Find the minimum diameter of a 54.3-m-long nylon string that will stretch no more than 1.09 cm when a load of 71.1 kg is suspended from its lower end. Assume that Ynylon = 3.51⋅
109 N/m2.
To solve this problem, we can use the formula for elongation of a string:
Elongation = (F * L) / (A * Y)
where
Elongation = 1.09 cm = 0.0109 m (converted to meters)
F = force applied = 71.1 kg * 9.81 m/s^2 (force due to weight)
L = length of the string = 54.3 m
A = cross-sectional area of the string
Y = Young's modulus for nylon = 3.51 * 10^9 N/m^2
First, we'll find the force applied:
Force = F = 71.1 kg * 9.81 m/s^2 = 698.031 N
Next, we'll rearrange the formula to find the cross-sectional area:
A = (F * L) / (Elongation * Y)
A = (698.031 N * 54.3 m) / (0.0109 m * 3.51 * 10^9 N/m^2)
A ≈ 1.216 * 10^-6 m^2
Now, we'll assume the string is circular and find the diameter. The formula for the cross-sectional area of a circle is:
A = π * (d/2)^2
We'll rearrange the formula to solve for the diameter:
d = 2 * sqrt(A / π)
d = 2 * sqrt(1.216 * 10^-6 m^2 / 3.14159265...)
d ≈ 0.00622 m = 6.22 mm
So, the minimum diameter for the nylon string is approximately 6.22 mm.