Question

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

Answer 1

We need to find the relation which best connects the lengths of the wire, the elongation caused, the type of materials used and the force acting on them to solve this problem. The value of Young’s modulus can be used from the textbook.

Complete answer:
Any material can undergo a deformation in the form of elongation or compression when a force is applied to the material. A longitudinal elongation is attained by applying a force along the direction of the wire.
The ratio of the elongation of the wire to the original length of the wire is the strain experienced by the material. It is given as –
$Strain,\delta =\dfrac{\Delta l}{l}$
We know that the force experienced by the material per unit area is defined as the stress applied to it. The ratio of longitudinal stress applied to the longitudinal strain of the system gives Young's modulus.
The stress applied is given as –

& Strain,\delta =\dfrac{\Delta l}{l} \\\ & Stress,S=\dfrac{F}{A} \\\ & \Rightarrow S=\dfrac{F}{\pi {{r}^{2}}} \\\ \end{aligned}$$ The Young’s modulus is given by the relation – $$\begin{aligned} & Y=\dfrac{Stress}{Strain} \\\ & \Rightarrow Y=\dfrac{\dfrac{F}{\pi {{r}^{2}}}}{\dfrac{\Delta l}{l}} \\\ & \Rightarrow Y=\dfrac{Fl}{\pi {{r}^{2}}\Delta l} \\\ \end{aligned}$$ Let us consider the situation given to us. Two materials are applied to an equal force at the same time as shown below. We know that the total elongation experienced by the materials can be given as the sum of the individual elongations. This can be written as – $$\Delta l=\Delta {{l}_{copper}}+\Delta {{l}_{steel}}$$  Now, let us rewrite the Young’s modulus equation in terms of the elongation as – $$\begin{aligned} & Y=\dfrac{Fl}{\pi {{r}^{2}}\Delta l} \\\ & \Rightarrow \Delta l=\dfrac{Fl}{\pi {{r}^{2}}Y} \\\ \end{aligned}$$ Now, we know the Young’s modulus for both the materials as – $$\begin{aligned} & {{Y}_{c}}=1.3\times {{10}^{11}}N{{m}^{-2}} \\\ & {{Y}_{S}}=2.0\times {{10}^{11}}N{{m}^{-2}} \\\ \end{aligned}$$ So, we can find the individual elongations as – $$\begin{aligned} & \Delta {{l}_{C}}=\dfrac{{{F}_{C}}{{l}_{C}}}{\pi {{r}_{C}}^{2}{{Y}_{C}}} \\\ & \Rightarrow \Delta {{I}_{C}}=\dfrac{{{F}_{C}}(2.2)}{\pi {{(1.5\times {{10}^{-3}})}^{2}}1.3\times {{10}^{11}}} \\\ & \Rightarrow \Delta {{l}_{C}}=2.394\times {{10}^{-6}}{{F}_{C}}m \\\ \end{aligned}$$ $$\begin{aligned} & \Delta {{l}_{S}}=\dfrac{{{F}_{S}}{{l}_{S}}}{\pi {{r}_{S}}^{2}{{Y}_{S}}} \\\ & \Rightarrow \Delta {{I}_{C}}=\dfrac{{{F}_{S}}(1.6)}{\pi {{(1.5\times {{10}^{-3}})}^{2}}2.0\times {{10}^{11}}} \\\ & \Rightarrow \Delta {{l}_{C}}=1.1318\times {{10}^{-6}}{{F}_{S}}m \\\ \end{aligned}$$ Now, we can add these two to get the force applied for this elongation as – $$\begin{aligned} & \Delta l=2.394\times {{10}^{-6}}{{F}_{C}}+1.1318\times {{10}^{-6}}{{F}_{S}} \\\ & \text{but,} \\\ & {{F}_{C}}={{F}_{S}}=F \\\ & \Rightarrow \Delta l=3.5258F \\\ & \Rightarrow 7\times {{10}^{-4}}=3.5258\times {{10}^{-6}}F \\\ & \Rightarrow F=\dfrac{7\times {{10}^{-4}}}{3.5258\times {{10}^{-6}}} \\\ & \therefore F=1.98\times {{10}^{2}}N \\\ \end{aligned}$$ The force applied on the system is approximately 200N. **Note:** The Young’s modulus is applicable only for forces that are applied along the direction of the wire and not in any other direction. We have the Shear stress which applies for the tangential forces applied on a body which can be almost as three times as much as Young’s modulus.