A cube of 10 N rests on a rough inclined plane of slope 3 in 5. If the
coefficient of friction is 0.6 , the minimum force required the cube
moving up the plane is
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Explanation:
According to the problem the weight of the cube is 10 N
The slop in sinθ=3/5
θ=37°
cos37°=4/5
Therefore the force required is
F=mgsinθ+μmgcosθ [ m is the mass of the cube and μ is the coefficient of friction]
F= 10×3/5+0.6×10×4/5
F=6+6×4/5
F=6(1+4/5)
=6×9/5
=54/5
=10.8 N
hence the minimum force necessary to start the cube move up the plane is is 10.8 N
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