1.05 mlong rod of negligible weight is supported at its ends by wires Aand Bof equal length (Given figure). The cross-sectional area ofis 2.0 m${\mathbf{m}}^{\mathbf{2}}$and that ofB is$\mathbf{4}\mathbf{.}\mathbf{00}{\mathbf{mm}}^{\mathbf{2}}$. Young’s modulus for wireis$\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{Pa}$; that for Bis$\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{Pa}$. At what point along the rod should a weightbe suspended to produce.

1. equal stresses in Aand B
2. equal strains in Aand B.

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 $1.80\times {10}^{11}Pa$

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

(a)

Now, from condition of equilibrium:

$\begin{array}{r}\sum {\tau }_A=0\\ 0=F_A\left(0\right)-W\left(I_A\right)+F_B\left(L\right)\\ W\cdot I_A=F_B\cdot L\\ F_B=\frac{W.I_A}{L}\end{array}$...........(2)

For point B ,

$\begin{array}{r}\sum {\tau }_B=0\\ 0=F_B\left(0\right)-W\left(I_B\right)+F_A\left(L\right)\\ W\cdot I_B=F_A\cdot L\\ F_A=\frac{W\cdot I_B}{L}\end{array}$ ..............(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.

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

Now, from equation (4).

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

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:

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

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.