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:

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

Tardigrade Admin · Accepted Answer

T prime sin 45°=0.2 g and T prime cos 45°=T <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e70731422a11578d176bb22520111256-.png" /> ∴ T =0.2 g =0.2 × 9.8=1.96 N . f lim =μ N=0.3 × 1 g =2.94 N As external force is less than limiting friction, so friction =1.96 N

Q. 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:

9.8 N

2.04 N

1.96 N

0.59 N

$T^{'} sin 4 5^{\circ} = 0.2 g$
and $T^{'} cos 4 5^{\circ} = T$

$∴ T = 0.2 g$
$= 0.2 \times 9.8 = 1.96 N .$
$f_{l i m} = μ N = 0.3 \times 1 g = 2.94 N$
As external force is less than limiting friction, so friction $= 1.96 N$

Q. 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: