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 equation, which relates the force applied to a spring or string to the amount it stretches or compresses.
Hooke's law equation is given by:
F = k * ΔL
Where:
F is the force applied (weight of the load)
k is the spring constant (also known as the Young's modulus or the modulus of elasticity)
ΔL is the change in length (the amount it stretches)
In this case, the force applied is the weight of the load, which can be calculated using the equation:
F = m * g
Where:
m is the mass of the load
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Let's calculate the force applied (F):
m = 68.9 kg (mass of the load)
g = 9.8 m/s^2 (acceleration due to gravity)
F = m * g
F = 68.9 kg * 9.8 m/s^2
F = 674.62 N
Now, let's calculate the change in length (ΔL):
ΔL = 0.0101 m (the maximum stretch allowed)
Now we can substitute the values of F and ΔL into Hooke's law equation to find the spring constant (k):
F = k * ΔL
674.62 N = k * 0.0101 m
To find the minimum diameter, we need to calculate the spring constant (k) of the nylon string using its Young's modulus (Ynylon) and the formula for the spring constant:
k = Ynylon * A / L
Where:
Ynylon is the Young's modulus of nylon (given as 3.51 * 10^8 N/m^2)
A is the cross-sectional area of the string
L is the length of the string
To find the cross-sectional area (A), we can use the diameter (D) of the string:
A = (π/4) * D^2
Let's substitute the values into the formula:
674.62 N = (3.51 * 10^8 N/m^2) * (π/4) * D^2 / 51.5 m
Now we can solve for the diameter (D):
D^2 = (674.62 N * 4 * 51.5 m) / (3.51 * 10^8 N/m^2 * π)
D^2 ≈ 0.000012814
Taking the square root of both sides gives:
D ≈ 0.003582 m
Finally, we convert the diameter from meters to centimeters:
D ≈ 0.003582 m * 100 cm/m
D ≈ 0.3582 cm
The minimum diameter of the nylon string that will stretch no more than 1.01 cm when a load of 68.9 kg is suspended from its lower end is approximately 0.3582 cm.
To find the minimum diameter of the nylon string, we need to consider the elastic deformation of the string. The formula that relates the stretch (ΔL) of a string to its length (L), diameter (D), material properties, and the applied load (F) is:
ΔL = (F * L) / (π * D^2 * Y)
where:
ΔL is the stretch of the string,
F is the applied load,
L is the length of the string,
D is the diameter of the string, and
Y is the Young's modulus of the material.
In this case, we are given:
F = 68.9 kg (the weight of the load),
L = 51.5 m (the length of the string),
Ynylon = 3.51 · 10^8 N/m^2 (the Young's modulus of nylon), and
ΔL = 1.01 cm (the maximum allowable stretch).
First, we need to convert the mass (F) into force (N) using the formula: F = m * g, where g is the acceleration due to gravity (approximately 9.81 m/s^2). Therefore, F = 68.9 kg * 9.81 m/s^2.
Now, let's rearrange the formula to solve for the diameter (D) of the string:
D = √((F * L) / (π * ΔL * Y))
Substituting the given values into the formula:
D = √((68.9 kg * 9.81 m/s^2 * 51.5 m) / (π * 1.01 cm * 10^-2 m/cm * 3.51 * 10^8 N/m^2))
Calculating this expression will give you the minimum diameter (D) of the nylon string.
Hookes Law:
elongation= force/Y * length/area
.00101=68.9*9.8/3.51E8 * 51.5/area
solve for area, then solve for diameter