Chapter 4, Problem 61P

A 1.00 × 10³ car is pulling a 300.-kg trailer. Together, the car and trailer have an acceleration of 2.15 m/s² in the positive x-direction. Neglecting frictional forces on the trailer, determine (a) the net force on the car, (b) the net force on the trailer, (c) the magnitude and direction of 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 61P

The net force on the car is 2150 N acting forward.

Explanation of Solution

Given Info: Mass of the car is $1.00\times {10}^3\text{kg}$. Mass of the trailer is 300 kg. The acceleration is $2.15{\text{ms}}^{-2}$.

Formula for the net force on the car is,

${\left(F_{net}\right)}_{car}=m_{c}a$

• a is the acceleration.
• $m_c$ is the mass of the car.

Substitute $1.00\times {10}^3\text{kg}$ for $m_c$ and $2.15{\text{ms}}^{-2}$ for a in the above expression to get a.

$\begin{array}{c}{\left(F_{net}\right)}_{car}=\left(1.00\times {10}^3\text{kg}\right)\left(2.15{\text{ms}}^{-2}\right)\\ =2150\text{N}\end{array}$

Conclusion:

The net force on the car is 2150 N acting forward.

To determine

The net force on the trailer.

Answer to Problem 61P

The net force on the trailer is 645 N acting forward.

Explanation of Solution

Given Info: Mass of the car is $1.00\times {10}^3\text{kg}$. Mass of the trailer is 300 kg. The acceleration is $2.15{\text{ms}}^{-2}$.

Formula for the net force on the trailer is,

${\left(F_{net}\right)}_{truck}=m_{t}a$

• a is the acceleration.
• $m_t$ is the mass of the trailer.

Substitute 300 kg for $m_t$ and $2.15{\text{ms}}^{-2}$ for a in the above expression to get a.

$\begin{array}{c}{\left(F_{net}\right)}_{truck}=\left(300\text{kg}\right)\left(2.15{\text{ms}}^{-2}\right)\\ =645\text{N}\end{array}$

Conclusion:

The net force on the trailer is 645 N acting forward.

To determine

The magnitude and direction of the force exerted by the trailer on the car.

Answer to Problem 61P

The force exerted by the trailer on the car is 645 N acting backwards.

Explanation of Solution

Given Info: Mass of the car is $1.00\times {10}^3\text{kg}$. Mass of the trailer is 300 kg. The acceleration is $2.15{\text{ms}}^{-2}$.

According to Newton’s law, the action and reaction forces are equal in direction and opposite in magnitude. Action force is the net force on the trailer.

From (b), the net force on the trailer is 645 N acting forward. Therefore, the force exerted by the trailer on the car is 645 N acting backwards.

Conclusion:

The force exerted by the trailer on the car is 645 N acting backwards.

To determine

The resultant force exerted by the car on the road.

Answer to Problem 61P

The resultant force exerted by the car on the road is 10190.78 N directed at an angle of ${74.08}^o$ below the horizon and backwards.

Explanation of Solution

Given Info: Mass of the car is $1.00\times {10}^3\text{kg}$. Mass of the trailer is 300 kg. The acceleration is $2.15{\text{ms}}^{-2}$.

The total horizontal force is,

$F_h={\left(F_{net}\right)}_{car}+{\left(F_{net}\right)}_{truck}$ (I)

The vertical force equals the weight of the car.

$F_v=m_{c}g$ (II)

The formula for the resultant force is,

$F_R=\sqrt{F_h^2+F_v^2}$

Substitute Equations (I) and (II) in the above equation.

$F_R=\sqrt{{\left[{\left(F_{net}\right)}_{car}+{\left(F_{net}\right)}_{truck}\right]}^2+{\left(m_{c}g\right)}^2}$

Substitute 2150 N for ${\left(F_{net}\right)}_{car}$, 645 N for ${\left(F_{net}\right)}_{truck}$, $1.00\times {10}^3\text{kg}$ for $m_c$ and $9.8{\text{ms}}^{-2}$ for g in the above expression to get $F_R$.

$\begin{array}{c}{F}_R=\sqrt{{\left[\left(2150\text{N}\right)+\left(645\text{N}\right)\right]}^2+{\left[\left(1.00\times {10}^3\text{kg}\right)\left(9.8{\text{ms}}^{-2}\right)\right]}^2}\\ =10190.78\text{N}\end{array}$

The direction of $F_R$ is,

$\theta ={\tan }^{-1}\left(\frac{F_v}{F_h}\right)$

Substitute Equations (I) and (II) in the above equation.

$\theta ={\tan }^{-1}\left(\frac{m_{c}g}{{\left(F_{net}\right)}_{car}+{\left(F_{net}\right)}_{truck}}\right)$

Substitute 2150 N for ${\left(F_{net}\right)}_{car}$, 645 N for ${\left(F_{net}\right)}_{truck}$, $1.00\times {10}^3\text{kg}$ for $m_c$ and $9.8{\text{ms}}^{-2}$ for g in the above expression to get $F_R$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(1.00\times {10}^3\text{kg}\right)\left(9.8{\text{ms}}^{-2}\right)}{\left(2150\text{N}\right)+\left(645\text{N}\right)}\right)\\ ={74.08}^o\end{array}$

Conclusion:

The resultant force exerted by the car on the road is directed at an angle of ${74.08}^o$ below the horizon and backwards.