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Let's tackle this problem step by step. We're given the force applied, the dimensions of the copper piece, and the shear modulus of elasticity.

First, let's calculate the cross-sectional area of the copper piece using the given dimensions:

Area (A) = length × width = 15.2 mm × 19.1 mm = 290.72 mm²

Now, we can use Hooke's Law to find the strain (ε), which relates stress to strain via the modulus of elasticity:

Hooke's Law: Stress (σ) = Shear Modulus (G) × Strain (ε)

We know that stress (σ) is force (F) divided by area (A), so:

σ = F / A

Plugging in the values, we get:

σ = 44500 N / 290.72 mm²

Let's convert the area to square meters for consistency:

1 mm² = 1 × 10^(-6) m²

So, 290.72 mm² = 290.72 × 10^(-6) m²

Now, let's calculate stress:

σ = 44500 N / (290.72 × 10^(-6) m²)

Now, we can use Hooke's Law to find strain:

ε = σ / G

Plugging in the values:

ε = (44500 N / (290.72 × 10^(-6) m²)) / (42 × 10^9 N/m²)

Calculating this gives us the resulting strain.

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When dealing with a situation like this, it's important to first understand the equilibrium condition of the bar. Since it's symmetrically supported by three wires, each wire carries a portion of the weight of the bar.

To ensure each wire experiences the same tension, we must consider the properties of each wire material, copper and iron, as they affect the tension they can withstand.

Let's denote the tension in each wire as TT, and the lengths of the wires as LL. The tension in each wire can be given by T=m⋅g3T=3m⋅g, where mm is the mass of the bar and gg is the acceleration due to gravity.

Now, let's consider the material properties. The tension in a wire is directly proportional to its cross-sectional area (AA) and inversely proportional to its length (LL).

For a given tension, the equation becomes:

T=FAT=AF

Where FF is the force (weight of the bar) and AA is the cross-sectional area.

Since each wire has the same tension, we can set up the following equation:

T=FAcopper=FAironT=AcopperF=AironF

Given that the lengths of all wires are the same and the tension is constant, the key factor differentiating the wires is their material properties, specifically their Young's Modulus and their density.

Young's Modulus (EE) relates stress (σσ) to strain (εε) in the wire material:

σ=E⋅εσ=E⋅ε

For a wire under tension, σ=FAσ=AF and ε=ΔLLε=LΔL, where ΔLΔL is the change in length and LL is the original length.

Now, let's consider the density (ρρ) of each material. Density multiplied by volume (VV) gives mass (mm). For a cylindrical wire, volume is given by V=A⋅LV=A⋅L.

With these considerations, we can set up ratios involving Young's Modulus, density, and tension to solve for the ratios of their diameters. Let's denote the diameter of the copper wire as dcopperdcopper and the diameter of the iron wire as dirondiron.

TAcopper=TAironAcopperT=AironT

Tπ(dcopper2)2=Tπ(diron2)2π(2dcopper)2T=π(2diron)2T

(dcopper2)2(diron2)2=ρironρcopper×EcopperEiron(2diron)2(2dcopper)2=ρcopperρiron×EironEcopper

(dcopperdiron)2=ρironρcopper×EcopperEiron(dirondcopper)2=ρcopperρiron×EironEcopper

dcopperdiron=ρironρcopper×EcopperEirondirondcopper=ρcopperρiron×EironEcopper