Question

A cube of weight 10 Newton rest on a rough incline plane of slope 3 in 5 . The coefficient of friction is 0.6. the minimum force necessary to start the cube move up the plane is

Answer 1

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

Answer 2

Answer:

Slope 3 in 5 means 3 vertical distance in 5 horizontal motion:

F=(mg)sinθ+μ

k

mg cosθ

θ=sin

−1

(3/5);μ=0.6;mg=10N

F=mgsinθ+μmgcosθ=10.8

Explanation:

hope it helps