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

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 $2N_{1}\,cos\,30^{\circ}=W$
or $N_{1}=\frac{W}{\sqrt{3}} \ldots$ (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
$N_{2}\,sin\,\theta=N_{1}\,sin\,30^{\circ}$
or $N_{2}\,sin\,\theta=\frac{W}{2\sqrt{3}} \ldots$ (2)
$N_{2}\,cos\,\theta=W+N_{1}\,cos\,30^{\circ}=\frac{3W}{2}\ldots$(3)
Dividing (2) by (3) we get
$tan\,\theta=\frac{1}{3\sqrt{3}} $
or $\theta=tan^{-1}\left(\frac{1}{3\sqrt{3}}\right)$