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  • Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Q5   Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires. 

 

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Tension in the steel wire is F1

\\F_{1}=(4+6)\times 9.8\\ F_{1}=98N

Length of steel wire l1 = 1.5 m

The diameter of the steel wire, d = 0.25 cm

\\A=\pi \left ( \frac{d}{2} \right )^{2}\\ A=\pi \times \left ( \frac{0.25\times 10^{-2}}{2} \right )^{2}\\ A=4.9\times 10^{-6}m^{2}

Area od the steel wire, A=4.9\times 10^{-6}\ m^{2}

Let the elongation in the steel wire be \Delta l_{1}

Young's Modulus of steel, Y1 = 2\times 10^{11}Nm^{-2}

\\Y_{1}=\frac{F_{1}l_{1}}{\Delta l_{1}A }\\ \Delta l_{1}=\frac{F_{1}l_{1}}{Y_{1}A}\\ \Delta l_{1}=\frac{98\times 1.5}{2\times 10^{11}\times 4.9\times 10^{-6}}\\ \Delta l_{1}=1.5\times 10^{-4}\ m

Tension in the Brass wire is F2

\\F_{2}=(6)\times 9.8\\ F_{2}=58.8N

Length of Brass wire l2 = 1.5 m

Area od the brass wire, A=4.9\times 10^{-6}\ m^{2}

Let the elongation in the steel wire be \Delta l_{2}

Young's Modulus of steel, Y2 = 0.91\times 10^{11}Nm^{-2}

\\ \Delta l_{2}=\frac{F_{2}l_{2}}{Y_{2}A}\\ \Delta l_{2}=\frac{58.8\times 1}{0.91\times 10^{11}\times 4.9\times 10^{-6}}\\ \Delta l_{2}=1.32\times 10^{-4}\ m

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