Answer to Question #284854 in Physics for Cute

Question

A uniform rod 2 m long and weighing 85 N is pivoted to the wall as shown. A weight of 25 N is suspended at the other end of the rod. Find the magnitude of vertical and horizontal components of the force exerted on the pivot as well as the tension in the string.

Answer 1

θ=30° below is the angle between the horizontal rod and the string. W is the weight suspended, mg is the mass of the rod.

Equilibrium of forces:

$$mg+W-R_y+T\sin\theta=0,\\ R_x=T\cos\theta.$$

Equilibrium of torques:

$$-WL-mg\dfrac L2+LT\sin\theta=0.$$

Solution of the system:

$$R_x=117\text{ N},\\ R_y=177.5\text{ N},\\ T=135\text{ N}.$$