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Homework Help: Force by table leg

  1. Apr 29, 2015 #1
    1. The problem statement, all variables and given/known data
    https://courses.edx.org/static/content-mit-mrev~2013_Summer/problems/MIT/rayyan/check_points/Pictures/BK85.png [Broken]

    A horizontal uniform bar of mass 220 kg and length L = 3.0 m is placed on two supports, labeled A and B, located as shown in the diagram. A block of mass 30 kg is placed on the right end of the bar.

    Find the normal forces F_A and F_B exerted on the bar by the supports.

    2. Relevant equations
    I'm sure we'll use torque, T = rFsinθ. We'll probably use center of mass, Σmx / M. Because the bar is uniform, though, we know the center of mass will be the mid-point of the bar. Torque will be determined by the gravity acting on the center of mass, which has equation F = mg. Since the object is static the sum of all torques will be 0.

    3. The attempt at a solution
    I treat the point B as if it were a pivot point, in which case the distance to the center of mass is 0.5m to the left. Then the torque applied is T1 = 0.5 * 220g = 110g. The block has distance 1m to the right of the pivot and so exerts a torque T2 = -1*30g in the opposite direction. The normal force due to support A will then satisfy 110g-30g+N=0 and this implies N = -80g.

    However, this answer turns out to be wrong. Any idea what my mistake is?
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 29, 2015 #2


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    Are you defining N to be the force at A or the torque it exerts about B?
  4. Apr 29, 2015 #3
    Oh wow, can't believe how simple my mistake was. Thanks for the hint, haruspex!
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