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Homework Help: Support Column Compression

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data

    A support column is compressed 2.46 x 10^ -4m under a weight of 5.42 X 10^5N. how much is the column compressed under a weight of 4.8 X10^6N?

    2. Relevant equations


    3. The attempt at a solution

    By using that equation, I get 542000 = .000246K

    Solve for K, and I get a huge number, 2,203,252,032.52

    Then by using that number as my new K, and solving for the compressed distance with the new weight, I get 0.0022 m.

    Is that the right answer?

    Similarly, I figured I could just make a ratio proportion of:


    ------------- = ----------------


    Which also gives me an answer of 0.0022 m (2.2 X 10^ -3). Is this correct?
    Last edited: Dec 7, 2008
  2. jcsd
  3. Dec 7, 2008 #2


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    Gold Member

    Yes, either method gives you the correct solution, as long as the column is still behaving elastically in accordance with Hooke's law.
  4. Dec 7, 2008 #3
    Thanks Jay, when I first did this problem, I was looking at the answer of 2.2e-3 and shaking my head going how can this thing compress less when there is a heavier weight on it? However it wasn't after I was typing it in the computer that I realized that the first compression is 2.46e-4, and I went, duh. =)

    Thanks for letting me know i'm correct.
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