Look at the point where the bar is attached to the wall. Since the system is in static
equilibrium, the net torque about this point must be zero. The weight of the bar
produces a clockwise torque, of magnitude (L/2) Mg sin θ, and the tension in the
string produces a counterclockwise torque, of magnitude LT sin θ. Since these
torques balance, we have
There is a force acting on the rod by the wall, but since that point is taken to be the
pivot, its torque = 0. This “wall force” could, in theory, have both an x- and a y-
component, depending on what is needed to make sure the rod doesn’t move. Since
there are no other x forces in this example, there will be no x-component of the