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Mind boggled by a tension problem.

  1. Nov 4, 2006 #1

    Hi everyone! I am mind boggled by this tension problem.
    Here's what I grasp. If the clown pulls downward, the rope exerts an opposite force with the same magnitude upward. I don't understand how exactly the tension going upward exerts a tension along the three pulleys. Can anyone draw a free body diagram of the tensions throughout the pulleys and do his feet? Everytime I do it, the tension from last pulley points toward his feet, and that doesn't make any sense of course, since his feet is getting pulled backwards. I'm sure once I get the concept of tension I can finish the problem. Thanks for checking this out!
  2. jcsd
  3. Nov 4, 2006 #2


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    Homework Helper

    Brr, a clown [:yuck:]. I refuse to thing about anything. [:smile:]
  4. Nov 4, 2006 #3
    Try drawing the forces on his arms and the force on his legs only.
  5. Nov 4, 2006 #4

    Doc Al

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    Staff: Mentor

    Absolutely correct.
    Assuming the rope is massless and the pulleys frictionless, the tension created in the rope will be the same throughout the rope. If the clown pulls down on the rope with a force of X Newton's, that means the tension force exerted by the rope (at both ends) is X Newtons: The rope pulls up on the clown (at one end) and to the left on his feet (at the other end) with the same force.
    The tension is the same throughout the rope.
    Why doesn't that make any sense?
  6. Nov 4, 2006 #5
    well, if the clown is pulling downward on the rope, and the tension is the reaction of the clown's action. Then the clown is stationary since the forces are at equilibrium right? But while all this is happening, the clown's feet is touching the floor, where there is friction which in turn opposes the clown's action. So I don't understand how can there be two forces (the tension of the rope and the friction) opposing one force (the pulling down)? I thought for every force applied, there has to be an opposing force to it. I guess what I'm confused at then is how all the forces work in the problem.
  7. Nov 4, 2006 #6

    Doc Al

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    The clown pulling down on the rope, and the rope pulling up on the clown are "action-reaction" pairs, yes.

    Careful: Action-reaction pairs never produce equilibrium--they act on different objects. (This is important to understand.)

    The clown remains stationary as long as the net force on the clown is zero.

    What horizontal forces act on the clown? I count two: The rope pulling to the left and friction pulling to the right. As long as they balance, the clown will not be pulled off his feet.

    The rope exerts two different forces on the clown: It pulls up on the clown's hands (reaction: the clown's hands pull down on the rope) and it also pulls sideways on the clown's feet (reaction: the clown's feet pull back on the rope). These two forces happen to have the same magnitude, since they are both produced by the tension in the rope, but they are different forces.

    The friction is another force acting on the clown; it's a force of the ground pulling on the clown's feet--the reaction to the friction is the clown's feet pulling back on the ground.

    To find out what happens to the clown, analyze the forces acting on the clown.

    What forces act on the clown? (Consider vertical and horizontal forces separately.) I count 3 vertical forces and 2 horizontal forces acting on the clown.
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