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Really need some help-proof

  1. Nov 16, 2005 #1
    hi

    If net force on the body is zero and also sum of torques is zero,then object doesn't rotate and wherever we put an axis, the sum of torques is always zero.

    I know it's true and it does make sense, but can you show me mathematical proof (or vector proof ) that no matter where the axis is, the sum of torques is always zero?

    In the following example no matter where we put an axis the sum of torques is zero(I already solve it, so no need for help there)

    Code (Text):
                                                 
                 
         _____|C|__________                                
         |A|           |B|              
     
    Board( weight of the board is negligible ) length is 4 meters and lays on top of boxes A and B. Third box C is on top of board and has mass 16 kilos. Length from A to C is 0.5 m.

    thank you very much
     
    Last edited: Nov 16, 2005
  2. jcsd
  3. Nov 18, 2005 #2

    Astronuc

    User Avatar

    Staff: Mentor

    For the system to be static, i.e. [itex]\Sigma[/itex]F = 0 and [itex]\Sigma[/itex]M = 0 (sum of moments).

    The sum of forces is straight forward = FA+FB-FC = 0 => FA + FB = FC

    OK then one does [itex]\Sigma[/itex]M = 0

    Take the moment about A and B separately.

    At A, MA = 0, because the moment arm of FA is zero, and I am assuming no net torque applies, and one has MC + MB = 0. Take the + moment to be counter-clockwise, -lC*FC + lB*FB = 0.

    The same method can be applied at B.

    One needs one equation for each unknown otherwise the system is indeterminate.
     
  4. Nov 19, 2005 #3
    but how does that prove that no matter where the axis is the sum of torques is zero?

    Subquestion if I may...when we say the axis can be anywhere and sum is still zero...do we mean laying anywhere inside the body or can axis also lay outside the body?
     
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