Question

A force is applied on a rigid rod of mass 12 kg lying on a table as shown. The net torque on the rod about point A will be? (O is the centre of the rod)

• A. 100 N-m, clockwise
• B. 60 N-m, clockwise
• C. 60 N-m, counter-clockwise
• D. 160 N-m, counter-clockwise

Answer 1

The correct option is B 60 N-m, clockwise

As point A will accelerate, it is a non-inertial frame.

There will be a pseudo force acting at the centre as shown in figure.

For translatory motion,
F = Ma
20 = 12a
$a=\frac{5}{3}\frac{m}{s^2}$

$\Rightarrow F_{pseudo}=12\times \frac{5}{3}=20N,$ backwards.

Taking torque about A and assuming counter clockwise to be positive,
${\tau }_A=20\times 2-20\times 5$
= 60 N-m, clockwise.
(-ve sign indicates clockwise torque)