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Homework Help: 2 masses connected via pulley system.

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data

    The system shown below consists of a block of mass M = 3.6 kg resting on a frictionless horizontal ledge. This block is attached to a string that passes over a pulley, and the other end of the string is attached to a hanging block of mass m = 1.9 kg. The pulley is a uniform disk of radius 7.4 cm and mass 0.55 kg. If the (frictionless) ledge were adjustable in angle, at what angle would it have to be tilted upward so that once the system is set into motion the blocks will continue to move at constant speed?

    If anyone has trouble understanding how the pulley is set up, its basically 1 mass resting on a completely horizontal table with a pulley at the edge of the table leading to a hanging mass. Pretty much your typical pulley system problem.

    2. Relevant equations

    3. The attempt at a solution

    Basically I realize that the two tensions have to equal each other, and the tension on the mass hanging off the table is mg. This tension has to equal the tension on the inclined mass as well right? Wont when the tensions are equal the masses will move with constant velocity(no acceleration)?

    I ended up getting ~25 degrees, but its wrong according to webassign. I guess the pulley's moment of inertia has to be used somehow.
  2. jcsd
  3. Sep 26, 2010 #2
    I don't think inertia should affect the answer at all; inertia just tells you how fast something accelerates relative to applied force, and a constant motion situation involves no acceleration (linear or rotational).

    The tension will always be equal no matter what the situation (assuming a taught rigid string). What has to be equal is the gravitational force projected onto the direction of motion for each block.

    So, depending on the way the angles are set up, you want to solve something like:

    [tex]3.6 \sin\theta = 1.9[/tex]
    Last edited: Sep 26, 2010
  4. Sep 26, 2010 #3
    yeah thats exactly what I had. Solve for theta by arcsin(1.9/3.6) gives me 35.39 degrees, which also isn't right :/
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