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See-saw, force exerted up by pivot?

  1. Nov 5, 2006 #1
    see-saw, force exerted up by pivot???

    A child weighing 36 N climbs onto the right end of a see-saw, little knowing that it is tired down with a rope at its left end. The see-saw is 2.8 m long, weighs 11 N, and is tilted at an angle of 25 degrees from the horizontal. The center of the mass of the see-saw is half way along its length and lies right above the pivot.

    There are 5 questions total and I've sucesfully answered 4 of them so far yet this one is odd:

    Q: What is the total positive force exerted upwards by the pivot?

    I first tried simply adding the two weights together (47 N) but this was wrong; then I added the two forces regarding the torque plus the total weight of objects on the see saw (45.6779 *2 + 36 + 11), but again that was wrong. How do I put these together? thanx.
  2. jcsd
  3. Nov 5, 2006 #2
    Is the right end of the seesaw touching the ground? If it was I would be inclined to say that only the weight of the seesaw is on the pivot.
  4. Nov 5, 2006 #3
    Please explain how you obtained 45.6779 *2. What you should try to calculate is the force that the rope exerts downwards on the see-saw. Can you take it from here?
  5. Nov 5, 2006 #4
    thanks for the responses! the right end (with the child on it) is way up in the air and the rope is holding the left end on the ground.

    I came up with 45.6779 * 2 because that is the torque the child exerts on the right hand side and the torque the rope applies (they're the same since they're in equilibrium).
  6. Nov 5, 2006 #5
    oh I see! ok i'll try it.
  7. Nov 5, 2006 #6
    great! thanks! I got it: (36 + 11 + 36 [the rope's force {not torque}])
  8. Nov 5, 2006 #7
    45.6779 * 2. I don't think this is the actual torque, you must have got some calculations mixed up.

    You're welcome. Good job in figuring it out.
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