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.

To determine

The net force on the car.

Answer to Problem 42P

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

Explanation of Solution

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

Write the expression for the net force along horizontal direction.

${\left(F_{net}\right)}_{\text{car}}=m{a}_x$        (I)

Here, ${\left(F_{net}\right)}_{\text{car}}$ is the net force on car, $m$ is the mass of car and $a_x$ is the acceleration along horizontal direction.

The car is having only horizontal acceleration.

Conclusion:

Substitute, $1000\text{kg}$ for $m$, $2.15{\text{m/s}}^{\text{2}}$ for $a_x$ in Equation (I) to find ${\left(F_{net}\right)}_{\text{car}}$.

  $\begin{array}{c}{\left(F_{net}\right)}_{\text{car}}=\left(1000\text{kg}\right)\left(2.15{\text{m/s}}^{\text{2}}\right)\\ =2.15\times {10}^3\text{N}\end{array}$

The net force is acting along forward direction.

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

To determine

The net force on the trailer.

Answer to Problem 42P

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

Explanation of Solution

Write the expression for the net force along horizontal direction.

${\left(F_{net}\right)}_{\text{trailer}}=m_{\text{trailer}}{a}_x$        (II)

Here, ${\left(F_{net}\right)}_{\text{trailer}}$ is the net force, $m_{\text{trailer}}$ is the mass of the trailer and $a_x$ is the acceleration along horizontal direction.

Conclusion:

Substitute, $300\text{kg}$ for $m_{\text{trailer}}$, $2.15{\text{m/s}}^{\text{2}}$ for $a_x$ in Equation (II) to find ${\left(F_{net}\right)}_{\text{trailer}}$.

  $\begin{array}{c}{\left(F_{net}\right)}_{\text{trailer}}=\left(300\text{kg}\right)\left(2.15{\text{m/s}}^{\text{2}}\right)\\ =645\text{N}\end{array}$

The net force is acting along forward direction.

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

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 $\underset{\_}{645\text{}\text{N}}$.

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 $\underset{\_}{645\text{}\text{N}}$.

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 $\underset{\_}{1.02\times {10}^4\text{N}}$ 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.

$R_{car}=R_x\widehat{i}+R_y\widehat{j}$        (III)

Here, $R_{car}$ is the net force exerted by the road on the car, $R_x,R_y$ are the x and y component of force.

Write the expression for the x-component of force.

$R_x=\left[{\left(F_{net}\right)}_{\text{trailer}}+{\left(F_{net}\right)}_{\text{car}}\right]$        (IV)

Write the expression for the y-component of force.

$R_y=mg$        (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).

$\begin{array}{c}\left|R\right|=\sqrt{R_x^2+R_y^2}\\ =\sqrt{{\left[{\left(F_{net}\right)}_{\text{trailer}}+{\left(F_{net}\right)}_{\text{car}}\right]}^2+{\left(mg\right)}^2}\end{array}$        (VI)

Here, $\left|R\right|$ 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).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)\\ ={\tan }^{-1}\left(\frac{\left(mg\right)}{\left[{\left(F_{net}\right)}_{\text{trailer}}+{\left(F_{net}\right)}_{\text{car}}\right]}\right)\end{array}$        (VII)

Here, $\theta$ is the angle of direction.

Conclusion:

Substitute, $1000\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $645\text{N}$ for ${\left(F_{net}\right)}_{\text{trailer}}$, $2.15\times {10}^3\text{N}$ for ${\left(F_{net}\right)}_{\text{car}}$ in Equation (VI) to find $\left|R\right|$.

  $\begin{array}{c}\left|R\right|=\sqrt{{\left[\left(645\text{N}\right)+\left(2.15\times {10}^3\text{N}\right)\right]}^2+{\left(\left(1000\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\right)}^2}\\ =1.02\times {10}^4\text{N}\end{array}$

Substitute, $1000\text{kg}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$, $645\text{N}$ for ${\left(F_{net}\right)}_{\text{trailer}}$, $2.15\times {10}^3\text{N}$ for ${\left(F_{net}\right)}_{\text{car}}$ in Equation (VII) to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(\left(1000\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\right)}{\left[\left(645\text{N}\right)+\left(2.15\times {10}^3\text{N}\right)\right]}\right)\\ =74.1°\end{array}$

Here, the force with magnitude $1.02\times {10}^4\text{N}$ 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 $\underset{\_}{1.02\times {10}^4\text{N}}$ and directed at an angle $74.1°$ below the horizontal and rearward.