Physics Mechanical Properties of Solids

a = $\frac{3}{9}g$ , For C, 3g – T = 3a = (3) $\frac{3}{g}$ g

T = 2g = 20 N

$\frac{\sigma }{\in }=Y$

$\frac{\sigma }{Y}=\in \Rightarrow \frac{20}{5*10^{-8}*2*10^{11}}$

$=2*10^{-3}$  

$=20*10^{-4}\Rightarrow 20$  

Young's modulus is property of material of wire and it is independent of geometrical factors.

            E =   $\frac{Stress}{Strain}$

As temperature increases, strain increases

Elasticity decreases

Elongation in the wire is given as

$?I=\frac{Fl}{AY}$

$T_{max}=\frac{m{v}^2}{l}+mg$

$T_{max}=3mg$

$?l_{max}=\frac{3mgl}{\pi r^{2}Y}$

Energy stored in unit volume of wire

Stress = Y * strain

-> $\frac{T_1}{A}=Y*-\frac{\left({\mathcal{l}}_1-\mathcal{l}\right)}{\mathcal{l}}\left(i\right)$  

  $\frac{T_2}{A}=Y*\frac{\left({\mathcal{l}}_2-\mathcal{l}\right)}{\mathcal{l}}\left(ii\right)$                          

$\Rightarrow$ $\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}$                      

$\Rightarrow \mathcal{l}=\frac{T_1{\mathcal{l}}_2-T_2{\mathcal{l}}_2}{T_1-T_2}=\frac{T_2{\mathcal{l}}_1-T_1{\mathcal{l}}_2}{T_2-T_1}$            

              $\Delta L$ (change in length) $=\frac{F\mathcal{l}}{AE}=\frac{mg\mathcal{l}}{AE}$  

              $\Delta \mathcal{l}\propto g$  

              $\frac{\Delta {\mathcal{l}}_1}{g_1}=\frac{\Delta {\mathcal{l}}_2}{g_2}\Rightarrow \frac{10^{-4}}{10}=\frac{6*10^{-5}}{g_1}$  

              $g_1=\frac{6*10^{-4}}{10^{-4}}=$ 6 m/sec²

$Y=\frac{\sigma }{\in }=\frac{20}{10^{-10}},$ Y → young's modules of elasticity.

              Stress $\left(\sigma \right)$ at strain $\left(\in \right)$  5 * 10^-4

              $\sigma =\in Y$  

              $=5*10^{-4}*20*10^{10}$  

              = $10^2*10^6$  

              = 10⁸

              $U_d\left(Energydensity\right)=\frac{1}{2}\sigma \in$  

              $=\frac{1}{2}*10^8*5*10^{-4}$  

              $\begin{array}{l}=2.5*10^4\\ =25*10^3=25kJ/m^3\end{array}$