Question

A wheel of radius $20$ $cm$ has forces applied to is as shown in the figure. The torque produced by the forces $4$ $N$ at A, $8$ $N$ at B, $6$ $N$ at C and $9$ $N$ at D at angles indicated is

• A. 5.4 N-m (anticlockwise)
• B. 1.8 N-m (clockwise)
• C. 2.0 N-m (clockwise)
• D. 5.4 N-m (clockwise)

Answer 1

The correct option is B 1.8 N-m (clockwise)

Given: A wheel of radius 20 cm has forces applied to is as shown in the figure.

To find the torque produced by the forces 4 N at A, 8 N at B, 6 N at C and 9 N at D at angles indicated

Solution:

We know,

$\tau =\overrightarrow{r}\times \overrightarrow{F}=r\left(F\sin \theta \right)...........\left(i\right)$

where $\tau$ is the torque, $r$ is the radius and $F$ is the force applied, $\theta$ is the angle at which the force is applied.

Given, $r=20cm=0.2m$

Now the torque produced by the forces 4 N at A is

$\tau =4N\times \sin \left({90}^{\circ }\right)\times 0.2=0.8Nm$ (by using the eqn(i) and substituting corresponding values from the figure)

in the anticlockwise direction.

The torque produced by the forces 8 N at B is

$\tau =8N\times \sin 30\times 0.2=1.6\times \frac{1}{2}=0.8Nm$ (by using the eqn(i) and substituting corresponding values from the figure)

in the clockwise direction.

The torque produced by the forces 6 N at C is

$\tau =0$ as force and radius vectors are in the same direction

The torque produced by the forces 9 N at D is

$\tau =9N\times \sin 90\times 0.2=1.8Nm$ (by using the eqn(i) and substituting corresponding values from the figure)

in the clockwise direction.

So Total torque is

$\tau =0.8-0.8-1.8=-1.8$

Hence 1.8 Nm torque is produced in clockwise direction (negative sign indicates this)