Auniform 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

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 

- The vertical component of F (so, ) applied in Y, so at a distance L from X. Its torque is


The weight points downwards (so, the torque is clockwise), while the torque of F points anti-clockwise, so the equilibrium of torques is

and since the weight is mg=10 N, re-arranging the equation we find

Ais the