Basic Physics Ramp & Friction Question

[olwyn]

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.
a. when the ramp is frictionless.
b) when the coefficient of kinetic friction is 0.18.

Fnet = ma
Fg= mg
Ff=uk(Fn)

a) With upwards and right as positive and downwards and left as negative
i) After drawing a diagram, I broke the Fnet into its x and y components.
I determined Ffrictionx= Ffriction(cos28), Ffrictiony= Ffriction(sin28). I will not include these in the horizontal and vertical components as it is a frictionless surface.

Fnet y = ma
Fnormal - Fg = ma
Fn= m(a-g)
The object is not accelerating in the horizontal direction so a=0
Fn= m (-g)
Fn = 90kg (-(-9.8m/s^2))= 882N.
So the vertical force is 882N in the upward direction.

Fnet x= ma
Fpush= ma
Fp=(90kg)(a).

I don't know what the acceleration is in the horizontal direction so I'm not sure where to go from here.

b). Again, break into horizontal and vertical components with up/right as positive, down/left as negative.
Fnet y = ma
Fpush - Ffrictionx = ma
Fp - Ff(Cos28) = ma
Fp - (cos28) (uk)(Fn)= ma
Fp - (cos28(uk)(-mg))= ma
Fp= ma +140.2
... I'm not sure if I'm on the right track here, or if I am, where to go from here.

 

[mfb]

I don't know what the acceleration is in the horizontal direction so I'm not sure where to go from here.

The speed is constant, what do you expect?

Fpush= ma

No. Why should it? There is more than one force involved.

I don't understand your approach at (b), but try to solve (a) first.