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physics

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A meter stick is pivoted at its 50-cm mark but does not balance because of non-uniformities in its material that cause its center of gravity to be displaced from its geometrical center. However, when masses of 150 and 200 grams are placed at the 10-cm and 75-cm marks respectively, balance is obtained. The masses are then interchanged and balance is again obtained by shifting the pivot point to the 43-cm mark. Find the mass of the meter stick and the location of its center of gravity.

I know there is going to be 2 unknowns and x+7 is the new distance from the pivot to the c of mass of the meter stick, but I am so lost as to how to put this together, please help. thanks

  • physics -

    First balance
    ...150................M.........200......
    ...................^.....................
    ....10............50..x..........75......

    second balance
    ...150................M.........200......
    ..............^...........................
    ....10.......43.......x..........75......

    M=mass
    x=distance from left (=graduation)

  • physics -

    diagrams help, i had these almost exactly drawn out, but i do not know how to create the equations to solve for the unknowns. thanks

  • physics -

    First balance
    ...150................M.........200......
    ...................^.....................
    ....10.........50.....x..........75......

    second balance
    ...200................M.........150......
    ..............^...........................
    ....10....43..........x..........75......

    third line: distance from left
    M=mass
    x=distance from left (=graduation)

    For first balance, take moments about pivot point (=50), clockwise = positive, counterwise = negative:

    -150(50-10)+M(x-50)+200(75-50)=0
    For second balance:
    -200(43-10)+M(x-43)+150(75-43)=0

    Solve for M and Mx.
    Divide Mx by M to get x.
    I get
    M=800/7=114.3
    x=58.75 cm from zero.

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