Question

A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force.
when the ramp is frictionless.
when the coefficient of kinetic friction is 0.18.

Answer 1

Downward component of weight (along incline)
= mg(sin(θ))
Normal component of weight
= mg(cos(&theta));

Upward component of horizontal force (along incline)
= F(cos(θ))
Normal component of horizintal force
= F(sin(θ))

Coefficient of kinetic friction = μ

Total normal reaction
N= mg(cos(θ)+F(sin(θ))
Frictional resistance
=μ(N)

For equilibrium along the inclined plane
upward force = downward force

mg(sin(θ))+μN=F(cos(θ))
or
mg(sin(θ))+μ(mg(cos(θ)+F(sin(θ)))=F(cos(θ))
Substituting
m=90 kg
g=9.8 m/s²
θ=28°
μ=0.18
Solve for F
With μ=0
solve for F.

I get approximately 7*10^2N and 4.7*10^2N for the two cases.