Determine the reactions at the supports:

A smooth sphere weighing 200 N is resting as shown in Figure. Determine the reactions at the supports.

Solution

Let us first ascertain the directions of reactions. As the wall is vertical and smooth, the reaction at A shall be horizontal, i.e. normal to the wall. Similarly, the reaction at B shall be normal to the line inclined at 60^o to the horizontal. Both these reactions shall pass through O, the centre of sphere, because these are normal drawn to the tangents of the sphere at A and B. The weight of the sphere may be supposed concentrated at O. Therefore; there concurrent forces keep the sphere at rest. Let us apply conditions of equilibrium:

∑ F_x = 0

∴ R_A - R_B cos θ = 0

θ = 30^o by the geometry of the figure

And, ∑ F_y = 0

 ∴ R_A  = 0.866 R_B

R_B  sin 30^o  - W = 0

∴ R_B  (0.5) = 200

∴ R_B  = 200 /0.5 = 400 N

∴ R_A  = 0.866 × 400 = 346.4 N

The same problem may be solved using Lami's theorem which states, "If three of the forces (coplanar) acting on a body keep it at rest then each force is proportional to the sin of the angle between the other two forces"

            R_A/ sin (90^o + θ) =       R_B /sin 90^o  =   W/ sin (180^o  - θ)