Question

At the end A of the homogeneous rod with a mass of 400g, which has a point O of rotation, the body with a mass of 800g is suspended, fig. 4.24. What must be the mass of the suspended body at point B so that the bar is in equilibrium?

Answer 1

the case requires rotational equilibrium, for which the torque about O has to be 0.

The length of the rod is unclear, so i'll answer it according to the divisions in rod.

force at A = 0.8g

force at b = xg

0.8g*2 = xg*4

x = 0.4 = 400g