91Ó°ÊÓ

The young’s modulus is defined as the ratio of stress experienced by the object and the simultaneous strain acting on it.

Consider the formula for the Young’s modulus is shown below:

$\mathbf{Y}\mathbf{=}\frac{\mathbf{Stress}\mathbf{}\mathbf{of}\mathbf{}\mathbf{object}}{\mathbf{Strain}\mathbf{}\mathbf{of}\mathbf{}\mathbf{object}}$ …… (1)

Here, Y is Young modulus of the object.

Length of rod is L = 1.05 m.

Cross sectional area of A is $2.00{\mathrm{mm}}^2$.

Cross sectional area of B is$4.00{\mathrm{mm}}^2$

Young’s modulus for wire A is $1.80×{10}^{11}Pa$

Young’s modulus for wire B is$1.20×{10}^{11}Pa$

(a)

Now, from condition of equilibrium:

$\begin{array}{r}∑{\mathrm{Ï„}}_A=0\\ 0=F_A\left(0\right)−W\left(I_A\right)+F_B\left(L\right)\\ Wâ‹\dots I_A=F_Bâ‹\dots L\\ F_B=\frac{W.I_A}{L}\end{array}$...........(2)

For point B ,

$\begin{array}{r}∑{\mathrm{Ï„}}_B=0\\ 0=F_B\left(0\right)−W\left(I_B\right)+F_A\left(L\right)\\ Wâ‹\dots I_B=F_Aâ‹\dots L\\ F_A=\frac{Wâ‹\dots I_B}{L}\end{array}$ ..............(3)

Here $I_A$, and $I_B$are distances from point where acts W , A and B. Note that connection between those 2 is:

$I_A=L-l_B$ ................(4)

Now, let stress at both point is same.

$\begin{array}{r}{\mathrm{s}}_{\mathrm{A}}={\mathrm{s}}_{\mathrm{B}}\\ \frac{{\mathrm{F}}_{\mathrm{A}}}{{\mathrm{A}}_{\mathrm{A}}}=\frac{{\mathrm{F}}_{\mathrm{B}}}{{\mathrm{A}}_{\mathrm{B}}}\\ \mathrm{From}\mathrm{equation}\left(2\right)\mathrm{and}\left(3\right)\\ \frac{\mathrm{W}â‹\dots {\mathrm{I}}_{\mathrm{B}}}{\mathrm{L}}=\frac{\mathrm{W}â‹\dots {\mathrm{I}}_{\mathrm{A}}}{{\mathrm{A}}_{\mathrm{A}}}\\ \frac{{\mathrm{I}}_{\mathrm{B}}}{{\mathrm{A}}_{\mathrm{A}}}=\frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{A}}_{\mathrm{B}}}\end{array}$

Now, from equation (4).

$\begin{array}{r}\frac{{\mathrm{I}}_{\mathrm{B}}}{{\mathrm{A}}_{\mathrm{A}}}=\frac{\mathrm{L}−{\mathrm{I}}_{\mathrm{B}}}{{\mathrm{A}}_{\mathrm{B}}}\\ {\mathrm{I}}_{\mathrm{B}}{\mathrm{A}}_{\mathrm{B}}=\left(\mathrm{L}−{\mathrm{I}}_{\mathrm{B}}\right){\mathrm{A}}_{\mathrm{A}}\\ {\mathrm{I}}_{\mathrm{B}}\left({\mathrm{A}}_{\mathrm{B}}+{\mathrm{A}}_{\mathrm{A}}\right)={\mathrm{LA}}_{\mathrm{A}}\\ {\mathrm{I}}_{\mathrm{B}}=\frac{{\mathrm{LA}}_{\mathrm{A}}}{\left({\mathrm{A}}_{\mathrm{B}}+{\mathrm{A}}_{\mathrm{A}}\right)}\\ \mathrm{Now},\mathrm{by}\mathrm{substituting}\mathrm{all}\mathrm{the}\mathrm{numerical}\mathrm{values}\mathrm{in}\mathrm{above}\mathrm{equation}\mathrm{and}\mathrm{solve}\mathrm{as}:\\ {\mathrm{I}}_{\mathrm{B}}=\frac{(1.05\mathrm{m})\left(2×{10}^{−6}\mathrm{m}\right)}{\left(4×{10}^{−6}\mathrm{m}+2×{10}^{−6}\mathrm{m}\right)}\\ {\mathrm{I}}_{\mathrm{B}}=0.35\mathrm{m}\end{array}$

Substitute the value of in equation (4).

$\begin{array}{r}{I}_A=L−0.35\mathrm{m}\\ I_A=1.05\mathrm{m}−0.35\mathrm{m}\\ I_A=0.7\mathrm{m}\end{array}$

(b)

Now, let weight produce equal strain in A and B.

$$\begin{gathered} Strai{n}_A=Strai{n}_B \\ \frac{Stres{s}_A}{Y_A}=\frac{Stres{s}_B}{Y_B} \end{gathered}$$

From equation (2)

$\frac{\frac{F_A}{A_A}}{Y_A}=\frac{\frac{F_B}{A_B}}{Y_B}$

Now, from equation (2) and (3)

$\begin{array}{r}\frac{{\mathrm{I}}_{\mathrm{B}}}{{\mathrm{A}}_{\mathrm{A}}}=\frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{Y}}_{\mathrm{A}}}\\ \frac{{\mathrm{I}}_{\mathrm{B}}}{{\mathrm{Y}}_{\mathrm{B}}}\\ \frac{{\mathrm{Y}}_{\mathrm{A}}}{}=\frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{Y}}_{\mathrm{B}}{\mathrm{A}}_{\mathrm{B}}}\\ \mathrm{By}\mathrm{equation}\left(4\right)\mathrm{solve}\mathrm{as}:\\ \frac{{\mathrm{I}}_{\mathrm{B}}}{{\mathrm{Y}}_{\mathrm{A}}{\mathrm{A}}_{\mathrm{A}}}=\frac{\mathrm{L}−{\mathrm{I}}_{\mathrm{B}}}{{\mathrm{Y}}_{\mathrm{B}}{\mathrm{A}}_{\mathrm{B}}}\\ \frac{{\mathrm{A}}_{\mathrm{B}}{\mathrm{Y}}_{\mathrm{B}}}{{\mathrm{Y}}_{\mathrm{A}}{\mathrm{A}}_{\mathrm{A}}}=\frac{\mathrm{L}−{\mathrm{I}}_{\mathrm{B}}}{{\mathrm{I}}_{\mathrm{B}}}\\ \frac{\mathrm{L}}{{\mathrm{I}}_{\mathrm{B}}}−1=\frac{{\mathrm{A}}_{\mathrm{B}}{\mathrm{Y}}_{\mathrm{B}}}{{\mathrm{Y}}_{\mathrm{A}}{\mathrm{A}}_{\mathrm{A}}}\\ \frac{\mathrm{L}}{{\mathrm{I}}_{\mathrm{B}}}=\frac{{\mathrm{A}}_{\mathrm{B}}{\mathrm{Y}}_{\mathrm{B}}}{{\mathrm{Y}}_{\mathrm{A}}{\mathrm{A}}_{\mathrm{A}}}+1\end{array}$

Substitute the numerical values in above equation:

$\frac{L}{I_B}=\frac{\left(4×{10}^{−6}{\mathrm{m}}^2\right)\left(1.20×{10}^{11}\mathrm{Pa}\right)}{\left(4×{10}^{−6}{\mathrm{m}}^2\right)\left(1.80×{10}^{11}\mathrm{Pa}\right)}+1$

Solve further as:

$\begin{array}{r}{\mathrm{I}}_{\mathrm{B}}=\frac{\mathrm{L}}{2.33}\\ {\mathrm{I}}_{\mathrm{B}}=\frac{1.05\mathrm{m}}{2.33}\\ {\mathrm{I}}_{\mathrm{B}}=0.45\mathrm{m}\\ \mathrm{Now},\mathrm{substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{equation}\left(4\right).\\ {\mathrm{I}}_{\mathrm{A}}=\mathrm{L}−0.45\mathrm{m}\\ =1.05\mathrm{m}−0.45\mathrm{m}\\ =0.6\mathrm{m}\end{array}$

Hence, at${\mathrm{I}}_{\mathrm{A}}=0.45\mathrm{m}$ and${\mathrm{I}}_{\mathrm{B}}=0.6\mathrm{m}$ weight W be suspended to produce when equal strains in A and B.