Physics hard...
posted by richard on .
Young's Modulus (Y) 20 X 1010 N/m2
Shear Modulus (S) 8.1 X 1010 N/m2
Bulk Modulus (B) 16 X 1010 N/m2
The table to the above represents various properties of steel. You have steel wire 4.9 meters in length that stretches 0.16 cm when subjected to a force of 400 N.
What would the diameter of the wire be if you wanted the wire to stretch 0.06 cm less when subjected to this same force?

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