Question

A cube of 10N rests on a rough inclined plane of slope 3 in 5.the coefficant of the friction is 0.6.The minimum force nessary to start the cube moving up the plane?

Answer 1

Answer:

tan 3/5 = 37°

the minimum force required to move up the plane should be more than the total downward force.

fs= 0.6*10 = 6N

mgsin37°=10*3/5= 6N

total downward force = 6+6 = 12N

Fmin should be 12N.

Answer 2

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