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Q4.6-29PE

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Found in: Page 162

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# 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?

(a) The total resistance force is 910 N.

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

See the step by step solution

## 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:

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

Here, Fnet is the net force acting on the system, mb is the mass of the boat, mt is the mass of the trailer, mc 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 (mb+mt), 1100 kg for mc, 1900 N for F, and 0.550 m/s2 for a in the above expression, and we get,

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

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:

${\mathbit{F}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{0}}{\mathbf{.8}}{\mathbit{f}}{\mathbf{=}}\left({m}_{b}+{m}_{t}\right){\mathbit{a}}$

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

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

localid="1655713865611" $\begin{array}{c}F\text{'}-0.8×910\text{ N}=700\text{ kg}×{\text{0.55 m/s}}^{2}\\ F\text{'}-728\text{ N}=\text{385 N}\\ \mathit{\text{F'}}=\left(385+728\right)\text{ N}\\ \mathit{\text{F'}}=1113\text{ N}\end{array}$

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