A lift cable which is 50m long ,is made from high tensile steel of Young Modulus 2.0x 10^11 Pa. the cable is made from 100 wires each of radius 2.0mm . Calculate the extra extension in the cable when 10 passengers pile into the lift ,the total mass of the passenger is 500kg .

I think I would use Hooke's law here. Note the wires are in parallel, so the effective

Modulus is 100*2*E11 Nt/m^2, so

deltaL=FL/AE where F is force, L is length, A is area of each wire, and E is the modulus of Elasticity.

deltaL=500kg*9.8N/kg*50m/PI*(2E-3)^2*200E11N/m^2

check that.

To calculate the extra extension in the cable when the passengers pile into the lift, we can use Hooke's Law, which states that the extension of a spring or cable is directly proportional to the applied force.

First, let's calculate the force applied by the 10 passengers. The total mass of the passengers is 500 kg, and we can find the force using the equation F = mass x gravity.

F = 500 kg x 9.8 m/s^2
F = 4900 N

Next, let's find the cross-sectional area of the cable. The cable is made from 100 wires, and each wire has a radius of 2.0 mm. The total cross-sectional area of the cable can be calculated using the equation:

A = π x r^2
A = π x (2.0 mm)^2

Now, we need to convert the radius to meters:

r = 2.0 mm / 1000
r = 0.002 m

Substituting the values into the equation:

A = π x (0.002 m)^2

Next, we can calculate the stress applied to the cable using the equation:

Stress = Force / Area

Stress = 4900 N / (π x (0.002 m)^2)

Now, we can calculate the extension in the cable using Hooke's Law. The formula for elongation due to stress is:

Elongation = (Stress x Length) / (Young's Modulus)

The length of the cable is given as 50 m, and the Young Modulus of high tensile steel is 2.0 x 10^11 Pa. Substituting the values into the equation:

Elongation = (Stress x Length) / Young's Modulus
Elongation = (Stress x 50 m) / (2.0 x 10^11 Pa)

Now, you can plug in the value of stress that we calculated earlier to find the elongation of the cable.