Question

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

Short answer

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

Step-by-step solution

Defining torque

The force that makes an object rotate is a torque.

$\tau =rF\sin \theta$

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

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

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

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,

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

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

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

$$\begin{gathered} r=0.1 m, \\ F=12 N, \\ \theta =30° \end{gathered}$$

Therefore

$$\begin{gathered} {\tau }_{12N}=12\cos 30°\times 0.1 \\ {\tau }_{12N}=\left(12\times \frac{\sqrt{3}}{2}\times 0.1\right) \\ {\tau }_{12N}=\left(12\times 0.866\times 0.1\right) \\ {\tau }_{12N}=1.0392 Nm \end{gathered}$$

It rotates in anticlockwise direction.

Hence, the total torque can be calculated as

$$\begin{gathered} \tau ={\tau }_{10N}+{\tau }_{9N}-{\tau }_{12N} \\ \tau =2.5 Nm+2.25 Nm-1.039 Nm \\ \tau =3.71 Nm \end{gathered}$$

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