Chapter 10, Problem 10.61P

To determine

The angle ${\theta }_1$ and ${\theta }_2$ at equilibrium.

Answer to Problem 10.61P

The angle ${\theta }_1$ and ${\theta }_2$ at equilibrium are, $\begin{array}{l}{\theta }_1=-{4.6}^0\\ {\theta }_2={34.1}^0\end{array}$.

Explanation of Solution

Given information:

Load $W=\frac{kL}{10}$.

A diagram with all required dimensions has been given.

  

Calculation:

Consider the deflection of each spring is $s_B,s_C,s_D$.

  

By using the geometry, deflection of each spring

  $\begin{array}{l}{s}_B=\frac{L}{2}\mathrm{Sin}{\theta }_1\\ s_C=L\mathrm{Sin}{\theta }_1\\ s_D=L\mathrm{Sin}{\theta }_1+\frac{L}{2}\mathrm{Sin}{\theta }_2\\ h=L\mathrm{Sin}{\theta }_1+L\mathrm{Sin}{\theta }_2\end{array}$

The total potential energy = gravitational potential energy + potential energy of springs

  $V=V_g+V_e$......1

Where,

  $\begin{array}{l}{V}_g=Wh\\ V_g=-\frac{kL}{10}\times \left(L\sin {\theta }_1+L\sin {\theta }_2\right)\\ V_g=-\frac{k{L}^2}{10}\times \left(\sin {\theta }_1+\sin {\theta }_2\right)(2)\end{array}$

V_e = Sum of potential energy of all springs

(Each value of potential energy of spring $V_e=\frac{1}{2}k{s}^2$, so we have to add all the three)

  $\begin{array}{l}{V}_e=\frac{1}{2}k{s}_B{}^2+\frac{1}{2}k{s}_C{}^2+\frac{1}{2}k{s}_D{}^2\\ V_e=\frac{1}{2}k{\left(\frac{L}{2}\sin {\theta }_1\right)}^2+\frac{1}{2}k{\left(L\sin {\theta }_1\right)}^2+\frac{1}{2}k{\left(L\sin {\theta }_1+\frac{L}{2}\sin {\theta }_2\right)}^2\\ V_e=\frac{1}{2}k{L}^2\left(\frac{1}{4}{\sin }^2{\theta }_1+{\sin }^2{\theta }_1+{\sin }^2{\theta }_1+\frac{1}{4}{\sin }^2{\theta }_2+\sin {\theta }_1\sin {\theta }_2\right)\\ V_e=\frac{1}{2}k{L}^2\left(\frac{9}{4}{\sin }^2{\theta }_1+\frac{1}{4}{\sin }^2{\theta }_2+\sin {\theta }_1\sin {\theta }_2\right)(3)\end{array}$

Substituting, 2 and 3 in 1.

  $\begin{array}{l}V=V_g+V_e\\ V=-\frac{k{L}^2}{10}\times \left(\sin {\theta }_1+\sin {\theta }_2\right)+\frac{1}{2}k{L}^2\left(\frac{9}{4}{\sin }^2{\theta }_1+\frac{1}{4}{\sin }^2{\theta }_2+\sin {\theta }_1\sin {\theta }_2\right)\end{array}$

Now, differentiate V with respect to ${\theta }_1$ and ${\theta }_2$ separately and then finally equate them to zero to find the value of ${\theta }_1$ and ${\theta }_2$ respectively.

Differentiating with respect to ${\theta }_1$ ,

  $\begin{array}{l}\frac{dV}{d{\theta }_1}=-\frac{k{L}^2}{10}\times \left(\cos {\theta }_1+0\right)+\frac{1}{2}k{L}^2\left(\frac{9}{4}\times 2\sin {\theta }_1\cos {\theta }_1+\frac{1}{4}\times 0+\cos {\theta }_1\sin {\theta }_2\right)\\ \frac{dV}{d{\theta }_1}=-k{L}^2\cos {\theta }_1\left(\frac{1}{10}-\frac{18}{8}\sin {\theta }_1-\frac{1}{2}\sin {\theta }_2\right)\end{array}$

Now, substituting $\frac{dV}{d{\theta }_1}=0$ ,

  $\begin{array}{l}-k{L}^2\cos {\theta }_1\left(\frac{1}{10}-\frac{18}{8}\sin {\theta }_1-\frac{1}{2}\sin {\theta }_2\right)=0\\ \frac{1}{10}-\frac{18}{8}\sin {\theta }_1-\frac{1}{2}\sin {\theta }_2=0\\ 9\sin {\theta }_1+2\sin {\theta }_2=\frac{2}{5}(4)\end{array}$

Similarly, differentiating with respect to ${\theta }_2$ ,

  $\begin{array}{l}\frac{dV}{d{\theta }_2}=-\frac{k{L}^2}{10}\times \left(0+\cos {\theta }_2\right)+\frac{1}{2}k{L}^2\left(0+\frac{1}{4}\times 2\sin {\theta }_2\cos {\theta }_2+\cos {\theta }_2\sin {\theta }_1\right)\\ \frac{dV}{d{\theta }_2}=-k{L}^2\cos {\theta }_2\left(\frac{1}{10}-\frac{1}{4}\sin {\theta }_2-\frac{1}{2}\sin {\theta }_1\right)\end{array}$

Now, substituting $\frac{dV}{d{\theta }_2}=0$ ,

  $\begin{array}{l}-k{L}^2\cos {\theta }_2\left(\frac{1}{10}-\frac{1}{4}\sin {\theta }_2-\frac{1}{2}\sin {\theta }_1\right)=0\\ \frac{1}{10}-\frac{1}{4}\sin {\theta }_2-\frac{1}{2}\sin {\theta }_1=0\\ 2\sin {\theta }_1+\sin {\theta }_2=\frac{2}{5}(5)\end{array}$

Solving linear equations 4 and 5 simultaneously we get,

  $\begin{array}{l}\sin {\theta }_1=-.08\\ \sin {\theta }_2=0.56\end{array}$

So taking inverse,

  $\begin{array}{l}{\theta }_1=-{4.6}^0\\ {\theta }_2={34.1}^0\end{array}$