Hookes Law:
elongation= force/Y * length/area
.00101=68.9*9.8/3.51E8 * 51.5/area
solve for area, then solve for diameter
Find the minimum diameter of a 51.5-m-long nylon string that will stretch no more than 1.01 cm when a load of 68.9 kg is suspended from its lower end. Assume that Ynylon = 3.51·108 N/m2
1 answer
1 answer