Question

Steel wire rope is used to lift a heavy object. We use a 3.1m steel wire that
is 6.0mm in diameter and lift a 1700kg object. Then, the wire elongates
0.17m. Calculate the Young’s modulus for the rope material.

Answer 1

Answer:

Young's modulus for the rope material is 20.8 MPa.    

 

Explanation:

The Young's modulus is given by:

$E = \frac{FL_{0}}{A\Delta L}$

Where:

F: is the force applied on the wire

L₀: is the initial length of the wire = 3.1 m

A: is the cross-section area of the wire

ΔL: is the change in the length = 0.17 m

The cross-section area of the wire is given by the area of a circle:

$A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2}$

Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:

$F = T_{w} = W_{o}$

Where:

$T_{w}$: is the tension of the wire

$W_{o}$: is the weigh of the object = mg

m: is the mass of the object = 1700 kg

g: is the acceleration due to gravity = 9.81 m/s²

$F = mg = 1700 kg*9.81 m/s^{2} = 16677 N$

Hence, the Young's modulus is:

$E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa$          

Therefore, Young's modulus for the rope material is 20.8 MPa.                

I hope it helps you!