The two wires shown in figure are made of the

same material which has a breaking stress of 8 × 10⁸ N m⁻². The area of cross section of the upper wire is 0.006 cm² and that of the lower wire is 0.003 cm². The mass m₁ = 10 kg, m₂ = 20 kg and the hanger is light. (a) Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break first if the load is increased? (b) Repeat the above part if m₁ = 10 kg and m₂ = 36 kg.

(a) Given:
$$\begin{gathered} \mathrm{Breaking}\mathrm{stress}\mathrm{of}\mathrm{wire}=8\times {10}^8\mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{Area}\mathrm{of}\mathrm{cross}-\mathrm{section}\mathrm{of}\mathrm{upper}\mathrm{wire}\left(A_{\mathrm{u}}\right)=0.006{\mathrm{cm}}^2=6\times {10}^{-7}\mathrm{m} \\ \mathrm{Area}\mathrm{of}\mathrm{cross}-\mathrm{section}\mathrm{of}\mathrm{lower}\mathrm{wire}\left(A_{\mathrm{l}}\right)=0.003{\mathrm{cm}}^2=3\times {10}^{-7}\mathrm{m} \\ {m}_1=10\mathrm{kg},m_2=20\mathrm{kg} \end{gathered}$$

Tension in lower wire $T_l=m_{1}g+w$
Here: g is the acceleration due to gravity
w is the load
∴ Stress in lower wire$=\frac{{\mathrm{T}}_l}{{\mathrm{A}}_l}=\frac{m_{1}g+w}{{\mathrm{A}}_l}$
$$\begin{gathered} \Rightarrow \frac{m_{1}g+w}{A_l}=8\times {10}^8 \\ \Rightarrow w=\left[\left(8\times {10}^8\right)\times \left(3\times {10}^{-7}\right)\right]-100 \\ \Rightarrow w=140\mathrm{N}\mathrm{or}14\mathrm{kg} \end{gathered}$$

Now, tension in upper wire $T_2=m_{1}g+m_{2}g+w$

∴ Stress in upper wire$=\frac{{\mathrm{T}}_u}{{\mathrm{A}}_u}=\frac{m_{\mathit{2}}g+m_{1}g+w}{{\mathrm{A}}_u}$
$$\begin{gathered} \Rightarrow \frac{m_{2}g+m_{1}g+w}{{\mathrm{A}}_u}=8\times {10}^8 \\ \Rightarrow w=180\mathrm{N}\mathrm{or}18\mathrm{kg} \end{gathered}$$

For the same breaking stress, the maximum load that can be put is 140 N or 14 kg. The lower wire will break first if the load is increased.

(b) $\mathrm{If}{m}_1=10\mathrm{kg}\mathrm{and}{m}_2=36\mathrm{kg}$:
Tension in lower wire $T_l=m_{1}g+w$
Here: g is the acceleration due to gravity
w is the load
∴ Stress in lower wire:

$$\begin{gathered} \Rightarrow \frac{{\mathrm{T}}_l}{{\mathrm{A}}_l}=\frac{m_{1}g+w}{{\mathrm{A}}_l}=8\times {10}^5 \\ \Rightarrow w=140\mathrm{N} \end{gathered}$$

Now, tension in upper wire $T_2=m_{1}g+m_{2}g+w$

∴ Stress in upper wire:

$$\begin{gathered} \Rightarrow \frac{{\mathrm{T}}_u}{{\mathrm{A}}_u}=\frac{m_{2}g+m_{1}g+w}{{\mathrm{A}}_u}=8\times {10}^5 \\ \Rightarrow w=20\mathrm{N} \end{gathered}$$

For the same breaking stress, the maximum load that can be put is 20 N or 2 kg. The upper wire will break first if the load is increased.