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Q12P

Expert-verifiedFound in: Page 346

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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 $8.3\times {10}^{3}\text{N}$ .

F = 550 N

Displacement of the center of the rope is* 0.30 m* .

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:

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

** **

FBD of the rope:

From the figure,

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

* *

Hence,

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

* *

By solving for T:

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

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

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