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Found in: Page 335

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# Question: Find the net torque on the wheel in Figure P10.27 about the axle through ${\mathbit{a}}{\mathbf{=}}{\mathbf{10}}{\mathbf{.}}{\mathbf{0}}{\mathbf{ }}{\mathbit{c}}{\mathbit{m}}$, taking and ${\mathbit{b}}{\mathbf{=}}{\mathbf{25}}{\mathbf{.}}{\mathbf{0}}{\mathbf{ }}{\mathbit{c}}{\mathbit{m}}$.

The net torque on the wheel about the axle through O taking $a=10.0 \text{cm}$ and $b=25.0 \text{cm}$ is $\tau =3.71$.

See the step by step solution

## Step 1: Defining torque

The force that makes an object rotate is a torque.

$\tau =rF\mathrm{sin}\theta$

## Step 2: Calculating the net torque

Consider the given figure.

Find the torque for 10.0N.

${\tau }_{10N}=rF$

Since the radius of the torque 10N is b.

Substitute $r=0.25m$, $F=10 N$

${\tau }_{10N}=0.25×10$

${\tau }_{10N}=2.5 Nm$

Thus, the torque rotates in clockwise direction.

Calculate the torque for 9.00N.

Let the radius of 9.00N is b,

Substitute, $r=0.25m$,

$F=9N$

Therefore,

${\tau }_{9N}=rF$

${\tau }_{9N}=0.25×9$

${\tau }_{9N}=2.25N$, rotates in clockwise direction.

Determine the torque for $12.0N$

Since it has an angle, the force is perpendicular.

Thus, the radius is from the figure,

Substitute the following values:

$r=0.1 m,\phantom{\rule{0ex}{0ex}}F=12 N,\phantom{\rule{0ex}{0ex}}\theta =30°$

Therefore

${\tau }_{12N}=12\mathrm{cos}30°×0.1\phantom{\rule{0ex}{0ex}}{\tau }_{12N}=\left(12×\frac{\sqrt{3}}{2}×0.1\right)\phantom{\rule{0ex}{0ex}}{\tau }_{12N}=\left(12×0.866×0.1\right)\phantom{\rule{0ex}{0ex}}{\tau }_{12N}=1.0392 Nm$

It rotates in anticlockwise direction.

Hence, the total torque can be calculated as

$\tau ={\tau }_{10N}+{\tau }_{9N}-{\tau }_{12N}\phantom{\rule{0ex}{0ex}}\tau =2.5 Nm+2.25 Nm-1.039 Nm\phantom{\rule{0ex}{0ex}}\tau =3.71 Nm$

Therefore, the net torque exerted on the wheel is $\tau =3.71 Nm$.