physics

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.

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  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.

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