Compression of an object under it's own weight.

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  • Thread starter LordGfcd
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  • #1
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Main Question or Discussion Point

Let's first consider a cube side length a, mass m, Young's modulus of the block is E. How do we calculate the decrease of the height of the center of mass of that cube ?
 

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  • #2
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What are your thoughts so far on how to approach this problem?
 
  • #3
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Actually I solved this problem assuming the cube is deform before the normal force make balance with gravity. I eventually find an acceptable result (a=10cm,m=1kg,E=10^7 Pa) : 2,27 μm. But I still think my approaching is wrong because I didn't consider the normal force (action equal minus reaction ofcourse). So I must consider the normal force too, but I don't know how . And, if the normal force balance with gravity, isn't the cube will stop deforming ?
 
  • #4
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The differential force balance on the section of the cube between z and z + ##\Delta z## (z is measured downward from the top) is $$a^2\frac{d\sigma}{dz}=\rho g a^2$$ where A is the cross sectional area, ##\sigma## is the compressive stress, and ##\rho## is the density of the material ##m/(a^2L)##. Do you see how this result is derived?
 
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  • #5
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Thank you very much, I was completely wrong with my argument.
 

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