The applied weight is
M g = 675.2 N
The maximum allowable strain is
deltaL/L = 0.0101/51.5 = 1.96*10^-4
The maximum allowable stress is
Ynylon*(strain) = 6.88*10^4 N/m^2
That equals 675.2 N divided by the minimum allowable area. Solve for that area and the string 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