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Where to place a weight on a bar so that the net torque is 0

  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data
    A 0.100-kg, 59.9-cm-long uniform bar has a small 0.055-kg mass glued to its left end and a small 0.100-kgmass glued to the other end. You want to balance this system horizontally on a fulcrum placed just under its center of gravity.

    no diagram given
    2. Relevant equations
    Στ = 0 N*m
    τ=F×d
    3. The attempt at a solution
    Massbar=.100 kg
    L=0.599m
    m1=0.055kg
    m2=0.100kg

    I am confused about how to set up my diagram. I have made a bar. The only forces acting on the bar are the two weights from the two masses.

    I didn't include the weight of the bar itself because since its uniform I wanted to make the center the pivot axis.

    SO my torque equation looks like this:
    Στ=0=-m1g(d`) + m2g(d2)

    the first torque is negative because it is going in the clockwise direction.

    But from this equation i can get infinite many distances depending on what d2 is.
     
  2. jcsd
  3. Dec 10, 2016 #2
    What are the rules/restrictions for this problem? Are either weights' location predetermined?
     
  4. Dec 10, 2016 #3

    NascentOxygen

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    Staff: Mentor

    There's your mistake. It's not the CoM of the bar where you will pivot, it's to be the CoM of the assembly.
     
  5. Dec 11, 2016 #4
    the 0.05 kg mass is said to be on the left end, the 0.10 kg mass is said to be on the right end. That is all that is specified. And I assume they do not mean at the very end otherwise the bar would be in motion.
     
  6. Dec 11, 2016 #5

    gneill

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    Staff: Mentor

    I think they mean that the weights are indeed affixed to the very ends of the bar. Your job is to locate the fulcrum so that the system balances horizontally.
     
  7. Dec 11, 2016 #6
    Ok so then I have another force that is the weight of the bar, which is at d=0.295m.
    Ok using this information and the previous post this is what I did.

    Στ=0=0.05gd+0.10g(d-0.295)-0.10g(0.599-d)

    rearranging I got to

    0.15d - 0.0295 = -0.10d + 0.0599
    0.25d = 0.0894
    d = 0.3576 =35.8 cm

    which my online homework took as the correct answer even though it said exact answer was 35.2 cm.

    Thank you for your guidance on clearing up my confusion.
     
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