A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. total

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6.System of Particles and Rotational Motion

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$A$ 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 $BD = 1m.$

Solution

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

$\Longrightarrow 2 \times 1.5+6 x=4 \times 1.5$

$\Longrightarrow 6 x=3$

$\Longrightarrow x=0.5 m$

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

Thus, the force will be downwards at $D$ such that $BD = 1$ $\mathrm{m}$

Standard 11

Physics

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