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Torque Meter Stick Question

  1. Jun 24, 2007 #1
    1. The problem statement, all variables and given/known data

    A metre stick is found to balance at the 49.5cm mark when placed on a fulcrum. When a 47.7 gram mass is attached at the 1.5cm mark, the fulcrum must be moved to the 38.6cm mark for balance. What is the mass (in grams) of the metre stick.

    2. Relevant equations

    Torque = Force x Distance

    F = ma (?)

    3. The attempt at a solution

    Ok, well, I thought the torque equation would be good for this, but then again, what is the point of finding torque, when I can't use it to find the mass? Or can I?

    I was thinking about finding Force (0.0477 x 9.8), but that just leaves me with Force.

    Am I going completely wrong about this question or am I on the right track? Any help would be greatly appreciated.
     
  2. jcsd
  3. Jun 24, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The total torque about any point- for example, the left end of the meter stick, must be 0. And you certainly can "use it to find mass" since it involves mass!

    There are three sources of torque here- the 47.7 gram mass, the meterstick itself, and the fulcrum.

    Since I am calculating torque around the left end of the meter stick, and the force due to the 47.7 gram mass is downward, the torque due to it is clockwise which I will take to be negative: that torque is -47.7g(1.5)= -71.55g dyne-cm.

    If we let m be the mass of the meter stick, its downward force is -mg and again the torque is negative. Since we are told that its center of mass is at 49.5 cm, that torque is -(49.5)mg.

    Finally, in order to support the meterstick, the fulcrum must be pushing upward (positive force) equal to the total weight of the meterstick and added mass: g(47.7+ m) and is applied at distance 36.6 cm from the left end. The torque from that is 36.6g(47.7+ m) dyne-cm.

    All those must total 0: -71.55g- 49.5mg= 36.6g(47.7+ m). Solve that equation for m (of course, "g" cancels easily).
     
  4. Jun 24, 2007 #3
    Ok, thanks for the very descriptive reply, but could you explain a couple things. First of all, what is it meant by the term 'dyne-cm'? Also, when you talk about for example, '-(49.5)mg', thats force right? You also say 36.6cm from the left end, do you mean 38.6?

    Edit: Sorry, but I have completely lost myself. I can understand the concept that the force downwards has to equal the force upwards, but the equation is just messing me up.
     
    Last edited: Jun 24, 2007
  5. Jun 24, 2007 #4

    Doc Al

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    Staff: Mentor

    A dyne is a unit of force, so a dyne-cm is a unit of torque.
    No, that's a force times distance and thus a torque.
    I'm sure that's just a typo.

    You might find the equations easier to understand if you always choose the fulcrum as your reference point for calculating torques. When the fulcrum is at the 38.6cm mark, compare clockwise and counter-clockwise torques. (What torque-producing forces act on the meterstick? At what distance from the fulcrum does each force act?)
     
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