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Question

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


Solution

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 $$\theta=0^0\implies F_g=0, f_l=mg=20N$$
as, $$F_g<f_l$$, frictional force is zero.

B. when $$\theta=90^0\implies F_g=mgsin90^0=20N, f_l=mgcos90^0=0$$
frictional force is zero.

C. when $$\theta=30^0\implies F_g=mgsin30^0=10N, f_l=mgcos30^0=10\sqrt{3}N$$
as, $$F_g<f_l$$, frictional force is $$F_g=10N$$.

D. when $$\theta=60^0\implies F_g=mgsin60^0=10\sqrt{3}, f_l=mgcos60^0=10N$$
as, $$F_g>f_l$$, frictional force is $$f_l=10N$$.

Physics

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