On Torque...Where will a load of 10N will be placed on the uniform rod?

Question

A uniform rod 1m and weighing 30N is supportd in a horizontal position by a fulcrum with weighs of 40N and 50N suspended from its ends. Where will a load of 10N will be placed on the uniform rod?

Answer 1

If the rod's fulcrum is in the middle and the rod is currently unbalanced - due to a 40N at one end and a 50N at t'other - then the 10N would go on top of the 40N to balance it.

If the rod is already balanced as it is with the weights on - then the 10N would go on top of the fulcrum to maintain equilibrium.

Answer 2

If the fulcrum is situated at the center, then the torque produced by the 50N weight equals 50*0.50m, and the torque produced by the 40N weight equals 40*0.50m. I presume that you are asking where will the 10N load be placed so that the rod will be in equilibrium. By inspection, we can immediately say that it should be placed at the point where the 40N weight is. But to give you a clearer picture, let's write the equation:

Torque of 50N weight=25Nm
Torque of 40N weight=20Nm
Balancing torque = 25-20=5Nm

Since the load is given as 10N, and if x =the distance from the fulcrum, then

Balancing torque=5Nm=10Nx

x=5/10m
=0.5m at the point where 40N weight is.