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

To find the minimum diameter of the nylon string, we can use Hooke's law, which states that the force (F) applied to a spring or elastic material is directly proportional to the displacement or stretch (ΔL) it undergoes, and inversely proportional to its spring constant (k).

The formula for Hooke's law is: F = k * ΔL

In our case, the force applied is the weight of the load, which is given as 68.9 kg. Since weight is calculated as W = m * g (where m is mass and g is the acceleration due to gravity), we can find the force by multiplying the mass by the acceleration due to gravity.

W = m * g
F = 68.9 kg * 9.8 m/s^2

Next, we need to convert the stretch (1.01 cm) to meters.

ΔL = 1.01 cm * (1 m / 100 cm)

Now, we have all the necessary values to calculate the spring constant (k) using Hooke's law.

F = k * ΔL
68.9 kg * 9.8 m/s^2 = k * (1.01 cm * (1 m / 100 cm))

Let's solve for k:

k = (68.9 kg * 9.8 m/s^2) / (1.01 cm * (1 m / 100 cm))

Now, we can use the formula for the spring constant (k) of a cylindrical spring:

k = (Y * A) / L

where Y is the Young's modulus (elasticity) of the material, A is the cross-sectional area of the string, and L is the length of the string.

We are given Ynylon = 3.51·10^8 N/m^2 and L = 51.5 m. We need to find A, the cross-sectional area.

k = (Y * A) / L
A = (k * L) / Y
A = [(68.9 kg * 9.8 m/s^2) / (1.01 cm * (1 m / 100 cm))] * 51.5 m / (3.51·10^8 N/m^2)

Finally, we can find the minimum diameter (d) of the string using the formula for the area of a circle:

A = π * (d/2)^2

d = sqrt((4 * A) / π)

After substituting the value of A into the formula, we can calculate the minimum diameter of the nylon string.