Question

A uniform solid cylinder of mass $m$ rests on two horizontal planks. A thread is wound on the cylinder. The hanging end of the thread is pulled vertically down with a constant force $F$ (figure shown above). Find the maximum magnitude of the force $F$ which still does not bring about any sliding of the cylinder, if the coefficient of friction between the cylinder and the planks is equal to $k$.

• A. $F_{max}=3\frac{3kmg}{(2-3k)}$
• B. $F_{max}=\frac{3kmg}{(3k-2)}$
• C. $F_{max}=\frac{3kmg}{(2-3k)}$
• D. $F_{max}=3\frac{3kmg}{(3k-2)}$

Answer 1

Answer: (C)

$F_{max}=\frac{3kmg}{(2-3k)}$

Let a = magnitude of acceleration
Let $\alpha$ = magnitude of angular acceleration
Let N = magnitude of normal force at either of the planks
$\tau$ = magnitude of torque about axis of cylinder
F will be maximum when frictional force, f, is maximum => $f=kN$
Since there is no sliding of the cylinder,
a = R$\alpha$
From Newton's Second Law, $ma=2f$
Since there is no motion in y-direction, $2N=F+mg$
$\tau$ = $I\alpha$
$(F-2f)R=\frac{m{R}^{2}a}{2R}=\frac{maR}{2}$
$F=3f$
$F_{max}$ = 3$f_{max}$ =$3kN=3k$ $\frac{F_{max}+mg}{2}$
=> $F_{max}$ = $\frac{3kmg}{2-3k}$