Question

A couple of $10\text{N}$ acting on a rod of length $2\text{m}$ pivoted about its centre $O$ as shown in figure. Find the magnitude of resultant torque acting on rod about point $O$.

• A. $10\text{N.m}$
• B. $30\text{N.m}$
• C. $20\text{N.m}$
• D. Can't be determined

Answer 1

The correct option is A $10\text{N.m}$

Magnitude of torque due to force $F$ is given as:
$|\tau |=F\times r_{\perp }$
$r_{\perp }=$ Perpendicular distance of force from the axis of rotation

$\Rightarrow$The component forces of couple tend to produce rotation about $O$ in anticlockwise direction, so torque due to both forces will add up.

$\Rightarrow$Since $10\cos {30}^{\circ }$ component passes through point $O$, hence its torque vanishes.

Net torque on rod due to couple will be:
$|{\tau }_{\text{net}}|=(10\sin {30}^{\circ }\times r_{\perp })+(10\sin {30}^{\circ }\times r_{\perp })$
$\because r_{\perp }=1\text{m}$
$\Rightarrow |{\tau }_{\text{net}}|=(10\sin {30}^{\circ }\times 1)+(10\sin {30}^{\circ }\times 1)$
$|{\tau }_{\text{net}}|=10\sin {30}^{\circ }\times 2=2\times 10\times \frac{1}{2}$
$\therefore |{\tau }_{\text{net}}|=10\text{N.m}$