A uniform bar AB of mass m and length L is fixed to a vertical wall. The bar can rotate about end A in the vertical plane. A supporting string has one end fixed to the well and the other end to the midpoint of the bar making an angle $θ$ with it as shown in Fig. A block of mass M hangs from end B of the bar. The system is in static equilibrium. What is the contact force F in the case M = m and $θ$ = 30°?

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$4 \sqrt{2} m g$

$3 \sqrt{3} m g$

$3 \sqrt{5} m g$

$2 \sqrt{7} m g$

answer is D.

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Detailed Solution

The bar is in translational as well as rotational equilibrium. For translational equilibrium, the net horizontal and vertical force must be zero, i.e.

$F_{x} - T_{x} = 0 \Rightarrow F_{x} - T \cos θ = 0 (1) a n d F_{y} + T_{y} - m g - M g = 0 o r F_{y} + T \sin θ - (m + M) g = 0 (2)$

For rotational equilibrium, the net clockwise torque about A = net anti-clockwise torque about A, i.e.

$M g \times L + m g \times \frac{L}{2} = T_{y} \times \frac{L}{2} \Rightarrow (M + \frac{m}{2}) g = \frac{T \sin θ}{2} \Rightarrow T = \frac{(2 M + m) g}{\sin θ} (3)$

Notice that we do not need Eqs. (1) and (2) to find T. Putting M = m and θ= 30° in Eq. (3), we get T= 6mg. So the correct choice is (c).

$F_{x} = T \cos θ = 6 m g \times \cos 30^{0} = 3 \sqrt{3} m g$

From Eq. (2), we have

$F_{y} = (m + M) g - T \sin θ = (m + M) g - \frac{(2 M + m) g}{\sin θ} \times \sin θ = - M g = - m g (∵ M = m) \Rightarrow F = \sqrt{{F_{x}}^{2} + {F_{y}}^{2}} = \sqrt{{(3 \sqrt{3} m g)}^{2} + {(- m g)}^{2}} = m g \times \sqrt{27 + 1} = \sqrt{28} m g = 2 \sqrt{7} m g$