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Found in: Page 346

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# Question: In Fig 12-30, trying to get his car out of mud, a man ties one end of a rope around the front bumper and the other end tightly around a utility pole 18 m away. He then pushes sideways on the rope at its midpoint with a force of 550 N , displacing the center of the rope 0.30 m , but the car barely moves. What is the magnitude of the force on the car from the rope? (The rope stretches somewhat.)

Answer:

Magnitude of force on the car from the rope is $8.3×{10}^{3}\text{N}$ .

See the step by step solution

## Step 1: Understanding the given information

F = 550 N

Displacement of the center of the rope is 0.30 m .

## Step 2: Concept and formula used in the given question

By drawing an FBD of the rope, you can calculate the angle by using the given geometry. By applying static equilibrium conditions, you can get the tensions on the both sides of rope. Then by solving the torque equation, you can calculate the tension on the rope by using the formula given below.

Static Equilibrium conditions:

$\sum {F}_{x}=0\phantom{\rule{0ex}{0ex}}\sum {F}_{y}=0\phantom{\rule{0ex}{0ex}}\sum \tau =0$

## Step 3: Calculation for the magnitude of the force on the car from the rope

FBD of the rope:

From the figure,

$\theta ={\text{tan}}^{-1}\left(\frac{0.3}{9}\right)\phantom{\rule{0ex}{0ex}}=1.{9}^{0}\phantom{\rule{0ex}{0ex}}\sum {F}_{x}=0\phantom{\rule{0ex}{0ex}}{T}_{1}×\mathrm{cos}\left(\theta \right)-{T}_{2}×\mathrm{cos}\left(\theta \right)=0\phantom{\rule{0ex}{0ex}}$

Hence,

${T}_{1}={T}_{2}=T\phantom{\rule{0ex}{0ex}}\sum {F}_{y}=0\phantom{\rule{0ex}{0ex}}2T×\mathrm{sin}\left(\theta \right)-F=0\phantom{\rule{0ex}{0ex}}2T×\mathrm{sin}\left(1.9\right)-550=0\phantom{\rule{0ex}{0ex}}$

By solving for T:

$T=8300\text{N}\phantom{\rule{0ex}{0ex}}=8.3×{10}^{3}\text{N}\phantom{\rule{0ex}{0ex}}$

Hence, the Magnitude of force on the car from the rope is $8.3×{10}^{3}\text{N}$ .

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