Two equal heavy spheres,each of radius r,are | vedclass

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(B) Let the center of the cup be $O$ and the centers of the two spheres be $C_1$ and $C_2$. The distance from the center of the cup to the center of e

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$2$

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(B) Let the center of the cup be $O$ and the centers of the two spheres be $C_1$ and $C_2$. The distance from the center of the cup to the center of each sphere is $R - r = 3r - r = 2r$.The distance between the centers of the two spheres is $r + r = 2r$.Consider the triangle formed by the center of the cup and the centers of the two spheres. This is an equilateral triangle with sides $2r, 2r, 2r$.Thus,the angle $	heta$ that the normal reaction $N_1$ (from the cup) makes with the vertical is $30^\circ$.Resolving forces on one sphere:Vertical equilibrium: $N_1 \cos 	heta = mg$Horizontal equilibrium: $N_1 \sin 	heta = N_2$,where $N_2$ is the reaction between the two spheres.Given $	heta = 30^\circ$,we have $\sin 30^\circ = 1/2$.Therefore,$N_1 (1/2) = N_2$,which implies $N_1 / N_2 = 2$.

Two equal heavy spheres,each of radius $r$,are in equilibrium within a smooth cup of radius $3r$. The ratio of the reaction between the cup and one sphere to that between the two spheres is:

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