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Horizontal force with a crate at a 27 degree angle

  1. Sep 14, 2006 #1
    A 100 kg crate is pushed at constant speed up the frictionless 27° ramp shown in the figure. What horizontal force F is required?
    What force is exerted by the ramp on the crate?

    I can't paste the figure up so I hope this question describes the scenario well enough.

    What's giving me problems is how am I suppose to find F when I don't even know n? I figured I use the angle to help solve for others. I know Fg=mg which is (100)(9.8) right? Would that give me one side of the triangle?
     
  2. jcsd
  3. Sep 14, 2006 #2
    The problem says at a constant speed, therefore a=0. Also, there is no friction slowing you down. F=ma, a=0, so the net force is zero. You need to find the vertical and horizontal components of the applied force. So F(applied)*cos(27)=F(applied,horizontal), and F(applied)*sin(27)=F(applied, Vertical).

    That should get you off to a good start, remember you don't know F(applied), so you'll have to solve for two variables (substitution in otherwords).
     

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    Last edited: Sep 14, 2006
  4. Sep 15, 2006 #3
    I'm still confused. First of all did you post a figure cause it's not showing up. I understand that I need to use substitution to solve for one and then solve for the other but I still feel like I don't have enough information. This is what I have down:

    Xcomponent
    ∑Fx= max
    mgsin27 = ma
    a=gsin27

    Ycomponent
    ∑Fy= may
    N-mgcos27= 0
    N= mgcos27

    Is this at all right?
     
  5. Sep 15, 2006 #4

    radou

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    Homework Helper

    In the y-sum you forgot to add F, so you have mgcos27 + Fsin27 = N.
     
  6. Sep 15, 2006 #5

    Doc Al

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    Staff: Mentor

    Is the applied force horizontal? or parallel to the ramp?
     
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