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Torque due to gravity when T'net' = 0

  1. Apr 2, 2010 #1
    a ladder of length, 'L', leans against a wall with its base, 'd' metres, from the wall. the base doesn't slide or move. the ladder has a mass 'm'. how would i go about finding the torque of the ladder due to gravity, about the pivot point at its base.. and consequently the torque of the wall on the ladder. I am presuming that the torque due to gravity is equal to the torque from the wall on the ladder?

    Is it as simple as using cos or sine of the overall weight of the ladder to find the force on the wall?

    i know how to do friction sort of things so we can just assume for this there is no friction..

    any help is appreciated, thanks :)
  2. jcsd
  3. Apr 2, 2010 #2
    Apply the definition of Torque = the cross product of the Force vector and the Radius vector (directed from the pivot point to the point of application).

    Also, consider the force of gravity to be acting on the center-of-mass of the ladder, and we have:

    [tex]Torque = \textbf{F}\times\textbf{R}[/tex]

    [tex]\tau = <0, -mg>\times<d/2, \frac{\sqrt{L^2 - d^2}}{2}>[/tex]

    [tex]\tau = mgd/2[/tex]​

    Assuming you're familiar with that vector notation. It's pretty straight-forward.

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