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Friction-less pulley homework

  1. Oct 29, 2003 #1
    A block weighing 130. N is on an incline. It is held back by a weight of 45.0 N hanging from a cord that passes over a frictionless pulley and is attached to the block on the incline. Find the angle theta at which the block will slide down the plane at constant speed. coefficient of friction is equal to 0.620

    My physics teacher gave me that problem thursday and I still haven't had any luck figuring it out. He said that I would have to use the quadratic formula and a trigonometric identity to figure it out. Any help would be appreciated
  2. jcsd
  3. Oct 29, 2003 #2


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    You need to find where the forces parallel to the incline balance out, which is where the blocks do not accelerate. The best way for you to find the solution is to draw a free body diagram. Since you are solving for the case where the acceleration is 0, you know that the tension in the string is equal to the weight of the hanging block (45 N). So to balance the forces along the incline, you account for the parallel component of the weight, the friction, and the tension.

    You'll end up with an equation with a constant term, a term proportional to sinθ, and a term prportional to cosθ. I suppose there are a number of ways to solve this. For one, you can change variables to say x = sinθ, which would make cosθ=√(1-x^2). You can rearrange the equation and square both sides to get a quadratic equation in x. Solve for x, then use arcsin to solve for θ
  4. Oct 30, 2003 #3
    Last edited by a moderator: Oct 30, 2003
  5. Oct 30, 2003 #4


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    You know there is a force parallel to the incline and pointing upward of 45 N (the weight on a pully).

    You know there is a force directly downward of 130 N (the block's weight).

    The incline forms a right triangle in which the incline itself is the hypotenuse and the vertical is a leg. Use that to find the component of the block's weight parallel to the incline.

    There is another right triangle in which the normal to the incline is the hypotenus and the vertical is a leg. Use that to find the Normal component of the blocks weight. The static friction force is that normal component times the coefficient of static friction.

    You now have three forces: the component of the block's weight DOWN the incline, the 45 N force UP the incline, and the static friction force UP the incline. The block will just start to move when those add to 0.
  6. Oct 30, 2003 #5
    Atleast this is an extra credit thing. I've looked at my diagram and both of your instructions and still have no idea where to get the force of the block parallel to the incline (Fn in the diagram if ya looked)

    I had a problem awhile back similar to this but I just can't remember how to start off. He said I will need to use the quadratic formula, and I'm not understanding how to get a quadratic equation out of cos(theta) and sin(theta). Also, the force of the block is throwing me off too. Wouldn't the Force of friction be a segment of the force parallel(45 N) to the incline?

    I've been staring at this problem for awhile, but I just can't get my foot stuck in the door. No light bulbs are coming on.
    Last edited by a moderator: Oct 30, 2003
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