Question

For the rod to be in equilibrium about C, the point of application of force is at a distance of from point A. Consider the rod is massless.

• A. 30 cm
• B. 90 cm
• C. 50 cm
• D. 70 cm

Answer 1

The correct option is D 70 cm

Net force acting on the rod $F_{net}$ = 10 N acting downwards.
Net torque acting on the rod about point $C$
${\tau }_C$ = (20 $\times$ 0) + (30 $\times$ 20) = $600N-cm$clockwise
For the rod to be in equilibrium at C, the net force acting on the rod should be zero and the net torque about C should be zero.
This implies, a force of $10N$ acts on the rod in the upward direction at some point.
Let the point of application be at a distance $x$ from point $C$
$600=10x$ $\Rightarrow$ $x$ = $60cm$
$\therefore$ $70cm$ from A is the point of application of the force.