Posted by richard on Thursday, May 6, 2010 at 7:59pm.
You want the amount of stretching to decrease from 0.16 cm to 0.10 cm. That is 5/8 of the orginal stetching. To make that happen, the stress nust be decreased by a factor of 5/8. Since the load is the same, the area must increase by a factor of 8/5. The diamenter must increase by a factor sqrt(8/5) = 1.265
Now we have to figure out the original wire diameter -- the one that stretched 0.16 cm.
The wire that stretched 0.16 cm had an area A given by
F/A = Y * (deltaL)/L = 20*10^10*0.16*10^-2/4.9 = 6.53*10^7 N/m^2
A = 6.13*10^-6 m^2
D = sqrt[4A/pi} = 2.79*10^-3 m = 2.79 mm
The new wire diamneter is 1.265 times 2.79 mm, or about 3.5 mm
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