Question

6. At the ends of the homogeneous rod that rests on the point O, are hung two bodies weighing 55 N and 20N. Find the weight of the rod, knowing that it is in balance.

Answer 1

Torques to the left and right of O:

$$20x+W·\frac xL·\frac x2=55(L-x)+W·\frac{L-x}{L}·\frac{L-x}{2},\\\space\\ 20x+\frac{Wx^2}{2L}=55(L-x)+\frac{W(L-x)^2}{2L}.$$

Forces to the left and right of O:

$$20+W\frac{x}{L}=55+W\frac{L-x}{L}.$$

As we see, we have two equations and three undefined values: x, L, W. Thus, this systems does not have a solution. A rod of any corresponding weight and length can be in equilibrium with weights 20 and 55 N.