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Trig problems (relating to FBD)

  1. Jul 11, 2007 #1
    1. The problem statement, all variables and given/known data

    Car being towed up incline. It makes 20deg angle to the horizontal. Goes at constant velocity. No friction. Tow rope rated at 6000N max..
    will it break?

    2. Relevant equations

    I dont need help in whole question, its just the trig which i keep getting wrong. (i have taken trig).
    The answer is actually worked out in book but i dont understand why the trig is as it is.



    3. The attempt at a solution
    So in my FBD i orientate axis so that normal force points along pos y-axis, tension points along pos x-axis.
    I know that if incline angle is 20deg, then the angle weight makes with neg y axis will also be 20deg

    The book says that wsuby = -w cos theta.
    This is messing me up on alot of problems so heres what I understand, in very basic trig terms:

    The weight vector, neg y axis and horizontal make up a triangle.
    Theta is the angle (20deg) between the neg y axis and the weight.

    Therefore weight is the adgacent side
    Neg y-axis is the Hyp side
    the horizontal is the opposite side (to the angle).

    So, ,based on the triangle i have chosen, the y-component of the weight = the hyp side of the triangle and the adjacent side is the weight vector.

    To sum up, I have the adjacent, and the angle, and I want the hyp.
    So, according to trig, because cos theta = adj / hyp,
    then hyp = adj / cos theta.

    So why does the book say that the y component of the weight is -w * cos theta????
    Thanks for any help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 11, 2007 #2

    PhanthomJay

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    Why do you say that the weight is the adjacent side? And that the y axis is the hypotenuse? Draw your diagram again.
     
  4. Jul 11, 2007 #3
    Along an inclined plane, the incline is the x-axis and the y-axis is perpendicular to it.
     
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