Chapter 5, Problem 23P

(a)

To determine

The net force on the car.

Answer to Problem 23P

The net force on the car is $2.15\times {10}^3\text{N}$ in forward direction.

Explanation of Solution

The car pulls the trailer with a particular force which provides the acceleration to the car in the forward direction.

Write the expression for the force acting on the car as.

  $F_c=m_{c}a$                                                                                                       (I)

Here, $F_c$ is the force on the car, $m_c$ is the mass of the car and $a$ is the acceleration of the car.

Conclusion:

Substitute $1000\text{kg}$ for $m_c$ and $2.15\text{m}/{\text{s}}^{\text{2}}$ for $a$ in equation (I).

  $\begin{array}{c}{F}_c=\left(1000\text{kg}\right)\left(2.15\text{m}/{\text{s}}^{\text{2}}\right)\\ =2.15\times {10}^3\text{kg}.\text{m}/{\text{s}}^{\text{2}}\\ =2.15\times {10}^3\text{N}\end{array}$

Thus, the net force on the car is $2.15\times {10}^3\text{N}$ in forward direction.

(b)

To determine

The net force on the trailer.

Answer to Problem 23P

The net force on the car is $645\text{N}$ in forward direction.

Explanation of Solution

The trailer is connected to the car. It will also move with the same acceleration as that of the car in the forward direction.

Write the expression for the force acting on the trailer as.

  $F_t=m_{t}a$                                                                                                       (II)

Here, $F_t$ is the force on the trailer, $m_t$ is the mass of the trailer and $a$ is the acceleration of the trailer.

Conclusion:

Substitute $300\text{kg}$ for $m_t$ and $2.15\text{m}/{\text{s}}^{\text{2}}$ for $a$ in equation (II).

  $\begin{array}{c}{F}_t=\left(300\text{kg}\right)\left(2.15\text{m}/{\text{s}}^{\text{2}}\right)\\ =645\text{kg}.\text{m}/{\text{s}}^{\text{2}}\\ =645\text{N}\end{array}$

Thus, the net force on the trailer is $645\text{N}$ in forward direction.

(c)

To determine

The force exerted by the trailer on the car.

Answer to Problem 23P

The force exerted by the trailer on the car is $645\text{N}$ in backward direction.

Explanation of Solution

The car pulls the trailer by applying a force in the horizontal direction. By newton’s third law of motion, this action of the car on the trailer is accompanied by a reaction force from the trailer acting on the car in the opposite direction.

Write the expression for the force exerted by the trailer on the car as.

  $F_{tc}=-F_t$                                                                                                      (III)

Here, $F_{tc}$ is the force exerted by the trailer on the car.

The negative sign signifies that the force is acting in the opposite direction.

Conclusion:

Substitute $645\text{N}$ for $F_t$ in equation (III).

  $F_{tc}=-\left(645\text{N}\right)$

Thus, the force exerted by the trailer on the car is $645\text{N}$ in backward direction.

(d)

To determine

The resultant force exerted by the car on the road.

Answer to Problem 23P

The force exerted by the car on the road is $\text{1}\text{.02}\times {\text{10}}^4\text{N}$ at $74.1°$ below horizontal in backward direction.

Explanation of Solution

The car is moving on the road. The forces exerted by the road on the car are the reaction force of the gravitational force and the forward force in the forward direction.

Write the expression for the net force acting on the car due to road as.

  $F=\left(m_c+m_t\right)a$                                                                                           (IV)

Here, $F$ is the net force acting due to road.

The normal reaction by the road on the car is opposite to the weight of the car acting in the downward direction.

Write the expression for the normal reaction on the car from road as.

  $n_c=m_{c}g$                                                                                                       (V)

Here, $n_c$ is the normal reaction on the car.

Write the expression for the net force acting on the car due to road as.

  $F_{net}=\sqrt{{\left(F\right)}^2+{\left(n_c\right)}^2}$                                                                                   (VI)

Here, $F_{net}$ is the net force acting on the car.

Write the expression for the angle made by the force with the horizontal as.

  $\theta ={\tan }^{-1}\left(\frac{n_c}{F}\right)$                                                                                          (VII)

Here, $\theta$ is the angle made by the resultant force with the horizontal.

Conclusion:

Substitute $1000\text{kg}$ for $m_c$, $300\text{kg}$ for $m_t$ and $2.15\text{m}/{\text{s}}^{\text{2}}$ for $a$ in equation (IV).

  $\begin{array}{c}F=\left(1000\text{kg}+300\text{kg}\right)\left(2.15\text{m}/{\text{s}}^{\text{2}}\right)\\ =2.795\times {10}^3\text{kg}\text{.m}/{\text{s}}^{\text{2}}\\ \approx 2.80\times {10}^3\text{N}\end{array}$

Substitute $1000\text{kg}$ for $m_c$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (V).

  $\begin{array}{c}{n}_c=\left(1000\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\\ =9.80\times {10}^3\text{kg}\text{.m}/{\text{s}}^{\text{2}}\\ =9.80\times {10}^3\text{N}\end{array}$

Substitute $\left(2.80\times {10}^3\text{N}\right)$ for $F$ and $\left(9.80\times {10}^3\text{N}\right)$ for $n_c$ in equation (VI).

  $\begin{array}{c}{F}_{net}=\sqrt{{\left(2.80\times {10}^3\text{N}\right)}^2+{\left(9.80\times {10}^3\text{N}\right)}^2}\\ =\sqrt{1.039\times {10}^8}\text{N}\\ \text{=1}\text{.019}\times {10}^4\text{N}\\ \approx \text{1}\text{.02}\times {10}^4\text{N}\end{array}$

Substitute $\left(2.80\times {10}^3\text{N}\right)$ for $F$ and $\left(9.80\times {10}^3\text{N}\right)$ for $n_c$ in equation (VII).

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{9.80\times {10}^3\text{N}}{2.80\times {10}^3\text{N}}\right)\\ ={\tan }^{-1}\left(3.5\right)\\ \approx 74.1°\end{array}$

Thus, the force exerted by the car on the road is $\text{1}\text{.02}\times {\text{10}}^4\text{N}$ at $74.1°$ below horizontal in backward direction.