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Center of Gravity problem

  1. Feb 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A partially loaded plane weighs 115,000 pounds. A famous actor's steamer trunk, weighing 300 pounds, is put into the rear cargo hold 250 inches behind the center of gravity. How much does this extra hold shift the center of gravity.

    2. Relevant equations
    W1 = W x (L2 / L)

    W2 = W x (L1 / L)

    W = 115300
    W1 = 115000
    W2 = 300
    L = 250

    3. The attempt at a solution
    I did this:
    W2 = W x (L1 / L)
    so....
    300 = 115300 x (L1 / 250)
    so...
    300 = 461.2L1
    divide both sides by 461.2

    L1 = .6504770165

    The center of gravity of the plane shifts .6504770165 inches toward the rear.


    Your input is encouraged.
     
  2. jcsd
  3. Feb 26, 2015 #2

    PhanthomJay

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    Looks good, but get rid of some of those figures after the decimal point, maybe just say 0.65 inches
     
  4. Feb 26, 2015 #3
    Thank you for the input. Would you mind taking a look at the more advanced center of gravity problem I posted. It's really frustrating me at the moment.
     
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