Chapter 4, Problem 41P

To determine

The force the car exerts on the trailer when it applies a force on the ground to accelerate.

Answer to Problem 41P

Solution:

The force exerted by the car on the trailer is 1200 N.

Explanation of Solution

The car is connected to the trailer; hence, both the car and the trailer have the same acceleration. The force exerted by the car on the trailer can be calculated using the formulae for the force of friction and Newton’s second law.

Given:

The mass of the car $m_c=1280\text{ kg}$

Mass of the trailer $m_t=350\text{ kg}$

Force exerted by the car on the ground $F_c=3.6\times {10}^3\text{ N}$

Formula used:

    $F_{fr}={\mu }_k{m}_{t}g$

    $a=\frac{F_c-F_{fr}}{m_t+m_c}$

    $F=F_{fr}+m_{t}a$

Calculation:

The trailer has a coefficient of kinetic friction ${\mu }_k$ between itself and the ground. Calculate the force of friction.

    F_fr=μ_km_tg

        =(0.15)(350 kg)(9.8 m/s²)

        =514.5 N

The car exerts a force $F_c$ on the ground. The force of friction acts in the opposite direction to it. The net force acting on the car and the trailer produces an acceleration an on both the car and the trailer.

Calculate the acceleration acting on the car and the trailer using the values of the quantities given in the problem.

    $a=\frac{F_c-F_{fr}}{m_t+m_c}$

        $=\frac{\left(3.6\times {10}^3\text{N}\right)-\left(514.5\text{ N}\right)}{\left(1280\text{ kg}\right)+\left(350\text{ kg}\right)}$

        =1.89 m/s²

The trailer also experiences the same acceleration a. If the car exerts a force F on the trailer, then the net force which produces the acceleration a is the difference between this force and the force of kinetic friction. Writing Newton’s second law expression for the trailer,

    $F-F_{fr}=m_{t}a$

Calculate the force exerted on the trailer using the value given in the problem.

     F=F_fr+m_ta

       =(514.5 N)+(350 kg)(1.89 m/s²)

       =1176 N