Question

$A,B$ and $C$ are the three forces each of magnitude $4N$ acting in the plane of paper as shown in Fig. The point $O$ lies in the same plane. What is the resultant torque about the point $O$?

• A. $4Nm$(clockwise)
• B. $4Nm$(anti clockwise)
• C. $4.4Nm$(clockwise)
• D. $4.4Nm$(anti clockwise)

Answer 1

The correct option is C $4.4Nm$(clockwise)

Moment is defined as $rXF$ Here r is nothing but perpendicular distance from the force to that point O. Given the distances as 0.9 m, 0.8m, 0.6m for A,B,C respectively from point O. All has same magnitude of force that is 4 N. Hence
For A
$moment=0.9\times 4=3.6N-m$
For B
$moment=0.8\times 4=3.2N-m$
For C
$moment=0.6\times 4=2.4N-m$.
Here A and B as both are in the clockwise direction where C gets subtracted as it is in an anti-clockwise direction. Hence resultant torque is
$3.6+3.2-2.4=4.4N-m$.
Hence $4.4N-m$ is the resultant torque.