   Chapter 9, Problem 66P

Chapter
Section
Textbook Problem

A 25.0-m long steel cable with a cross-sectional area of 2.03 times 10−3 m2 is used to suspend a 3.50 × 103-kg container. By how much will the cable stretch once bearing the load?

To determine

The stretch of a steel cable once bearing the load.

Explanation

Given info: The cross-sectional area of the steel cable is 2.03×103m2, Young’s modulus of the wire is 20.0×1010N/m2, length of the cabe is 25.0m, mass of the load is 3.50×103kg, and acceleration due to gravity is 9.80m/s2.

Yong’s modulus is defined as Y=stress/strain=(F/A)/(ΔL/L0) and from which, the increase in length of that wire is calculated as ΔL=(FL0)/(AY) where the force is tension force due to loads weight and it is given by F=mg.

The formula for the increase in length of the wire is,

ΔL=(mg)L0AY

• m is mass of the load
• g is acceleration due to gravity
• L0 is length of the wire
• A is cross-sectional area of the wire
• Y is Young’s modulus of the wire

Substitute 3.50×103kg for m, 9

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