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Incline Plane/Pulley Problem

  1. Dec 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Two packing crates of masses m1 = 10.0 kg and m2 = 3.00 kg are connected by a light string that passes over a frictionless pulley as in Figure P4.26. The 3.00 kg crate lies on a smooth incline of angle 42.0°. Find the acceleration of the 3.00 kg crate.

    2. Relevant equations

    3. The attempt at a solution

    i know exactly how to do this problem, and i got the right answer. I just cant figure out why you are allowed to use the same angle to the ground of the inclined plane to solve the problem, as in this picture:


    why are both theta's the same?
  2. jcsd
  3. Dec 3, 2007 #2
    anyone have an idea?
  4. Dec 3, 2007 #3
    It simply geometry. The weight vector in your picture makes a right triangle with the left end of the incline. The incline makes an angle theta with respect to the horizontal. Now within that right triangle, you have 3 angles; theta, a 90 degree angle, and an angle that is 180 degrees minus 90 degrees minus theta, lets call this angle beta. You have drawn a line perpendicular to the incline that completes another right triange along with the weight vector. This triangle has a 90 degree angle (right beside the first 90 angle), and an angle at the top of the triangle that, when added to the angle beta, must equal 90 degrees. In our first triangle we know that theta and beta must add to 90 degrees. So for our second triangle, we know that the top angle, when added to beta, must equal 90 degrees, and from the first triangle, we know that theta fits this requirement. That's the explanation in words, but I would suggest that you toy around with the angles to get a feel for it.
  5. Dec 3, 2007 #4
    Yes. Extend the blue line in your free body to the ground, such that it makes a ninety degree angle with it, and then you shoul see by some analysis of both triangles that the angles are equal.

    Fudge around with it and see what you get.

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