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Finishing off a force equation

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A 18.6-kg box rests on a frictionless ramp with a 13.7° slope. The mover pulls on a rope attached to the box to pull it up the incline. If the rope makes an angle of 38.8° with the horizontal, what is the smallest force F the mover will have to exert to move the box up the ramp?

    ____ N
    http://img93.imageshack.us/img93/7287/4figure44alt.gif [Broken]

    2. Relevant equations



    3. The attempt at a solution
    [tex]\sum\vec{F}=ma[/tex]

    [tex]\sum\vec{F}=0[/tex]

    [tex]\sum\ F_{x}=0 => F_{x}-W_{x}=0[/tex]

    [tex]\sum\ F_{y}=0[/tex]

    I made a graph from the ramp model and found:

    a=0
    m=18.6 kg
    θ=13.7°
    σ=38.8°
    B=25.1°
    F=?

    w=mg
    [tex]F_{x}=F\cos{B}[/tex]
    [tex]W_{x}=W\sin\theta[/tex]
    [tex]F_{y}=F[/tex][tex]\sin[/tex][tex]B[/tex]
    [tex]W_{y}=W\cos\theta[/tex]

    I thought 18.6(9.81)sinθ would give me the answer but that wasn't correct.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 11, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good.

    Good.

    How did you solve for F from the equations above?
     
  4. Oct 11, 2009 #3
    Oh, I think I see what I was missing. I need to find w=mg and plug that into Wx and solve right?

    -----

    Ok, I got [tex]43.2149=Fcos(25.1)[/tex]
     
    Last edited: Oct 11, 2009
  5. Oct 11, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Set up the equation Fx = Wx and solve for F. (Using your equations for Fx and Wx.)
     
  6. Oct 11, 2009 #5
    Alright, great! I got 43.2149=F*.9056

    so 43.2149/.9056 = 47.7196N

    : )
     
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