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Question: A copper wire of length 2.2 m and a steel wire of length 1.6 m. Both of diameter 3.0 mm, are connect...

A copper wire of length 2.2 m and a steel wire of length 1.6 m. Both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.

Explanation

Solution

The formula for Young's modulus for the elongation of a wire is given as Y=F/AΔL/LY = \dfrac{{F/A}}{{\Delta L/L}}, where A is the area of cross section, L is the length of the wire, F is the force applied or the load. Young's modulus for copper and steel is 1.1×1011Nm−21.1 \times {10^{11}}N{m^{ - 2}} and 2.0×1011Nm−22.0 \times {10^{11}}N{m^{ - 2}} respectively. Substituting these values we get the value of load (F).

Complete step by step answer:
Given, length of the copper wire, LC=2.2m{L_C} = 2.2m. The length of the Steel wire, LS=1.6m{L_S} = 1.6m. Diameter of both the wires, d=3.0mmd = 3.0mm.
The radius of both the wires,
r=d2 ⇒r=32 ⇒r=1.5mm ⇒r=1.5×10−3mr = \dfrac{d}{2}\\\ \Rightarrow r = \dfrac{3}{2}\\\ \Rightarrow r = 1.5mm \\\ \Rightarrow r = 1.5 \times {10^{ - 3}}m.
So we get the area of cross section of the wires as,
A=πr2 ⇒A=3.14×(1.5×10−3)2 ⇒A=7.065×10−6mA = \pi {r^2}\\\ \Rightarrow A = 3.14 \times {\left( {1.5 \times {{10}^{ - 3}}} \right)^2}\\\ \Rightarrow A = 7.065 \times {10^{ - 6}}m.

Young's modulus of steel is 2.0×1011Nm−22.0 \times {10^{11}}N{m^{ - 2}}. Young's modulus of copper is 1.1×1011Nm−21.1 \times {10^{11}}N{m^{ - 2}}. Young's modulus(Y) is defined as the ratio of stress to strain, Y=stressstrainY = \dfrac{{stress}}{{strain}}.

Where stress is force per unit area (stress=FA)\left( {stress = \dfrac{F}{A}} \right) and strain is the ratio of change in length to the original length(strain=ΔLL)\left( {strain = \dfrac{{\Delta L}}{L}} \right).
Therefore Young’s modulus is, Y=F/AΔL/LY = \dfrac{{F/A}}{{\Delta L/L}}. -----------(1)
Substituting the values for the Steel wire in equation (1)
2.0×1011=F7.065×10−6ΔLS1.6 ⇒2.0×1011=1.6F7.065×10−6ΔLS  2.0 \times {10^{11}} = \dfrac{{\dfrac{F}{{7.065 \times {{10}^{ - 6}}}}}}{{\dfrac{{\Delta {L_S}}}{{1.6}}}}\\\ \Rightarrow 2.0 \times {10^{11}} = \dfrac{{1.6F}}{{7.065 \times {{10}^{ - 6}}\Delta {L_S}}} \\\
⇒ΔLS=1.6F2.0×1011×7.065×10−6 ⇒ΔLS=0.1132×10−5F\Rightarrow \Delta {L_S} = \dfrac{{1.6F}}{{2.0 \times 1{0^{11}} \times 7.065 \times 1{0^{ - 6}}}}\\\ \Rightarrow\Delta {L_S} = 0.1132 \times {10^{ - 5}}F

Change in the length of Steel wire is 0.1132×10−5 \times {10^{ - 5}}F.
Substituting the values for copper wire in equation (1),
1.1×1011=F7.065×10−6ΔLC2.2 ⇒1.1×1011=2.2F7.065×10−6ΔLC1.1 \times {10^{11}} = \dfrac{{\dfrac{F}{{7.065 \times {{10}^{ - 6}}}}}}{{\dfrac{{\Delta {L_C}}}{{2.2}}}} \\\ \Rightarrow 1.1 \times {10^{11}}= \dfrac{{2.2F}}{{7.065 \times {{10}^{ - 6}}\Delta {L_C}}}
⇒ΔLC=2.2F1.1×1011×7.065×10−6 ⇒ΔLC=0.2831×10−5F\Rightarrow \Delta {L_C} = \dfrac{{2.2F}}{{1.1 \times 1{0^{11}} \times 7.065 \times 1{0^{ - 6}}}}\\\ \Rightarrow\Delta {L_C} = 0.2831 \times {10^{ - 5}}F.
Change in the length of copper wire is 0.2831×10−5 \times {10^{ - 5}}F.

Given the net change in the length of the wire is 0.70mm=0.7×10−3m = 0.7 \times {10^{ - 3}}m.
Net change in length,
ΔL=ΔLS+ΔLC ⇒0.7×10−3=(0.1132×10−5+0.2831×10−5)F  \Delta L = \Delta {L_S} + \Delta {L_C} \\\ \Rightarrow 0.7 \times 1{0^{ - 3}} = (0.1132 \times 1{0^{ - 5}} + 0.2831 \times 1{0^{ - 5}})F \\\
⇒F=0.3963×10−50.7×10−3 ⇒F=0.5661×10−2 ∴F=56.61N\Rightarrow F = \dfrac{{0.3963 \times {{10}^{ - 5}}}}{{0.7 \times {{10}^{ - 3}}}}\\\ \Rightarrow F = 0.5661 \times {10^{ - 2}}\\\ \therefore F = 56.61N

Therefore the force applied or the load is equal to 56.61N.

Note: Modulus of elasticity is the measure of the stress–strain relationship on the object. Modulus of elasticity is the prime feature in the calculation of the deformation response of concrete when stress is applied. Young’s Modulus (also referred to as the Elastic Modulus or Tensile Modulus), is a measure of mechanical properties of linear elastic solids like rods, wires.