Question

Three identical rigid circular cylinders $A,B$ and $C$ are arranged on smooth inclined surfaces as shown in figure The least value of $\theta$ that prevent the arrangement from collapse is

• A. $ta{n}^{-1}\left(1/2\right)$
• B. $ta{n}^{-1}\left(1/2\sqrt{3}\right)$
• C. $ta{n}^{-1}\left(1/3\sqrt{3}\right)$
• D. $ta{n}^{-1}\left(1/4\sqrt{3}\right)$

Answer 1

Answer: (C)

W = weight of each sphere



$N_2$ = normal reaction between $A$ and inclined plane

$N_1$ = normal reaction between A and C = normal

reaction between $B$ and $C$

Free body diagram of C:

Resolving vertically $2N_{1}cos30^{\circ }=W$

or $N_1=\frac{W}{\sqrt{3}}\dots$ (1)

When the arrangement is on the point of collapsing, the reaction between A and B is zero

Free body diagram of A:

Resolving horizontally and vertically

$N_{2}sin\theta =N_{1}sin30^{\circ }$

or $N_{2}sin\theta =\frac{W}{2\sqrt{3}}\dots$ (2)

$N_{2}cos\theta =W+N_{1}cos30^{\circ }=\frac{3W}{2}\dots$(3)

Dividing (2) by (3) we get

$tan\theta =\frac{1}{3\sqrt{3}}$

or $\theta =ta{n}^{-1}\left(\frac{1}{3\sqrt{3}}\right)$