A cube of 10N rests on plane of coefficient of friction 0.6 the slope of the plane is 3 in 5 the minimum force is requide to start the cube moving up the plane is?
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Given sinθ=35θ=37°cos37°=45Minimum force required is F=mgsinθ+μmgcosθF=10×35+0.6×10×45F=6+6×45F=6+6×45=6(1+45)=6×95=545=10.8 NRegards
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Answer:
Answer:
The answer will be 10.8 N
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
Explanation:
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