Figure 12-81 shows a $\mathbf{300}\mathbf{\text{\hspace{0.17em}kg}}$ cylinder that is horizontal. Three steel wires support the cylinder from a ceiling. Wires 1 and 3 are attached at the ends of the cylinder, and wire 2 is attached at the center. The wires each have a cross-sectional area of$\mathbf{2}\mathbf{.00}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$ . Initially (before the cylinder was put in place) wires 1 and 3 were$\mathbf{2.0000}{\mathbf{\text{\hspace{0.17em}m}}}^{\mathbf{\text{2}}}$ long and wire 2 was $\mathbf{6.00}\mathbf{\text{\hspace{0.17em}mm}}$longer than that. Now (with the cylinder in place) all three wires have been stretched. What is the tension in (a) wire 1 and (b) wire 2?

The mass of the cylinder,$\text{m}=300\text{\hspace{0.17em}kg}$

The cross-sectional area of each wire,$\text{A}=2.00\times {10}^{-6}{\text{\hspace{0.17em}m}}^2$

The original length of wires 1 and 3,${\text{L}}_{\text{1}}={\text{L}}_{\text{2}}=\text{L}=2.0000\text{\hspace{0.17em}m}$

The elongation in the wire 1 and 3 is more than in wire 2 by an amount of

$\mathrm{y}=6.00\text{\hspace{0.17em}mm}\left(\frac{{10}^{-3}\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}mm}}\right)=6.00\times {10}^{-3}\text{\hspace{0.17em}m}$

You draw the free body diagram. The system is at equilibrium, for such a system, the vector sum of the forces acting on it is zero. You can use the concept of elasticity for steel wires. There is Young’s modulus of elasticity for steel wire. The formulas used are given below.

$\begin{array}{c}\frac{\mathbf{F}}{\mathbf{A}}\mathbf{=}\mathbf{E}\frac{\mathbf{\Delta L}}{\mathbf{L}}\\ \mathbf{\Sigma }{\overrightarrow{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\mathbf{0}\end{array}$

Three steel wires support the cylinder from the ceiling as shown in the figure. Due to steel material, there is Young’s modulus of elasticity produced in these wires. This is a static equilibrium condition hence the wires$1$and$3$must be stretched thin wire$2$by an amount of$\mathrm{y}$. Then all the wires have the same length after elongation as,

role="math" localid="1661350203235" ${\mathrm{\Delta L}}_1={\mathrm{\Delta L}}_3={\mathrm{\Delta L}}_2+\mathrm{y}\left(1\right)$

According to the expression of Young’s modulus of elasticity$\mathrm{E}$as

$\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{E}\frac{\mathrm{\Delta L}}{\mathrm{L}}$

For the wire $1$as

$\begin{array}{l}\frac{{\mathrm{F}}_1}{\mathrm{A}}=\mathrm{E}\frac{{\mathrm{\Delta L}}_1}{{\mathrm{L}}_1}\\ {\mathrm{\Delta L}}_1=\frac{{\mathrm{EL}}_1{\mathrm{F}}_1}{\mathrm{A}}\end{array}$

For the wire $2$ as

$\begin{array}{c}\frac{{\mathrm{F}}_2}{\mathrm{A}}=\mathrm{E}\frac{{\mathrm{\Delta L}}_2}{{\mathrm{L}}_2}\\ {\mathrm{\Delta L}}_2=\frac{{\mathrm{EL}}_2{\mathrm{F}}_2}{\mathrm{A}}\end{array}$

For The wire $3$ as

$\begin{array}{l}\frac{{\mathrm{F}}_3}{\mathrm{A}}=\mathrm{E}\frac{{\mathrm{\Delta L}}_3}{{\mathrm{L}}_3}\\ {\mathrm{\Delta L}}_3=\frac{{\mathrm{EL}}_3{\mathrm{F}}_3}{\mathrm{A}}\end{array}$

Equation (1) becomes as

$\begin{array}{c}\frac{{\mathrm{EL}}_1{\mathrm{F}}_1}{\mathrm{A}}=\frac{{\mathrm{EL}}_3{\mathrm{F}}_3}{\mathrm{A}}=\frac{{\mathrm{EL}}_2{\mathrm{F}}_2}{\mathrm{A}}+\mathrm{y}\\ {\mathrm{F}}_1={\mathrm{F}}_3={\mathrm{F}}_2+\frac{\mathrm{yEA}}{\mathrm{L}}\end{array}$

This is the static equilibrium condition for the cylinder.

According to the static equilibrium condition, the sum of the vertical forces acting on the beam is zero.

Hence,

$\begin{array}{c}{\mathrm{\Sigma F}}_{\mathrm{y}\mathrm{net}}=0\\ {\mathrm{F}}_1+{\mathrm{F}}_3+{\mathrm{F}}_2-\mathrm{mg}=0\\ {\mathrm{F}}_1+{\mathrm{F}}_1+{\mathrm{F}}_1-\frac{\mathrm{yEA}}{\mathrm{L}}-\mathrm{mg}=0\\ {\mathrm{F}}_1=\frac{\left(\mathrm{mg}+\frac{\mathrm{yEA}}{\mathrm{L}}\right)}{3}\\ {\mathrm{F}}_1=\frac{\left(300\mathrm{kg}\times 9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}+\left(\frac{6.00\times {10}^{-3}\mathrm{m}\times 200\times {10}^9{\text{N/m}}^{\text{2}}\times 2.00\times {10}^{-6}{\mathrm{m}}^2}{2.0000\mathrm{m}}\right)\right)}{3}\\ {\mathrm{F}}_1=1380\text{\hspace{0.17em}N}\end{array}$

$\begin{array}{l}{\mathrm{F}}_1={\mathrm{F}}_2+\frac{\mathrm{yEA}}{\mathrm{L}}\\ {\mathrm{F}}_2={\mathrm{F}}_1-\frac{\mathrm{yEA}}{\mathrm{L}}\\ {\mathrm{F}}_2=1380\mathrm{N}-\left(\frac{6.00\times {10}^{-3}\mathrm{m}\times 200\times {10}^9{\text{\hspace{0.17em}N/m}}^{\text{2}}\times 2.00\times {10}^{-6}{\mathrm{m}}^2}{2.0000\mathrm{m}}\right)\\ {\mathrm{F}}_2=180\text{\hspace{0.17em}N}\end{array}$