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Basic Physics Ramp & Friction Question

  1. Nov 7, 2014 #1
    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.
     
  2. jcsd
  3. Nov 7, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The speed is constant, what do you expect?
    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.
     
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