   Chapter 9, Problem 69P

Chapter
Section
Textbook Problem

For safety in climbing, a mountaineer uses a nylon rope that is 50. m long and 1.0 cm in diameter . When supporting a 90.-kg climber, the rope clongates 1.6 m. Find its Young’s modulus.

To determine
The Young’s modulus of the nylon rope.

Explanation
Yong’s modulus is defined as Y=stress/strain=(F/A)/(ΔL/L0) and the cross-sectional area of the rope is A=πr2=π(d/2)2 . The tension in the rope is equal to the weight of the climber that is F=W=mg

Given info: The diameter of the rope is 1.0cm , length of the rope is 50m , mass of the climber is 90kg , the elongation of the nylon rope is 1.6m , and acceleration due to gravity is 9.80m/s2 .

The formula for the Young’s modulus of the rope is,

Y=4mgL0πd2ΔL

• m is mass of the climber.
• g is acceleration due to gravity.
• L0 is length of the rope.
• d is diameter of the rope.
• ΔL is elongation of the rope.

Substitute 90kg for m , 9

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