Q4.6-29PE - College Physics (Urone)

Question

A 1100-kg car pulls a boat on a trailer.

(a) What total force resists the motion of the car, boat, and trailer, if the car exerts a 1900-N force on the road and produces an acceleration of 0.550 m/s²? The mass of the boat plus trailer is 700 kg. 

(b) What is the force in the hitch between the car and the trailer if 80% of the resisting forces are experienced by the boat and trailer?

Step-by-Step Solution

Answer

(a) The total resistance force is 910 N.

(b) The force in the hitch between the car and the trailer is 1113 N.

Step 1: Given data

• Mass of the car = 1100 kg.
• Force exerted by the car = 1900 N.
• Mass of the boat and trailer = 700 kg.
• Acceleration = 0.550 m/s².

Step 2: (a) Determine the total resistance force

Apply Newton’s second law of motion as:

 

$\begin{array}{c}{\mathit{F}}_{\mathit{n}\mathit{e}\mathit{t}}\mathbf{=}\mathit{M}\mathit{a}\\ \mathit{F}\mathbf{-}\mathit{f}\mathbf{=}\mathbf{(}{\mathit{m}}_{\mathit{b}}\mathbf{+}{\mathit{m}}_{\mathit{t}}\mathbf{+}{\mathit{m}}_{\mathit{c}}\mathbf{)}\mathit{a}\end{array}$   ………….……………(i)

 

Here, F_net is the net force acting on the system, m_b is the mass of the boat, m_t is the mass of the trailer, m_c is the mass of the car, a is the acceleration, and F is the force exerted by the car, and f is the resistance force.

Substitute 700 kg for (m_b+m_t), 1100 kg for m_c, 1900 N for F, and 0.550 m/s² for a in the above expression, and we get,

 

$\begin{array}{c}1900\text{\hspace{0.33em}N}-f=\left(700+1100\right)\text{\hspace{0.33em}kg}\times {\text{0.55\hspace{0.33em}m/s}}^2\\ 1900\text{ N}-f=1800\text{\hspace{0.33em}kg}\times {\text{0.55\hspace{0.33em}m/s}}^2\\ f=\left(1900-990\right)\text{\hspace{0.33em}N}\\ f=910\text{\hspace{0.33em}N}\end{array}$

 

Hence, the total resistance force is 910 N.

Step 3: (b) Determine the force in the hitch between the car and the trailer

Apply Newton’s second law of motion as:

 

$\mathit{F}\mathbf{\text{'}}\mathbf{-}\mathbf{0}\mathbf{.8}\mathit{f}\mathbf{=}\left(m_b+m_t\right)\mathit{a}$

 

Here, F’ is the force in the hitch between the car and the trailer.

 

Substitute 900 N for f, 700 kg for (m_b+m_t), and 0.550 m/s² for  a in the above expression, and we get,

 

$\begin{array}{c}F\text{'}-0.8\times 910\text{\hspace{0.33em}N}=700\text{\hspace{0.33em}kg}\times {\text{0.55\hspace{0.33em}m/s}}^2\\ F\text{'}-728\text{\hspace{0.33em}N}=\text{385\hspace{0.33em}N}\\ \mathit{\text{F'}}=\left(385+728\right)\text{\hspace{0.33em}N}\\ \mathit{\text{F'}}=1113\text{\hspace{0.33em}N}\end{array}$

 

Hence, the force in the hitch between the car and the trailer is 1113 N.