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Center of gravity & mass of a nonconformed meterstick

  1. Oct 6, 2009 #1
    A meter stick is pivoted the 50-cm mark but does not balance because of nonconformities in material.
    150 g and 200 g weights ar placed at the 10-cm and 75-cm marks to balance the meterstick.
    When the weights are interchange, the pivot point is at the 43-cm mark.
    find the mass of the stick and the center of gravity.


    I've worked the problem seven ways from sunday. this keyboard is giving me a hard time, I can't really write it all out for you. :(
    keys are sticking.
     
  2. jcsd
  3. Oct 6, 2009 #2
    Yay! new keyboard;

    setup 1
    w1 d1 = w2 d2
    (1.47 N)(.5m -.1 m) = (1.96 N)(.5-.75 N)
    .588Nm = -.490Nm
    now what?

    setup 2
    w1 d1 = w2 d2
    (1.96 N)(.0m-.430m) = (1.47 N)(.750m -.430m)
    -.647 Nm = .470 Nm
    okay...
     
  4. Oct 7, 2009 #3

    Doc Al

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    You need the weight of the meterstick in those equations. Assume its weight W acts at position x. Set up two equations (like you did but including the torque due to the weight) and solve for those unknowns.
     
  5. Oct 7, 2009 #4
    What I did so far is this;
    since w1 d1 = w2 d2 I got the moments for each side on both setups, and the distance from the axis to the weight and set the equations up

    (1.47 N + x)(.4 m) = (1.96 + y )(.25 m) for first setup

    (1.96 N + x)(.33m)=(1.47 N + y)(.32m) for the second setup
    rearranged to get y= (.66 N + x(.33m))/.32 m
    and x = y(.32m)/(.66+.33m)

    subbed and solved to get x = 6.66N, .680 kg and y = 3.57 N, .364 kg

    .364 kg + .68 kg = 1.04 kg meter stick

    that's all I've got. I dont know if I did it right, and I don't know what comes next.
    Any ideas?
     
  6. Oct 7, 2009 #5

    Doc Al

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    w1 d1 = w2 d2 assumes there are only two forces, but here you have three. Use ΣT = 0 or clockwise torques = counterclockwise torques, but include all forces.

    Call the position of the meterstick's center of mass X and its weight W. From the first setup, can you tell if the meterstick's center of mass is above or below the 50 cm mark?
     
  7. Oct 7, 2009 #6
    I'm lost...
     
  8. Oct 7, 2009 #7

    Doc Al

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    Start by trying to answer my question:
    Hint: Compare the torques due to the two added weights. Which is greater? So on what side of the pivot (50 cm mark) must the stick's center of mass lie in order to balance things out?
     
  9. Oct 7, 2009 #8
    the torque on the 0-50 side is .588 N, its .490 on the 50-100 side. the 50-100 side of the ruler is greater, so the center of mass is somewhere below the 50 mark.
     
  10. Oct 7, 2009 #9

    Doc Al

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    OK. That means the added weights add more torque on the 0-50 side in order to balance out the torque due to the stick's weight.

    Just the opposite. In order to balance the torque from the added weights (which you just found) the stick's center of mass must be on the 50-100 side.
     
  11. Oct 7, 2009 #10
    alright. so if there's a difference of .098 N on the heavier side, and the distance on that side is .25 m, the heavier side has (0.098 N)(.25m) more torque, right?

    what next?
     
  12. Oct 7, 2009 #11

    Doc Al

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    What you just determined is that the stick's center of mass is on the 50-100 side of the stick. Let's say that it's at a point X meters above the 50 cm mark. Now write the torque equation for setup #1.
     
  13. Oct 7, 2009 #12
    +(1.47 N)(.5m -.1 m) + -(1.96 N)(.5-.75 N) + -(50m + x)(w) = 0 ?
     
    Last edited: Oct 7, 2009
  14. Oct 7, 2009 #13

    Doc Al

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    Almost. Try this: Identify the clockwise and counterclockwise torques. Set net clockwise torque = net counterclockwise torque. Be sure to measure distance from the pivot point for all torques.
     
  15. Oct 7, 2009 #14
    this?
    counterclockwise is +, clockwise is - right?

    +(1.47 N)(.5m -.1 m) + -(1.96 N)(.5-.75 N) + -(50m + x)(w) = 0 ?
     
  16. Oct 7, 2009 #15

    Doc Al

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    That's good.

    For the second term, careful with signs (and units). The distance from the pivot is .25m. The torque is -.

    For the third term, what's the distance of the stick's center of mass from the pivot.
     
  17. Oct 7, 2009 #16
    +(1.47 N)(.5m -.1 m) + -(1.96 N)(.75-.5 N) + -(50m - x)(w) = 0 ?
     
  18. Oct 7, 2009 #17

    Doc Al

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    Tell me how you are defining X.
     
  19. Oct 8, 2009 #18

    Doc Al

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    In case it wasn't clear what I was getting at:
    (1) If x is how far the stick's center of mass is from the midpoint (the 0.5m mark), then the torque about the pivot due to the stick's weight is -(w)(x).
    (2) If x is how far the stick's center of mass is from the zero mark, then the distance from the pivot is x-.5m, thus the torque term would be -(w)(x-.5).

    These remarks apply to setup #1.
     
  20. Oct 19, 2009 #19
    (1)
    ∑ T(from left end) = 0, => equilibrium condition
    0.15(0.1)9.8 – 0.5F + 0.2(0.75)9.8 = 0

    F = 9.8[0.015+0.15]/0.5 = 9.8(0.165/0.5) = 3.23 N

    (2) Interchange weights
    ∑ T(from left end) = 0, => equilibrium condition
    0.2(0.1)9.8 – 0.43F + 0.15(0.75)9.8 = 0

    F = 9.8[0.15(0.75) + 0.2(0.1)]/0.43 = 3.02 N

    F(1) – F(2) = 3.23 – 3.02 = 0.021 N => stick’s weight

    mass of stick: m = 0.021/9.8 = 0.0214 kg = 21.4 g

    Center of gravity: <x> = [0.5(3.23) + 0.43(3.02)]/6.25 = 0.466 m = 46.6 cm
     
  21. Oct 19, 2009 #20

    Doc Al

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    You included the upward support force of the pivot on the stick, which you called F. Good! But you forgot about the weight of the stick itself: you must include that force in your torque equation.

    It's easier if you find torques about the given pivot point, then you don't have to worry about the support force from the pivot. Let's call counterclockwise positive and clockwise negative. Call the mass of the stick M and its center of gravity at point x.

    Torque from 150 g mass about pivot: +(.15)g(.4)
    Torque from 200 g mass about pivot: -(.2)g(.25)
    Torque from stick mass M about pivot: -(M)g(x - .5)

    See if you can continue from here.
     
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