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Q4.6-29PE

Expert-verifiedFound in: Page 162

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/s ^{2}? 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.

- 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/s
^{2}.

**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/s^{2} 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.

**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/s^{2} for *a* in the above expression, and we get,

localid="1655713865611" $\begin{array}{c}F\text{'}-0.8\times 910\text{\hspace{0.33em}N}=700\text{\hspace{0.33em}kg}\times {\text{0.55\hspace{0.33em}m/s}}^{2}\\ F\text{'}-728\text{\hspace{0.33em}N}=\text{385\hspace{0.33em}N}\\ \mathit{\text{F'}}=\left(385+728\right)\text{\hspace{0.33em}N}\\ \mathit{\text{F'}}=1113\text{\hspace{0.33em}N}\end{array}$

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

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