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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鈥尘is8.0脳10鈭�4m .

b) Cable stretch when elevator hung by 362鈥尘is0.024鈥尘 .

Step by step solution

01

Understanding the given information

Mass of system is670鈥块驳.

The diameter of the wire is 2.5鈥塩尘(1鈥�m100鈥�cm)=0.025鈥�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,螖L=(wAY)L

Area of wire,A=蟺r2

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鈥�kg脳9.8鈥�m/s2=6566鈥塏

The area of the wire is

A=蟺r2=3.14脳(0.0125鈥�m)2=4.91脳10鈭�4m2

螖L=(6566鈥�N(4.91脳10鈭�4鈥�m2)(2.0脳1011鈥�N/m2))(12鈥�m)=8.0脳10鈭�4鈥尘
04

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

螖L=(6566鈥�N(4.91脳10鈭�4鈥�m2)(2.0脳1011鈥�N/m2))(362鈥�m)=0.024鈥尘

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