Answer:
Young's modulus for the rope material is 20.8 MPa.
Explanation:
The Young's modulus is given by:
[tex] E = \frac{FL_{0}}{A\Delta L} [/tex]
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:
[tex] A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2} [/tex]
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:
[tex] F = T_{w} = W_{o} [/tex]
Where:
[tex] T_{w} [/tex]: is the tension of the wire
[tex]W_{o} [/tex]: 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
[tex] F = mg = 1700 kg*9.81 m/s^{2} = 16677 N [/tex]
Hence, the Young's modulus is:
[tex] E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa [/tex]
Therefore, Young's modulus for the rope material is 20.8 MPa.
I hope it helps you!
