Determine the reactions at the supports:
A smooth sphere weighing 200 N is resting as shown in Figure. Determine the reactions at the supports.
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 60o 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:
∑ Fx = 0
∴ RA - RB cos θ = 0
θ = 30o by the geometry of the figure
And, ∑ Fy = 0
∴ RA = 0.866 RB
RB sin 30o - W = 0
∴ RB (0.5) = 200
∴ RB = 200 /0.5 = 400 N
∴ RA = 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"
RA/ sin (90o + θ) = RB /sin 90o = W/ sin (180o - θ)