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Force on an inclined plane

  1. Feb 2, 2009 #1
    A 1140 kg car is held in place by a light cable on a very smooth (frictionless) ramp, as shown in the figure. The cable makes an angle of 31.0 degrees above the surface of the ramp, and the ramp itself rises at 25.0 degrees above the horizontal.


    Find the tension in the cable. (T=?)
    How hard does the surface of the ramp push on the car? (N=?)


    I am having the most difficult time trying to put the equations together. When I finally think that I put them together correctly, I don't know how to solve for a specific variable. I know that in this problem I will have two unknowns, T and N. I just can't find the reasoning behind putting the equations together. Help!

    So far I think that:
    Fx = Tcos(31) - w*sin(25) = 0
    Fy = Tsin(31) - w*cos(25) +n = 0

    I think I need the most help setting up the equations and solving for the answer. Can you please be as detailed as possible?

    Thanks!
     

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  3. Feb 2, 2009 #2
    You seem to have got your resolving right.
    w=mg so get T from the first equation, and N from the second.
     
  4. Feb 2, 2009 #3
    ok so for Fx:
    I set it up as T = (mg)sin25
    cos 31
    My answer was: 5508.2 N

    For Fy: N = (mg)cos25 - Tsin31
    I plugged T into that equation and my answer was: 7288.3 N

    Is that correct? What's my next step?
     
  5. Feb 2, 2009 #4
    Sorry,the cos(31) was supposed to be under the underlined part in the Fx equation.
     
  6. Feb 2, 2009 #5
    I haven't checked your calculation, but there isn't a next step.
    You have the forces asked for.
     
  7. Feb 2, 2009 #6
    Thanks a lot. I got the answers!
     
  8. Feb 3, 2009 #7
    I deduced that your x axis was || to the slope, and y perpendicular to it.

    Here's a tip: you can choose any directions you like to resolve in.
    They don't have to be at right angles to each other.

    If you resolve in the direction perpendicular to T (the rope) you get
    an equation with N but no T.
     
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