Question

A horizontal uniform rod of mass ${}^{\prime }{m}^{\prime }$ has its left end hinged to the fixed incline plane, while its right end rests on the top of a uniform cylinder of mass ${}^{\prime }{m}^{\prime }$ which in turn is at rest on the fixed inclined plane as shown. The coefficient of friction between the cylinder and rod, and between the cylinder and inclined plane, is sufficient to keep the cylinder at rest.


The ratio of magnitude of frictional force on the cylinder due to the rod and the magnitude of frictional force on the cylinder due to the inclined plane is:

• A. $1:1$
• B. $2:\sqrt{3}$
• C. $2:1$
• D. $\sqrt{2}:1$

Answer 1

The correct option is A $1:1$

F.B.D of cylinder:


Where $f_r$: friction between rod and cylinder
$f_i$ = friction between cylinder and incline
$N_i$ = Normal reaction exerted by incline
$N_r$ = Normal reaction exerted by rod.
As cylinder is in equilibrium, balancing torque about point $\text{O}$,
$R\times f_r=R\times f_i$
$\Rightarrow f_f=f_i$
Torque of $Ni,Nr$ & $Mg$ will be zero as they pass through point $\text{O}$.
$\therefore f_f:f_i=1:1$