Question

A uniform rod 2 m long and weighing 65N is pivoted to the wall. A weight of 25 N is suspended at the other end of the rod. What is the magnitude of vertical and horizontal components of the force exerted on the pivot as well as the tension in the string?

Answer 1

Equilibrium of forces:

$$R+T-W-mg=0.$$

Equilibrium of torques:

$$TL-mgL+\dfrac W2·\dfrac L2=0.$$

The unknowns are R (reaction in the pivot) and T (tension in the string). Solve the system:

$$R=32.5\text{ N},\\ T=57.5\text{ N}.$$