Question

A uniform rod XY of weight 10.0N is freely hinged to a wall at X. It is held horizontal by a force F acting from Y at an angle 30° to the horizontal, as shown.

What is the value of F?
A- 5.0 N B- 8.7cm C- 10.0cm D-20.0cm

Answer 1

In order to solve the problem, we must require the equilibrium of all the torques acting on the rod. The fixed point is in X, so we have:
- The weight of the rod (mg) acting at the center of the rod (so, at a distance L/2 from X, where L is the length of the rod). So, the torque is 
$T_W = mg \frac{L}{2}$
- The vertical component of F (so, $F \sin 30^{\circ}$) applied in Y, so at a distance L from X. Its torque is
$F \sin 30^{\circ} L$

The weight points downwards (so, the torque is clockwise), while the torque of F points anti-clockwise, so the equilibrium of torques is
$F \sin 30^{\circ} L = mg \frac{L}{2}$
and since the weight is mg=10 N, re-arranging the equation we find
$F = \frac{10 N}{2 sin 30^{\circ}} = 10 N$

Answer 2

The value of the horizontal force acting on the rod is 10 N. Therefore option (C) is correct.:

Given data:

The weight of rod is, W = 10.0 N.

The angle made by force with respect to horizontal is, $\theta = 30^\circ$.

To maintain the steady position (equilibrium condition), the vertical component of force F must be balanced by the moment of force due to weight.

Therefore,

$(Fsin\theta) \times L = W \times \dfrac{L}{2} \\\\(F \times sin30^{\circ}) = \dfrac{10}{2} \\F = 10 \;\rm N$

Thus, the value of the horizontal force acting on the rod is 10 N. And option (C) is correct.

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