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Chapter 1: Q41E (page 360)

A steel cable with cross-sectional area 3.00 cm2 has an elastic limit of 2.40 x 108 Pa. Find the maximum upward acceleration that can be given a 1200-kg elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

Short Answer

Expert verified

The maximum upward acceleration that can be given in a 1200 kg elevator supported by the cable is10.2m/s2

Step by step solution

01

The given data

The elastic limit of the cable =2.40脳108Pa

The cross-sectional area of steel cable, A=3m2

Mass of elevator, m = 1200 kg

02

Formula used

Stress of wire 蠁=F鈯�A

Where F鈯�is the tensile force applied to an object (tension in the cable), A is the cross-sectional area

Net force, 鈭�F=ma

Where m is the mass of the object and a is the acceleration

03

Find the tension in the cable

Since stress cannot exceed one-third of the elastic limit, so stress

蠁=132.4脳108Pa=0.8脳108Pa

Now, tension F鈯�=础蠁

F鈯�=3脳10-4m20.8脳108Pa=2.4脳104N

Hence,F鈯�=2.4脳104N

04

Find the acceleration

Since net force is 鈭�F=ma

So

F鈯�-mg=ma a=F鈯�m-g

Hence acceleration is

a=2.4脳104N1200kg鈭�9.8m/s2=10.2m/s2

Therefore, the maximum upward acceleration that can be given in a 1200-kg elevator supported by the cable is 10.2m/s2

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