A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take ) [2025]

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Solution

Q.
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of $60 °$ with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (take $g = 10 m / s^{2}$ ) [2025]

1 200 N

2 200 $\sqrt{3}$

3 100 N

4 100 $\sqrt{3}$

Ans.
(4)

Given that: Mass of rod, $m = 20 kg,$

Length of rod, $l = 5 m$

Weight of the uniform rod = mg

$= 20 \times 10 = 200 N$

For vertical direction, $W = N_{2}$

For horizontal direction, $f = N_{1}$

(where $f$ is the frictional force)

Since rod is stationary, so net torque is zero (about B)

$\Rightarrow$ Distance from wall to the centre of mass of rod (L/2) is

$D = \frac{L}{2} \cos 30 ° = \frac{1}{2} (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}$

$⇒Now, distance O A = L \sin 30 ° = \frac{L}{2}$

Taking torque about point B to be zero

$\Rightarrow m g \times \frac{L}{2} \cos 30 ° = N_{1} \times L \sin 30 °$

$\Rightarrow m g \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = f \times \frac{1}{2} [As, N_{1} = f]$

$\Rightarrow 200 \times \frac{\sqrt{3}}{4} = \frac{f}{2} or f = 100 \sqrt{3} N$

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