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Q.42) A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60^{\circ}$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
A) $200 \sqrt{3} \mathrm{~N}$
B) $100 \mathrm{~N}$
C) $100 \sqrt{3} \mathrm{~N}$
D) $200 \mathrm{~N}$
Answers (1)
From the problem, $\theta = 30^\circ$
The frictional force balances the normal force on the wall which is given as-
$N = \frac{m g \cot \theta}{2} = \frac{20 \times 10 \sqrt{3}}{2} = 100 \sqrt{3} \mathrm{~N}$
Hence, the answer is the option 3.
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