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To which bodies can we apply the concept of torque?

  1. Dec 6, 2015 #1
    I am in introductory physics, and have just been introduced to the topic of toque. For forces in translational motion, I know that we idealize objects to be point particles. However, when it comes to torque, we don't deal with point particles, but with "lever arms." I guess essentially my question is, what does the typical free-body diagram look like for a torque problem? For regular forces the FBD is just a point with arrows.
     
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  3. Dec 6, 2015 #2

    DaveC426913

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    Toque is only applicable to Canadian bodies. :DD
     
  4. Dec 6, 2015 #3
    Did you also learn about the cross product? Torque is defined as the radius vector crossed with the force vector. You've learned about translational motion; F = ma (a constant force causes an object to accelerate translationally), analogously, a constant torque causes an object to accelerate about an axis of rotation (τ = I α) I = Moment of Inertia, α = angular acceleration

    The direction of this torque is determined by the cross product, it will be perpendicular to both the radius and force vectors. In a 2D plane, this translates to a direction of into or out of the page (do you know the "right hand rule"?). Small movement changes the direction of the radius vector and overall you'll observe rotational motion.

    -------

    Here's a cool video:

    If you define the radius vector from the center, and the force vector (gravity) pointing downward, the cross product is what causes it to precess around like that.
     
    Last edited: Dec 6, 2015
  5. Dec 6, 2015 #4

    SteamKing

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    Here is a cantilevered shaft which is loaded by three torques:

    images?q=tbn:ANd9GcThHY_AKTXAEUzyvC9XFbQRcTBHib4O6nEsnGugU7EpjJnD2soY.jpg
    Just like a force vector is an arrow (since it causes a body to translate in the direction of the force), a torque vector is a curly arrow (since it causes a body to rotate in the direction of the torque).
     
  6. Dec 7, 2015 #5

    CWatters

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    Here is a FBD for a 6m long crane arm that is hinged at the bottom left, it has mass 30kg and a 60kg mass hanging from the top end. There is a cable holding it up that has tension T. Your mission is to find the tension T...

    CH0810b.gif

    Even though the arm is subject to forces it can be treated as a torque problem. For example..

    The crane arm isn't accelerating (it's static) so the torques must sum to zero. We can write this equation for the torque about the pivot..

    Torque due to 30kg mass of arm + Torque due to 60kg mass + torque due to cable = 0

    If we define clockwise as +ve and assume g =10m/s/s that becomes..

    + 300*3*Sin(60) + 600*6*Sin(60) - T*4*Sin(30) = 0

    and we can solve for T.
     
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