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.
51.5(9.8)/(0.0101)3.14(3.51*108)
To find the minimum diameter of the nylon string, we need to use Hooke's law, which states that the elongation of a material is directly proportional to the force applied to it.
The formula for elongation (ΔL) is given as:
ΔL = (F * L) / (A * Y)
Where:
ΔL = Elongation (in meters)
F = Force applied (in Newtons)
L = Original length of the string (in meters)
A = Cross-sectional area of the string (in square meters)
Y = Young's modulus or elastic modulus of the material (in Pascal)
In this case, we are given the following values:
F = 68.9 kg * 9.8 m/s^2 (Converting kg to Newtons using acceleration due to gravity)
L = 51.5 m
ΔL = 0.0101 m
Ynylon = 3.51 * 10^8 N/m^2
To find the minimum diameter, we need to solve for A in the formula:
ΔL = (F * L) / (A * Y)
Rearrange the formula to solve for A:
A = (F * L) / (ΔL * Y)
Now, substitute the given values into the formula:
A = (68.9 kg * 9.8 m/s^2 * 51.5 m) / (0.0101 m * 3.51 * 10^8 N/m^2)
Calculate the value of A using a calculator:
A = 0.0001086 m^2 or 108.6 mm^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
Rearrange the formula to solve for d:
d = √((4 * A) / π)
Substitute the value of A into the formula:
d = √((4 * 0.0001086 m^2) / π)
Calculate the value of d using a calculator:
d ≈ 0.0111 m or 11.1 mm
Therefore, the minimum diameter of the nylon string should be approximately 11.1 mm to ensure it stretches no more than 1.01 cm when a load of 68.9 kg is suspended from its lower end.