# Support Column Compression

1. Dec 7, 2008

### kkmonte

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

F=KX

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.

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

.000246.................X

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

542,000.............4,800,000

Which also gives me an answer of 0.0022 m (2.2 X 10^ -3). Is this correct?

Last edited: Dec 7, 2008
2. Dec 7, 2008

### PhanthomJay

Yes, either method gives you the correct solution, as long as the column is still behaving elastically in accordance with Hooke's law.

3. Dec 7, 2008

### kkmonte

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.
Ken