Question

The two wires shown in figure are made of the same

material which has a breaking stress of $8{\times }^{8}N{M}^{-2}$. The area of cross section of the upper wire is $0.006c{m}^2$ and that of the lower wire is $0.003c{m}^2$. The mass $m_1=10kg,m_2=20kg$ 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 increaed ? (b) Repeat the above part if $m_1=10kg$ and $m_2=36kg.$

Answer 1

(a) Given that :

Stress = $8\times {10}^8\frac{1}{m^2}$

$A_u=0.006c{m}^3$

$A_1=0.003c{m}^2$

$m_1=10kg,m_2=20kg$

Therefore stress in lower rod

= $\frac{T_1}{A_1}=\frac{m_{1}g+wg}{A_1}$

(from free body diagram -2)

$\Rightarrow \frac{m_{1}g+wg}{A_1}=8\times {10}^8$

$\Rightarrow$ w = 14 kg

Stress in upper rod = $\frac{T_2}{A_2}=\frac{m_{2}g+m_{1}g+wg}{A_2}$

For FBD-2

$\Rightarrow \frac{m_{2}g+m_{1}g+wg}{A_2}=8\times {10}^8$

$\Rightarrow$ w = 18 kg

For same stress the maximum load that can be put is 14 kg. If the load is increased the lower wire will break first.

(b) If $m_1=10kg$

and $m_2=36kg$

$\Rightarrow \frac{T_1}{A_1}=\frac{m_{1}g+wg}{A_1}=8times{10}^5$

$\Rightarrow$ w = 14 kg

$\Rightarrow \frac{T_2}{A_2}=\frac{m_{2}g+m_{1}g+wg}{A_2}$

= $8\times {10}^5=2kg$

The maximum load that can be put is 2 kg. Upper wire will break first if load is increased.