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Chapter 1: Q29E (page 360)

In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5脳10-3cm2in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. For each wire, compute (a) the tensile strain and (b) the elongation.

Short Answer

Expert verified

(a) Upper wire: 3.1脳10-3

Lower wire: 2.0脳10-3

(b) Upper wire: 1.6 mm

Lower wire: 1.0 mm

Step by step solution

01

Given information

Length: l0=0.50m,

Area: A=2.5脳10-3cm2,

Aluminum sphere mass: mA=6kg,

Brass cube mass: mB=10kg.

02

Concept/Formula used

Y=l0F鈭�l

Where, Y is Young鈥檚 modulus, l0 is length of muscle, F is muscle force, A is cross-sectional area and 鈭�lis elongation.

03

Tension calculation in wires

Tension in the below steel wire

T2=mBg=(10kg)(9.80)=98N

Tension in the above steel wire

T1=mAg+T2=(6kg)(9.80)+98N=157N

04

(a) Calculation for Tensile strain

Strain in upper wire

Y=stressstrainstrain(蔚蠀)=stressY=T1AY蔚蠀=157N(2.5脳10-7m2)(2脳1011Pa)=3.1脳10-3

Strain in lower wire

strain(蔚L)=stressY=T2AY蔚L=98N(2.5脳10-7m2)(2脳1011Pa)=2.0脳10-3

05

(b) Calculation for elongation in wires

(b) Elongation in upper wire

鈭�lu=l0脳蔚u=(0.50m)(3.1脳10-3)=1.6脳10-3m=1.6mm

Elongation in Lower wire

鈭�lL=l0脳蔚L=(0.50m)(2.0脳10-3)=1.0脳10-3m=1.0mm

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