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Q.

A uniform disc is held in equilibrium on a rough horizontal surface (coefficient of friction $= μ$ ) by applying a force as shown. Then the maximum value of F for this to be possible is (weight of disc is W).

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a

zero

b

2 W 3

c

μ W 1 + 3 μ

d

2 μ W 1 + 3 μ

answer is D.

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Detailed Solution

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The disc is In complete equilibrium
$\frac{F}{2} = f$ (Rotational equilibrium and translation equilibrium)
$F \frac{\sqrt{3}}{2} + N = W$ (Translational equilibrium along vertical direction)

Also $f ≤ μ N$
$\frac{F}{2} \leq μ (- \frac{F \sqrt{3}}{2} + W) \Rightarrow \frac{F}{2} [1 + \sqrt{3 μ}] \leq W μ$
$F \leq \frac{2 W μ}{1 + \sqrt{3} μ}$

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