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A uniform sti...

Question

A uniform stick of mass M is placed in a frictionless well as shown in the figure. The stick makes an angle $θ$ with the horizontal. Then the force which the vertical wall exerts on right end of stick is :

A

\frac{M g}{2 c o t θ}

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B

\frac{M g}{2 t a n θ}

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C

\frac{M g}{2 c o s θ}

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D

\frac{M g}{2 s i n θ}

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Solution

The correct option is C $\frac{M g}{2 t a n θ}$
The force acting on the stick by the vertical wall is N, weight of the stick mg acts downward as shown.
Calculating the torque about point A on the ground( as shown) for rotational equilibrium, anti-clockwise torque by N is balanced by the clock-wise torque by weight of the stick.
$τ_{m g} = τ_{N}$
$∴ m g (\frac{l}{2} cos θ) = N (l sin θ)$ where l is the length of the stick.
$\frac{m g cot θ}{2} = N$
$\Rightarrow N = \frac{m g}{2 tan θ}$

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