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A light rod of length $4 m$ , can be maintained in equilibrium position as shown in the figure if we apply single force on it.
The required force

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would be appiled at a distance of $\frac{48}{17} m$ from the right end

the rod cannot be maintained in equilibrium under the action of a single force.

would have magnitude of $77$ N

Would have a line of action making an angle of $\tan^{- 1} (17 / 9)$ with negative x- axis

answer is A, B, C.

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Detailed Solution

For equilibrium of the rod, let us say force $R$ is appilied whose $X$ and $Y$ components are

$R_{x}$ and $R_{y}$ as shown in figure.

$R_{x} = 36 N$ , $R_{y} = 48 + 20 = 68 N$

For rotational equilibrium, $x \times 20 = 48 \times \times (4 - x)$

$\Rightarrow x = \frac{48}{17} m$ ; So, $R = \sqrt{36^{2} + 68^{2}} ≃ 77 N$ .

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