Question

Marsha is pushing down and to the right on a 12 kg box at an angle of 30 degrees below the horizontal. If the coefficient of kinetic friction with the box is 0.70, what force must she apply to move the box with constant velocity?

Answer 1

There are three possible ways to solve this problem:
The first one is:
0.70*[(12*9.81) - |F|*sin 30] = |F|*cos 30 or
The second one is:
|F|*[{0.70*(sin 30)}+(cos 30)] = 0.70*(12*9.81) or 
The third one is|F| = [{0.70*(12*9.81)}/{0.35 + cos 30}]


But this will all result to= 67.77 N