Question

A cylinder of weight W is shown in the figure. The coefficient of static friction for all surfaces is $\frac{1}{3}$. The applied force P = 2 W. (Assume counter-clockwise motion has just started.)Match the following list 1 with list 2.

Answer 1

The net force must be zero if there is no motion.
Horizontal force : $N_A-F_B=0$
Vertical force : $P+W-N_B-F_A=0$
For option (A):
Torque about COM,
P.d - r. ($F_A+F_B)=0$
$F_A=\mu N_A$
$F_B=\mu N_B$
$N_A=\mu N_B$
$N_B+\mu N_A=P+W$
$\mu R_o(N_A+N_B)=Pd$

$N_B(1+{\mu }^2)=P+W$

$\mu R_o{N}_B(1+\mu )$ = Pd

$N_B=\frac{Pd}{\mu R_o(1+\mu )}$

$\frac{Pd}{\mu R_o(1+\mu )}=\frac{(P+W)}{(1+{\mu }^2)}$

$d=(P+W)\mu R_o\frac{1+\mu }{P(1+{\mu }^2)}$

$P=2W,\mu =\frac{1}{3}$
$\Rightarrow \frac{d}{R_o}=\frac{3}{5}=0.6$

For (B), (C), (D) :
$N_B=\frac{Pd}{\mu R_o(1+\mu )}=2.7W$
$N_A=\frac{N_B}{3}=0.9W$
$F_A=\frac{N_A}{3}=0.3W$
$\Rightarrow \frac{F_A}{W}=0.3W$
$N_B=2.7W$
$N_A=0.9W$
Also, $F_B=N_A=0.9W$