Question

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.  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?

Answer 1

 Given: 

\[\text{ Breaking stress of wire }= 8 \times {10}^8 \text{ N/ m}^2 \]
\[\text{ Area of cross - section of upper wire } ( A_\text{u} ) = 0 . 006 {\text{ cm }}^2 = 6 \times {10}^{- 7} \text{m}\]
\[\text{ Area of cross - section of lower wire }( A_\text{l} ) = 0 . 003 {\text{cm }}^2 = 3 \times {10}^{- 7} \text{m}\]
\[ \text{m}_1 = 10 \text{ kg, m}_2 = 20 \text{ kg }\]

Tension in lower wire 

\[T_\text{l} = \text{m}_1 \text{ g + w}\]

Here : g is the acceleration due to gravity
           w is the load
∴ Stress in lower wire

\[= \frac{\text{T}_\text{l}}{\text{A}_\text{l}} = \frac{\text{m}_1 \text{g + w}}{\text{A}_\text{l}}\]

\[\Rightarrow \frac{\text{m}_1 \text{g + w}}{\text{A}_\text{l}} = 8 \times {10}^8 \]
\[ \Rightarrow \text{w} = \left[ \left( 8 \times {10}^8 \right) \times \left( 3 \times {10}^{- 7} \right) \right] - 100\]
\[ \Rightarrow \text{ w = 140 N or 14 kg }\]

Now, tension in upper wire 

\[T_2 = \text{ m }_1 \text{g + m}_2 \text{g + w}\]

∴ Stress in upper wire

 

\[= \frac{\text{T}_\text{u}}{\text{A}_\text{u}} = \frac{\text{m}_2 \text{g + m}_1 \text{g + w}}{\text{A}_\text{u}}\] 

\[\Rightarrow \frac{\text{m}_2 \text{g + m}_1 \text{g + w}}{\text{A}_\text{u}} = 8 \times {10}^8 \]
\[ \Rightarrow \text{w = 180 N or 18 kg}\]

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.