ch12-p082 - Since the mass does not change, either, then...

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82. The assumption stated in the problem (that the density does not change) is not meant to be realistic; those who are familiar with Poisson’s ratio (and other topics related to the strengths of materials) might wish to think of this problem as treating a fictitious material (which happens to have the same value of E as aluminum, given in Table 12-1) whose density does not significantly change during stretching.
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Unformatted text preview: Since the mass does not change, either, then the constant-density assumption implies the volume (which is the circular area times its length) stays the same: ( r 2 L ) new = ( r 2 L ) old L = L [ (1000/999.9) 2 1 ] . Now, Eq. 12-23 gives F = r 2 E L / L = r 2 (7.0 x 10 9 N/m 2 ) [ (1000/999.9) 2 1 ] . Using either the new or old value for r gives the answer F = 44 N....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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