upward. The objects are at rest so they have Solve: (a) The free-body diagram for the chandelier is shown in Figure (a) below. the tension in the bottom link of the chain, equals the magnitude of the upward force that the bottom of the chain exerts on the chandelier.

gives so

(b) Apply to the chain. The free-body diagram for the chain is shown in Figure (b) above. The

chandelier pulls downward on the chain with force is the tension in the top link of the chain. is the weight of the chain. We know from (a) that gives and

(c) Apply to the bottom half of the chain. The free-body diagram is shown in Figure (c) above.

is the weight of half the chain and is the tension in the middle link of the chain.

gives so Reflect: The tension increases from the top to the bottom of the chain.

4.38. Set Up: For simplicity, represent each acrobat as a point mass. The upper acrobat experiences the following forces: which are the force the trapeze exerts on the upper acrobat, the force that

the lower acrobat exerts on the upper acrobat, and the weight of the upper acrobat (respectively). Similarly, the lower acrobat experiences the following forces: which are the force that the upper acrobat exerts

on the lower acrobat and the weight of the lower acrobat (respectively). Note that and are equal in

magnitude by Newton’s third law. Solve: (a) The free-body diagram for the upper acrobat is shown below in Figure (a). (b) The free-body diagram for the lower acrobat is shown below in Figure (b). Reflect: A more realistic free-body diagram would show each acrobat as an extended object. The force on each leg or arm would be shown as a separate vector, which need not be perfectly vertical.

 

4.40. Set Up: Let the acceleration be in the direction. Apply to find the resultant horizontal force. Solve: (a) (b) The force is exerted by the blocks. The blocks push on the sprinter because the sprinter pushes on the blocks.  

4.46. Set Up: Let P be the magnitude of the upward force that the woman’s hand exerts on the briefcase. Take the direction upward so that and Also, since the elevator is accelerating downward, we have

Solve: The free-body diagram for the briefcase is shown in the figure below. Apply Newton’s Second Law to the briefcase: gives Solving for P we obtain:

By Newton’s third law, this is also the magnitude of the downward force that the briefcase exerts on the woman’s hand. Reflect: Since the acceleration is downward, P is less than the weight of the briefcase, which is

If the acceleration had been upward, P would be larger than the weight of the briefcase.

 

*4.51. Set Up: Let be upward. At his maximum height, While he is in the air,

Solve: (a) and gives

(b) For the motion while he is pushing against the floor, and gives

The acceleration is upward.

(c) The free-body diagram while he is pushing against the floor is given in the figure below. is the vertical force the floor applies to him.

(d) gives and The force he applied to the ground has this same magnitude and is downward. Reflect: The ground must push upward on him with a force greater than his weight in order to give him an upward acceleration.  

*4.59. Set Up: We can find the magnitude of the viscous force from Newton’s second law: The mass of the bacterium is

Solve: The viscous force needed to stop the bacterium in the required distance is

The correct answer is D.

4.60. Set Up: According to Newton’s second law Solve: If the bacterium is moving at a constant velocity, its acceleration must be zero. Thus, the net force on the bacterium must be zero. This means that the force exerted by the flagellum must be equal to the viscous force, which, from the previous problem, is The correct answer is B.