Question

A weightless rod is acted on by upward parallel forces of $2N$ and $4N$ ends $A$ and $B$ respectively.; The total length of the rod $AB=3m$. To keep the rod in equilibrium a force of $6N$ should act in the following manner.

• A. Downwards at any point between $A$ and $B$.
• B. Downwards at mid point of $AB$.
• C. Downwards at a point $C$ such that $AC=1m$.
• D. Downwards at a point $D$ such that $BD=1m$.

Answer 1

The correct option is D Downwards at a point $D$ such that $BD=1m$.

Let $x$ be the distance of the action of the force on the rod at point D from the mid-point of the rod.

Net torque on the mid-point of the rod

$⟹2\times 1.5+6x=4\times 1.5⟹6x=3⟹x=0.5m$

So, the distance of the position of force from A and B will be 2m and 1m.

Thus, the force will be downwards at D such that BD=1m.