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Homework Help: Tourqe of a sign on the wall

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data

    a uniform 2kg horizontal beam is 50cm long is bolted to a wall and supports a 5kg sign. calculate the tourqe produced by the combined weight of the beam and the sign about where it is mounted to the wall.

    2. Relevant equations

    tourqe = F x Perpendicular distance

    3. The attempt at a solution

    So im using the wall as the pivot point making the distance to the centre of mass of the beam 0.25m.

    the tourqe of the beam= (9.8x2)x(0.25)
    = 4.9Nm

    The tourqe of the sign= (9.8x5)x(0.5)

    And to get the total tourqe i have added these togther to get 29.4Nm

    But my course book says the answer is 118Nm?? i know this is a pretty simple question but i cant pick up my error :confused:
  2. jcsd
  3. Mar 28, 2010 #2


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    Gold Member

    looks like a book error, if you described the problem correctly.
  4. Mar 28, 2010 #3
    the only thing i changed from the book is that it says a light rather than a sign.. but i think it may be a book error to. This book has many errors
  5. Mar 28, 2010 #4
    Well, one error is that it is a torque :biggrin:
  6. Mar 28, 2010 #5
    you never know the book might spell it that way to??!?!!!??
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