A mineshaft has an elevator hung from a single steel-wire cable of diameter 2.5 cm. Young's modulus of the cable is 10×1010N/m2. When the cable is fully extended, the end of the cable is 500 m below the support. How much does the fully extended cable stretch when 1400 kg of ore is loaded into the elevator?
3.6
To calculate the stretch in the cable when ore is loaded onto the elevator, we need to use Hooke's Law, which states that the strain in a material is directly proportional to the stress applied to it.
The formula for calculating the stretch in a cable due to a point load is given by:
ΔL = (F * L) / (A * E)
Where:
ΔL = Change in length or stretch in the cable
F = Force or weight applied to the cable
L = Original length of the cable
A = Cross-sectional area of the cable
E = Young's modulus of the cable material
Now, let's calculate the stretch in the cable step by step:
Step 1: Calculate the cross-sectional area of the cable.
The diameter of the cable is 2.5 cm. The radius (r) is half the diameter. Therefore, r = 2.5 cm / 2 = 1.25 cm = 0.0125 m
The cross-sectional area (A) of the cable is given by the formula A = π * r^2.
So, A = 3.14 * (0.0125 m)^2 = 0.000490625 m^2
Step 2: Calculate the weight or force applied to the cable.
The force applied to the cable is the weight of the ore, which is 1400 kg, multiplied by the acceleration due to gravity, which is approximately 9.8 m/s^2.
So, F = 1400 kg * 9.8 m/s^2 = 13720 N
Step 3: Calculate the original length of the cable.
The difference in height between the end of the cable and the support is given as 500 m.
So, the original length (L) of the cable is 500 m.
Step 4: Calculate the stretch or change in length (ΔL) using the formula ΔL = (F * L) / (A * E).
ΔL = (13720 N * 500 m) / (0.000490625 m^2 * 10×10^10 N/m^2)
ΔL = (6860000 N*m) / (0.000490625 m^2 * 10×10^10 N/m^2)
ΔL = (6860000 N*m) / (0.00000490625 N/m^2)
ΔL = 1.39673 × 10^12 m
So, the fully extended cable will stretch by approximately 1.39673 × 10^12 meters when 1400 kg of ore is loaded onto the elevator.