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Question

(A) The weight of the sphere is $W = mg = 50 	imes 10 = 500\,N$,acting vertically downwards.Let $N_A$ be the normal force from the incline $A$ (at $6

Vedclass · Accepted Answer

$N_A = \frac{1000}{\sqrt{3}}\,N, \quad N_B = \frac{500}{\sqrt{3}}\,N$

Vedclass · Answer

(A) The weight of the sphere is $W = mg = 50 	imes 10 = 500\,N$,acting vertically downwards.Let $N_A$ be the normal force from the incline $A$ (at $60^{\circ}$ to the horizontal) and $N_B$ be the normal force from the vertical wall $B$ (acting horizontally).For equilibrium,the sum of vertical forces must be zero:$N_A \sin 60^{\circ} = W = 500\,N$$N_A \left( \frac{\sqrt{3}}{2} ight) = 500 \Rightarrow N_A = \frac{1000}{\sqrt{3}}\,N$For equilibrium,the sum of horizontal forces must be zero:$N_A \cos 60^{\circ} = N_B$$N_B = \left( \frac{1000}{\sqrt{3}} ight) \left( \frac{1}{2} ight) = \frac{500}{\sqrt{3}}\,N$Thus,$N_A = \frac{1000}{\sqrt{3}}\,N$ and $N_B = \frac{500}{\sqrt{3}}\,N$.

The $50\,kg$ homogeneous smooth sphere rests on the $30^{\circ}$ incline $A$ and bears against the smooth vertical wall $B$. Calculate the contact forces at $A$ and $B$.

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