A box with a mass of 4.0kg is sitting on a plane angled at 60degrees above the horizontal. The coefficient of friction between the box and the plane is 0.3. What is the horizontally applied force necessary to stop the box from sliding down the ramp?

Answer 1

mg = (4kg)(9.81m/s^2) = 39.24N. @ 60deg
m = mass in kg.
g = acceleration due to gravity.

Fp = mgsin60 = 39.24sin60 = 34N = Force parallell to plane.

Fv = mgcos60 = 39.24cos60 = 19.62N =
force Perpendicular to plane.

F = Fp + u*Fv = 34 + 0.3*19.62
F = 34 + 5.89 = 39.89N Parallel to
plane.

Fh = Fcos60 = 39.89cos60 = 19.95N =
The required hor. force.

Answer 2

CORRECTION!
F = Fp - uFv = 34 - 0.3*19.62,
F = 34 - 5.89 = 28.11N parallel to plane.

Fh = Fcos60 = 28.11cos60 = 14.06N =
The required hor. force.