Question

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

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

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{1}{3\sqrt{3}}\right)$

For the first cylinder(top) ,

$mg=2Ncos\beta$

For the second cylinder,

Along Y axis, $mg+Ncos\beta -N_{0}cos\theta =0$

Along X axis, $N_1+Nsin\beta -N_{0}sin\theta =0$

When the system starts to destroy, $N_1=0$

$\beta ={30}^{\circ }$

From first equation,

$N=\frac{mg}{2cos\beta }$

From second equation,

$N_0=\frac{3mg}{2cos\theta }$

From the third equation,

$\frac{mg}{2}tan\beta -\frac{3mg}{2}tan\theta =0$

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

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