A block of mass m=2kg is resting on a rough inclined plane of inclination 30∘ as shown. the coefficient of friction between the block and the plane μ=0.5 . What minimum force F should be applied perpendicular to the plane on the block so that the block does not slip on the plane (g=10m/s2)
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The perpendicular force F balances the perpendicular component of mass.
So
μF = mgcosβ
F= mgcosβ/μ
F=34.64N.
Cheers.
So
μF = mgcosβ
F= mgcosβ/μ
F=34.64N.
Cheers.
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