Question

A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of the stresses developed in the two wires .

Answer 1

Given:
Young's modulus of steel = 2 × 10¹¹ N m⁻²
Young's modulus of copper = 1.3 × 10 11 N m⁻²
Both wires are of equal length and equal cross-sectional area. Also, equal tension is applied on them.
 As per the question :

\[L_{\text{ steel}} = L_{\text{Cu}} \]
\[ A_{\text{ steel}} = A_{\text{Cu}} \] 
\[ F_{\text{Cu}} = F_{\text{Steel }} \]

Here: L_steel and L_Cu denote the lengths of steel and copper wires, respectively.
           A_steel and A_Cu denote the cross-sectional areas of steel and copper wires, respectively.
           F_steel and F_Cu  denote the tension of steel and cooper wires, respectively. 

\[\frac{\text{ Stress of Cu}}{\text{ Stress of Steel }} = \frac{F_{\text{Cu}}}{A_{\text{Cu }}}\frac{A_{\text{ Steel }}}{F_{\text{ Steel}}} = 1\]