Question

A sphere of mass $10\sqrt{3}$ kg is kept in equilibrium on a frictionless inclined plane with the help of a string as shown in the figure. Find the tensile force developed in the string and the force of normal reaction between the sphere and the inclined plane.

Answer 1

Tensile force developed in the string = 100 N Force of normal reaction = 100 N

Answer 2

Problem 4:

1. Forces: Weight ($mg = 100\sqrt{3}$ N, down), Normal ($N$, perpendicular to plane), Tension ($T$, at $30^\circ$ to plane, up-right). Plane is at $30^\circ$ to horizontal.
2. Equilibrium along plane: $T \cos(30^\circ) - mg \sin(30^\circ) = 0 \implies T(\sqrt{3}/2) - 100\sqrt{3}(1/2) = 0 \implies T = 100$ N.
3. Equilibrium perpendicular to plane: $N + T \sin(30^\circ) - mg \cos(30^\circ) = 0 \implies N + 100(1/2) - 100\sqrt{3}(\sqrt{3}/2) = 0 \implies N + 50 - 150 = 0 \implies N = 100$ N.