Question

The mass string system shown in the figure is in equilibrium. If the coefficient of friction between A and the table is $0.3$, the frictional force on A is:

• A. 9.8 N
• B. 2.04 N
• C. 1.96 N
• D. 0.59 N

Answer 1

Answer: (C)

$T^{\prime} \sin 45^{\circ}=0.2\, g$
and $T^{\prime} \cos 45^{\circ}=T$

$\therefore T =0.2\, g $
$=0.2 \times 9.8=1.96\, N . $
$f_{\lim } =\mu N=0.3 \times 1 g =2.94 \,N$
As external force is less than limiting friction, so friction $=1.96\, N$