Question

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?​

Answer 1

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

Answer 2

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: