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)