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A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 700 N is applied to each end of the wire. What minimum diameter is required for the wire?

Short Answer

Expert verified

Minimum diameter is : d = 1.9 mm .

Step by step solution

01

Given information:

Force: F= 700 N ,

Elongation: ∆l=0.25cm,

length: l0=2m.

02

Concept/Formula used:

Y=l0F∆l

Where, Y is Young’s modulus, l0 is length of muscle, F is muscle force, A is cross-sectional area and ∆lis elongation.

03

Calculation for diameter:

Y=I0F∆lA=I0F∆l

For steel: Y=2.0×1011Pa

A=(2.0m)(700N)(2.0×1011Pa)(0.25×10-2m)=2.8×10-6m2A=πr2r=Aπ=2.8×10-6m2π=9.44×10-4md=2r=2×9.44×10-4m=1.9mm

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