Question

A wheel of radius $20\text{cm}$ has four forces applied to it, as shown in the figure. The total torque produced by these forces about centre $O$ is

• A. $5.4\text{N-m},\text{Anticlockwise}$
• B. $1.8\text{N-m},\text{Clockwise}$
• C. $1.8\text{N-m},\text{Anticlockwise}$
• D. $5.4\text{N-m},\text{Clockwise}$

Answer 1

The correct option is B $1.8\text{N-m},\text{Clockwise}$


Net torque about centre $O$ is given by,

${\tau }_{\text{net}}={\tau }_{\text{clockwise}}-{\tau }_{\text{anticlockwise}}$

Clockwise torques:

${\tau }_D=9\times 20=180\text{N-cm}$

${\tau }_B=8\sin {30}^{\circ }\times 20=80\text{N-cm}$

Anti-clockwise torques:

${\tau }_A=4\times 20=80\text{N- cm}$

${\tau }_C=0$, since $F_C$ is a radial force.

Now,

${\tau }_{\text{net}}=(180+80)-(80+0)=180\text{N-cm}$

$\Rightarrow {\tau }_{net}=1.8\text{N-m}$ in clockwise direction.

Hence, option $\left(B\right)$ is the correct answer.