crate weighing 6.90 x 10^3 N is pulled up a 46° incline by a force parallel to the plane. If the coefficient of kinetic friction between the crate and the surface of the plane is 0.750, what is the magnitude of the applied force required to move the crate at a constant velocity?

Answer 1

Assuming the crate moves upwards.

The applied force P is upwards whether the crate moves upwards or downwards at a constant velocity.

coeff. of kinetic friction, μ = 0.75
angle of incline, θ = 46° (with horizontal)

If crate moves upwards with a constant velocity, then forces along plane sums to zero (no net force).

The resistive (friction) force acts down the plane in opposite direction to the motion.

Resolve weight normal to plane,
N=W*cos(θ)

Force down the plane due to weight (downwards)
F=W*sin(θ)

Frictional force (downwards)
Fr=μN = μW cos(θ)

Applied force, P (upwards)

Sum forces along plane:
P-Fr-F=0
P=μW cos(θ)+W sin(θ)

Substitute values and evaluate P.
I get about 8558N.