Question

A uniform bar of weight 250 newtons has a length of 6 meters. It is supported at the two ends. Where must a load of 300 newtons be placed so that the supporting force at the left end is twice that at the other end? What is the magnitude of each of the supporting forces?

Answer 1

The equilibrium conditions give:

$$\sum F_y=0:\quad R_A-250\:N-300\: N+R_B=0 \\ \sum M=0: (250\: N)*(x\: m)+(300\: N)*(3\: m)\\-R_B*(6\:m)=0$$

On the other hand

$$R_A=2R_B$$

Hence

$$2 R_B-250\:N-300\: N+R_B=0\\ R_B=183\: N\\ R_A=367\:N$$

$$x=\frac{183*6-300*3}{250}=0.792\:\rm m$$