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Chapter 12: Q62P (page 351)

A mine elevator is supported by a single steel cable2.5cmin diameter. The total mass of the elevator cage and occupants is670kg .By how much does the cable stretch when the elevator hangs by (a)12m of cableand (b) 362mof cable?(Neglect the mass of the cable.)

Short Answer

Expert verified

a) Cable stretch when elevator hung by 12 mis8.0×10−4m .

b) Cable stretch when elevator hung by 362 mis0.024 m .

Step by step solution

01

Understanding the given information

Mass of system is670 Kg.

The diameter of the wire is 2.5 cm(1 m100 cm)=0.025 m .

02

Concept and formula used in the given question

Using the given mass, you can find the weight of the elevator. Using Hooke's law formula, you can find the distance by which the cable is stretched. The formulas used are given below.

Hooke's law,ΔL=(wAY)L

Area of wire,A=πr2

W=mg

03

Calculation of how much the cable stretch when the elevator hangs by  12 m of cable

First, you found the weight of load from a given mass is

W=mg=672 kg×9.8 m/s2=6566 N

The area of the wire is

A=πr2=3.14×(0.0125 m)2=4.91×10−4m2

ΔL=(6566 N(4.91×10−4 m2)(2.0×1011 N/m2))(12 m)=8.0×10−4 m
04

(b) Calculation ofby how much the cable stretch when the elevator hangs by362 m   of cable

ΔL=(6566 N(4.91×10−4 m2)(2.0×1011 N/m2))(362 m)=0.024 m

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