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How does a net torque of 0 make a board snap?

  1. Apr 15, 2012 #1
    This is just a scenario I have been pondering.

    Imagine one took a board and placed it on a pivot in perfect equilibrium.

    Next they added two 10KG weights 1 inch out, on each side of the pivot.

    The board would probably not snap with the weights at this distance, but lets say one were to keep increasing the distance of the weights (in identicle increments) from the pivot.

    Experience leads me to the conclusion that eventually the board will break, but given my current knowledge in physics I could not model this with equations. I know that there is a net torque of 0 at the pivot. I know though that this does not mean that in reality the torque doesn't exist. However I am curious as to the exact explanation of why the board snaps, and a way to model this scenario.
     
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  3. Apr 15, 2012 #2

    Drakkith

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    There may not be a net torque on the board, but there is a force. That force is pulling downwards on the board, and once the force is greater than the boards ability to hold itself together it will snap.
     
  4. Apr 15, 2012 #3
    Thats right, but isn't the downwards force the same regardless of the distance of the weights? Or am I misremembering the properties of torque?
     
  5. Apr 15, 2012 #4

    Drakkith

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    My mistake, I misread and thought you were adding weights, not just moving them outwards. The downward force of the weights does remain the same. I believe it may have something to do with mechanical advantage, as each side of the board is acting like a lever.
     
  6. Apr 15, 2012 #5

    haruspex

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    The quantity you need to calculate is the bending moment.
    That the net torque is zero only means there's nothing to make the board tip.

    Because of the symmetry, it's the same as if one side were clamped in a vice (vise).
    The bending moment on the board is a maximum at the fulcrum, and it is the torque from one weight only.
     
  7. Apr 15, 2012 #6
    If the board is in equilibrium and rigid, not only do all the forces and torques on the entire board have to cancel, but also the forces and torques on any part of the board. This can be used to calculate tensions in the board.

    Consider the part of the board to the right of the pivot, The torques around the pivot on this part of the board should be 0. The board has a length L, and thickness d, and the weight a mass m.
    There's a torque (L/2)mg around the pivot. The pivot can't produce any torque around itself, so the torque has to be compensated by a force from the other half of the board. So the other half of the board has to exert a force F to the left.
    The torque produced by this force is at most Fd, if the force was all exerted at the top of the board, so you have Fd >= (L/2)mg
    Since L/2 is likely to be much bigger than d, this means F will be much bigger than the force of gravity on the weight, and it wil have to go up if L goes up.
    This force can only be produced if the board is stretched,
     
  8. Apr 15, 2012 #7

    PhanthomJay

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    Since the net torque on the board is 0 about any point, the applied torque about any point from the external force must be equal to the internal torque in the board at that point. That internal torque produces bending stresses in the board per well known egineering formulas. When the bending stress in the board exceeds its maximum stress capacity to withstand such torques (which depends on the material properties), the board cracks and breaks.
     
  9. Apr 15, 2012 #8
    PhantomJay I think that has helped me get it.. I did not know we had a way to measure the internal torque on an object, and this seems like a fairly basic scenario once you make use of those equations.
     
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