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Help with Torque and Force Please!

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A child weighing 38 N climbs onto the right
    end of a see-saw, little knowing that it is tied
    down with a rope at its left end. The see-saw
    is 2.6 m long, weighs 12 N and is tilted at an
    angle of 17◦ from the horizontal. The center
    of mass of the see-saw is half way along its
    length, and lies right above the pivot.

    1. What torque does the weight of the child
    exert about the pivot point? Take counter-
    clockwise to be positive. Answer in units of
    Nm.

    2. What torque does the weight of the see-saw
    exert about the pivot point? Take counter-
    clockwise to be positive. Answer in units of
    Nm.

    3. What torque does the rope exert about the
    pivot point? Take counter-clockwise to be
    positive. Answer in units of Nm.

    4. What force does the rope exert downwards on
    the see-saw? Give a positive answer. Answer
    in units of N.

    5. What is the total positive force exerted up-
    wards by the pivot? Answer in units of N.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 24, 2009 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    How would you think to start?
     
  4. Mar 24, 2009 #3
    Just a little bit short on work on your part! Please make an effort first.
     
  5. Mar 24, 2009 #4
    Well, try to look up for equations for Torque.Its a pretty straightforward problem
     
  6. Mar 24, 2009 #5
    Umm, I know it would probably seem like it would be straight forward, but I'm a complete idiot when it comes to physics. I also contribute that in part to the fact that my physics professor does not speak English very well and doesn't answer any of our questions during class so I've been trying to teach myself. That hasn't gone very well. Sorry. I'll post my attempt shortly... thank you!!!
     
  7. Mar 25, 2009 #6
    1. [tex]\Sigma[/tex][tex]\tau[/tex] = -mg - Fchild
    = -12N(.65m) - 38N(1.3m)
    = -57.2Nm

    I don't think that's right though... the answer shouldn't be negative should it?
     
  8. Mar 25, 2009 #7

    LowlyPion

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    Homework Helper

    When you take the sum of the Torques you take the moments acting about the pivot point. Since the see-saw itself has a center of mass at the pivot point it can't contribute to the net torque.

    The child does act at a distance along the see-saw of 1.3m, but you have failed to account for the 17 degree tilt. Since that moves the child effectively closer to the pivot, then the effective moment arm for the child's mass is incorrect.
     
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