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
$F = \frac{10 N}{2 sin 30^{\circ}} = 10 N$