Question 109 of 280

A heavy iron bar of weight $12kg$ is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle $60^{\circ}$ with the horizontal, the normal force applied by the man on bar is:

A$6kg-wt$B$12kg-wt$C$3kg-wt$D$6\sqrt{3}kg-wt$

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Answer: (C) $3kg-wt$

Let's consider the following diagram:

For the equilibrium of the system, the net torque, according to the above figure, on the system about point $A$ should be zero.
With reference to the above figure, the net anti-clockwise torque about $A$ is
${\tau }_{a}={F}_{2}L...\left(1\right)$
And, the net clockwise torque about the same point is
${\tau }_{c}=W\cos \theta \times \dfrac{L}{2}...\left(2\right)$
Equations (1) and (2) imply that
$W\cos \theta \times \dfrac{L}{2}={F}_{2}L \Rightarrow 12\times 10\times \cos 60^{\circ}\times \left(\dfrac{L}{2}\right)={F}_{2}L \Rightarrow {F}_{2}=\dfrac{120\times 10\times \cos 60^{\circ}}{2} =30$
Hence, the required weight can be expressed as
${F}_{2}=3kg-wt$