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(C) The forces acting on the sphere are: $1$. Weight $W = mg = 50 	imes 10 = 500 \, N$ acting vertically downwards. $2$. Normal reaction $N_A$ from t

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$\frac{1000}{\sqrt{3}} \, N$

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(C) The forces acting on the sphere are: $1$. Weight $W = mg = 50 	imes 10 = 500 \, N$ acting vertically downwards. $2$. Normal reaction $N_A$ from the incline $A$,perpendicular to the incline. $3$. Normal reaction $N_B$ from the vertical wall $B$,acting horizontally. Resolving the forces into horizontal and vertical components: For the vertical direction: $N_A \cos(30^{\circ}) = W = 500 \, N$. Therefore,$N_A = \frac{500}{\cos(30^{\circ})} = \frac{500}{\sqrt{3}/2} = \frac{1000}{\sqrt{3}} \, N$. For the horizontal direction: $N_B = N_A \sin(30^{\circ})$. Thus,the contact force at $A$ is $N_A = \frac{1000}{\sqrt{3}} \, N$.

$A$ $50 \, kg$ homogeneous smooth sphere rests on a $30^{\circ}$ incline $A$ and bears against a smooth vertical wall $B$. Calculate the contact force at $A$. (Take $g = 10 \, m/s^2$)

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