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Torque on a see-saw type object

  1. Jan 8, 2006 #1
    Torque on a "see-saw" type object

    I had this question on my Physics test last week. I missed it, and it's now time to do test corrections. I have read through the section once again, but I keep on getting the same answer.
    A uniform plank XY is supported by two equal 120-N forces at X and Y, as shown. the support at X is then moved to Z (half-way to the plank center). The supporting force at Y then becomes:
    A. 40
    B. 160
    C. 60
    D. 80
    E. 240
    I keep getting 60. I know that the net torque must remain 0, so as to keep the object from rotating. That being said, I cannot seem to come up with the correct answer (D).
    Here's what my logic was:
    If you move the point X, halfway to the center, it will produce 1/2 the amount of torque. So, the force at Y must decrease by 1/2 so as to keep the object in rotational equilibrium.
    Please help me cite my error. Thank you in advance
  2. jcsd
  3. Jan 8, 2006 #2


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    Homework Helper

    If the force at Z is half the distance (x) that the force at Y is (2x) from the centre,
    and the torques must be equal for balance,
    then the reaction at Z must be twice that of the reaction at Y.

    And, of course, the sum of these two reactions is 240 N.
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