A solid sphere of mass $10 ext{ kg}$ is placed over two smooth inclined planes as shown in figure.

Question

A solid sphere of mass $10 	ext{ kg}$ is placed over two smooth inclined planes as shown in figure. The normal reactions at 2 is $10x 	ext{ N}$. Find $x$ ($g = 10 	ext{ m/s}^2$)

Professor PC Thomas Classes · Answer

Let the normal reactions at surfaces 1 and 2 be $N_1$ and $N_2$. Surface 1 is inclined at $30^\circ$ to the horizontal, so $N_1$ makes $30^\circ$ with the vertical. Surface 2 is inclined at $60^\circ$ to the horizontal, so $N_2$ makes $60^\circ$ with the vertical. Balancing horizontal forces: $N_1 \sin 30^\circ = N_2 \sin 60^\circ \Rightarrow N_1 = \sqrt{3}N_2$. Balancing vertical forces: $N_1 \cos 30^\circ + N_2 \cos 60^\circ = mg \Rightarrow (\sqrt{3}N_2)(\sqrt{3}/2) + N_2(1/2) = 100 \Rightarrow 3N_2/2 + N_2/2 = 100 \Rightarrow 2N_2 = 100 \Rightarrow N_2 = 50 	ext{ N}$. Given $N_2 = 10x \Rightarrow 50 = 10x \Rightarrow x = 5$.