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 2.5×10-3cm2in 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:3.1×10-3

Lower wire: 2.0×10-3

(b) Upper wire: 1.6mm

Lower wire: 1.0mm

Step by step solution

01

Given information

Length:l0=0.50m,

Area:localid="1668006176532" A=2.5×10-3cm2,

Aluminum sphere mass: localid="1668006194126" mA=6kg,

Brass cube mass: localid="1668006203126" mB=10kg.

02

Concept/Formula used

Y=l0Fl

Where, Y is Young’s modulus,l0is length of muscle, F is muscle force, A is cross-sectional area and 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(ευ)=stressY=T1AYευ=157N(2.5×10-7m2)(2×1011Pa)=3.1×10-3

Strain in lower wire

strain(εL)=stressY=T2AYεL=98N(2.5×10-7m2)(2×1011Pa)=2.0×10-3

05

(b) Calculation for elongation in wires

(b) Elongation in upper wire

lu=l0×εu=(0.50m)(3.1×10-3)=1.6×10-3m=1.6mm

Elongation in Lower wire

lL=l0×εL=(0.50m)(2.0×10-3)=1.0×10-3m=1.0mm

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