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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