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Trigonometry slope problem please help.

  1. Aug 12, 2005 #1
    I am really stumped! Using only Trigonometry, I am suppose to calculate the amount of force that is needed to keep a 2000 lb car from rolling down a ramp of 30 degrees.

    Am I on the right track when I say: x=2000cos 30? Then what?
  2. jcsd
  3. Aug 12, 2005 #2
    Well the basic idea is good but I don't know how you came up with cos. If you look at a scetch you should be abel to find the right function for the force perralel to the slope. And just a small hint when your searching for a force it's usually a good idea to use weight instead of mass (I'm not very good at non-metric units soo I don't really know wheater lb steands for mass or weight).

    Hope this helps.

    PS. Maybe this thread would be more sutible for the Science Education Zone.
    Last edited: Aug 12, 2005
  4. Aug 12, 2005 #3
    Hmm...what is the coefficient of friction between the tires and the incline?

    Assuming the ramp is frictionless, your "force" must contain a component parallel to the ramp, with magnitude:

    [tex] F = \left( {2000 \, {\text{lb}}} \right)\sin \frac{\pi }{6} = 1000\,{\text{lb}} [/tex]

    Basically, gravity will exert 1000 lb down the ramp. To keep the car from rolling down, you must exert a force of 1000 lb UP the ramp (parallel to the ramp, antiparallel to the component of weight parallel to the ramp).

    However, if you do have friction, then to prevent rolling down the ramp (NOT skidding!), you must find the coefficient of 'static' friction between the tires and the ramp. Now, let [itex] \mu _S [/itex] represent this value. Thus, the force needed to prevent the car from rolling down is:

    [tex] F = 2000\,{\text{lb}}\left( {\sin \frac{\pi }{6} - \mu _S \cos \frac{\pi }{6}} \right) = 1000\,{\text{lb}}\left( {1 - \mu _S \sqrt 3 } \right) [/tex]

    **In the direction antiparallel to the weight component parallel to the ramp.

    :smile: What basic idea? Also, by "sketch", do you mean a free-body diagram?
    Anyway, those are always helpful (Unless you're in a pinch for time!...or the problem is just basic).
    Also "lb" stands for pounds, which are a customary unit of weight, I believe, according to
    Last edited: Aug 12, 2005
  5. Aug 12, 2005 #4
    Thanks for the explination.
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