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27 P

Expert-verifiedFound in: Page 335

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 ${\mathit{a}}{\mathbf{=}}{\mathbf{10}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\u200a}}{\mathit{c}}{\mathit{m}}$, taking and ${\mathit{b}}{\mathbf{=}}{\mathbf{25}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\u200a}}{\mathit{c}}{\mathit{m}}$.**

The net torque on the wheel about the axle through O taking $a=10.0\u200a\text{cm}$ and $b=25.0\u200a\text{cm}$ is $\tau =3.71$.

The force that makes an object rotate is a torque.

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

Consider the given figure.

Find the torque for 10.0N.

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

Since the radius of the torque 10N is b.

Substitute $r=0.25m$, $F=10\u200aN$

${\tau}_{10N}=0.25\times 10$

${\tau}_{10N}=2.5\u200aNm$

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,

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

${\tau}_{9N}=0.25\times 9$

${\tau}_{9N}=2.25N$, 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:

$r=0.1\u200am,\phantom{\rule{0ex}{0ex}}F=12\u200aN,\phantom{\rule{0ex}{0ex}}\theta =30\xb0$

Therefore

${\tau}_{12N}=12\mathrm{cos}30\xb0\times 0.1\phantom{\rule{0ex}{0ex}}{\tau}_{12N}=\left(12\times \frac{\sqrt{3}}{2}\times 0.1\right)\phantom{\rule{0ex}{0ex}}{\tau}_{12N}=\left(12\times 0.866\times 0.1\right)\phantom{\rule{0ex}{0ex}}{\tau}_{12N}=1.0392\u200aNm$

It rotates in anticlockwise direction.

Hence, the total torque can be calculated as

$\tau ={\tau}_{10N}+{\tau}_{9N}-{\tau}_{12N}\phantom{\rule{0ex}{0ex}}\tau =2.5\u200aNm+2.25\u200aNm-1.039\u200aNm\phantom{\rule{0ex}{0ex}}\tau =3.71\u200aNm$

Therefore, the net torque exerted on the wheel is $\tau =3.71\u200aNm$.

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