Question

Angle $\theta$ is gradually increased as shown in figure. For the given situation match the following two columns. (g = $10m{s}^{-2}$)

Answer 1

Normal force on the block is $N=mgcos\theta$

Limiting frictional friction is $f_l=\mu N=mgcos\theta$, in the upward the plane direction.

Component of gravitational force in downward the plane direction is,

$F_g=mgsin\theta$

A. when $\theta =0^0⟹F_g=0,f_l=mg=20N$

as, $F_g<f_l$, frictional force is zero.

B. when $\theta ={90}^0⟹F_g=mgsin{90}^0=20N,f_l=mgcos{90}^0=0$

frictional force is zero.

C. when $\theta ={30}^0⟹F_g=mgsin{30}^0=10N,f_l=mgcos{30}^0=10\sqrt{3}N$

as, $F_g<f_l$, frictional force is $F_g=10N$.

D. when $\theta ={60}^0⟹F_g=mgsin{60}^0=10\sqrt{3},f_l=mgcos{60}^0=10N$

as, $F_g>f_l$, frictional force is $f_l=10N$.