Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 4, Problem 42P

A 1 000-kg car is pulling a 300-kg trailer. Together, the car and trailer move forward with an acceleration of 2.15 m/s2. Ignore any force of air drag on the car and all frictional forces on the trailer. Determine (a) the net force on the car, (b) the net force on the trailer, (c) the force exerted by the trailer on the car, and (d) the resultant force exerted by the car on the road.

(a)

Expert Solution
Check Mark
To determine

The net force on the car.

Answer to Problem 42P

The net force on the car is 2.15×103N forward.

Explanation of Solution

The free body diagram for the system is given by figure 1.

Principles of Physics: A Calculus-Based Text, Chapter 4, Problem 42P

Write the expression for the net force along horizontal direction.

(Fnet)car=max        (I)

Here, (Fnet)car is the net force on car, m is the mass of car and ax is the acceleration along horizontal direction.

The car is having only horizontal acceleration.

Conclusion:

Substitute, 1000kg for m, 2.15m/s2 for ax in Equation (I) to find (Fnet)car.

  (Fnet)car=(1000kg)(2.15m/s2)=2.15×103N

The net force is acting along forward direction.

Thus, the net force on the car is 2.15×103N forward.

(b)

Expert Solution
Check Mark
To determine

The net force on the trailer.

Answer to Problem 42P

The net force on the car is 645N forward.

Explanation of Solution

Write the expression for the net force along horizontal direction.

(Fnet)trailer=mtrailerax        (II)

Here, (Fnet)trailer is the net force, mtrailer is the mass of the trailer and ax is the acceleration along horizontal direction.

Conclusion:

Substitute, 300kg for mtrailer, 2.15m/s2 for ax in Equation (II) to find (Fnet)trailer.

  (Fnet)trailer=(300kg)(2.15m/s2)=645N

The net force is acting along forward direction.

Thus, the net force on the car is 645N forward.

(c)

Expert Solution
Check Mark
To determine

The force exerted by the trailer on the car.

Answer to Problem 42P

The force exerted by the trailer on the car is 645N_.

Explanation of Solution

According to the Newton’s third each and every action has equalled and opposite reaction. In the given problem, the force acting on the trailer given by the car is 645N. This force is directed along forward.

Thus, according to Newton’s third law, the trailer will also give same amount of force on the car. Hence, The force exerted by the trailer on the car is 645N_.

(d)

Expert Solution
Check Mark
To determine

The resultant force exerted by the car on the road.

Answer to Problem 42P

The resultant force exerted by the car on the road is 1.02×104N_ and directed at an angle 74.1° below the horizontal and rearward.

Explanation of Solution

Write the expression for the net force exerted by the road acting on the car.

Rcar=Rxi^+Ryj^        (III)

Here, Rcar is the net force exerted by the road on the car, Rx,Ry are the x and y component of force.

Write the expression for the x-component of force.

Rx=[(Fnet)trailer+(Fnet)car]        (IV)

Write the expression for the y-component of force.

Ry=mg        (V)

Here, g is the gravitational acceleration.

Write the expression for the magnitude of the net force exerted by the road on the car by using (IV) and (V).

|R|=Rx2+Ry2=[(Fnet)trailer+(Fnet)car]2+(mg)2        (VI)

Here, |R| is the magnitude of the net force exerted by the road on the car.

Write the expression for the magnitude of the net force exerted by the road on the car by using (IV) and (V).

θ=tan1(RyRx)=tan1((mg)[(Fnet)trailer+(Fnet)car])        (VII)

Here, θ is the angle of direction.

Conclusion:

Substitute, 1000kg for m, 9.8m/s2 for g, 645N for (Fnet)trailer, 2.15×103N for (Fnet)car in Equation (VI) to find |R|.

  |R|=[(645N)+(2.15×103N)]2+((1000kg)(9.8m/s2))2=1.02×104N

Substitute, 1000kg for m, 9.8m/s2 for g, 645N for (Fnet)trailer, 2.15×103N for (Fnet)car in Equation (VII) to find θ.

θ=tan1(((1000kg)(9.8m/s2))[(645N)+(2.15×103N)])=74.1°

Here, the force with magnitude 1.02×104N exerted by the road on the car is acting at an angle 74.1° above the horizontal along forward direction.

Thus, according to Newton’s third law, the resultant force exerted by the car on the road is 1.02×104N_ and directed at an angle 74.1° below the horizontal and rearward.

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Chapter 4 Solutions

Principles of Physics: A Calculus-Based Text

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