Chapter 10, Problem 10.62P

The bar ABC is supported by three identical, ideal springs. Note that the springs are always vertical because the collars to which they are attached are free to slide on the horizontal rail. Find the angle $\theta$ at equilibrium if W = kL. Neglect the weight of the bar.

To determine

The angle $\theta$ at the equilibrium position.

Answer to Problem 10.62P

The value ofangle at the stable equilibrium positions is $\theta =6.15°$

Explanation of Solution

Given Information:

The weight of the load = W

The stiffness of each spring = k

Given that $W=kL$

The following figure is given,

Calculation:

Consider the following figure,

To calculate the angle for the equilibrium position, let us calculate the potential energy of the system. The total potential energy of the system consists of potential energy of the weight (V_g) and the potential energy of the springs (V_e)

The total potential energy = $V=V_g+V_e$

  $V=\left(W\times Y_g\right)+\left(\frac{1}{2}\times k\times s_1{}^2\right)+\left(\frac{1}{2}\times k\times s_2{}^2\right)+\left(\frac{1}{2}\times k\times s_3{}^2\right)$......(1)

  $\begin{array}{l}{Y}_g=\text{ The vertical distance of center of gravity of weight above the reference line}\\ \text{s = The elongation of the particular spring}\end{array}$

  $\begin{array}{l}\text{Here,}\\ Y_g=\frac{L}{2}\sin \theta \\ s_1=0\\ s_2=\left(\frac{L}{6}+\frac{2L}{3}\right)\sin \theta =\frac{5L}{6}\sin \theta \\ s_3=L\sin \theta \end{array}$

Putting the value of Y_g and s in equation (1)

  $\begin{array}{l}V=\left(W\times \left(\frac{L}{2}\sin \theta \right)\right)+\left(\frac{1}{2}\times k\times 0^2\right)+\left(\frac{1}{2}\times k\times {\left(\frac{5L}{6}\sin {\theta }_1\right)}^2\right)\\ +\left(\frac{1}{2}\times k\times {\left(L\sin \theta \right)}^2\right)\end{array}$

The potential energy of the system comes out to be

  $\begin{array}{l}V=\left(\frac{k{L}^2}{2}\sin \theta \right)+\left(\frac{25k{L}^2}{72}{\sin }^2\theta \right)+\left(\frac{k{L}^2{\sin }^2\theta }{2}\right)\\ \end{array}$

  $\begin{array}{l}V=\left(\frac{k{L}^2}{2}\sin \theta \right)+\left(\frac{61k{L}^2}{72}{\sin }^2\theta \right)\\ \end{array}$......(2)

Now, let us take the first derivative of the total potential energy of the system,

  $\begin{array}{l}\frac{dV}{d\theta }=\left(\frac{k{L}^2}{2}\cos \theta \right)+\left(\frac{122k{L}^2}{72}\sin \theta \cos \theta \right)\\ \end{array}$......(3)

Now, the principle of minimum potential energy can be used to find the value of angle $\theta$.

The roots of equation $\frac{dV}{d\theta }=0$ will provide the values of angles at equilibrium.

  $\begin{array}{l}\frac{dV}{d\theta }=0\\ \left(\frac{k{L}^2}{2}\cos \theta \right)+\left(\frac{122k{L}^2}{72}\sin \theta \cos \theta \right)=0\\ \frac{k{L}^2}{2}\times \cos \theta \left(1+\frac{122}{36}\sin \theta \right)=0\\ \text{The roots of this equation are}\\ \cos \theta =0and\sin \theta =\frac{-36}{122}\\ \text{The equlibrium positions are}\\ \cos \theta =0,\theta =\frac{\pi }{2}\\ \sin \theta =\frac{-36}{122},\theta ={\sin }^{-1}\left(\frac{-36}{122}\right)=6.15°\end{array}$

  $\theta =6.15°$

Now, differentiating equation 3 again to get $\frac{d^{2}V}{d{\theta }^2}$

  $\frac{d\left(\sin \theta .\cos \theta \right)}{d\theta }={\cos }^2\theta -{\sin }^2\theta =\cos 2\theta$

  $\begin{array}{l}\frac{d^{2}V}{d{\theta }^2}=-\frac{k{L}^2\sin \theta }{2}+\frac{122}{72}k{L}^2\left(\cos 2\theta \right)\\ \text{Now}\text{for}\theta =\frac{\pi }{2}\\ \frac{d^{2}V}{d{\theta }^2}=-\frac{k{L}^2}{2}\times 1-\frac{122}{72}k{L}^2\\ {\left(\frac{d^{2}V}{d{\theta }^2}\right)}_{\theta =\frac{\pi }{2}}=\frac{-79k{L}^2}{36}\left(-ve\right)\end{array}$

Thus, system is at unstable equilibrium at $\theta =\frac{\pi }{2}$.

  $\begin{array}{l}\text{Now}\text{for}\theta =6.15°\\ \frac{d^{2}V}{d{\theta }^2}=-\frac{k{L}^2\sin \theta }{2}+\frac{122}{72}k{L}^2\left(\cos 2\theta \right)\\ \frac{d^{2}V}{d{\theta }^2}=-\frac{k{L}^2\sin 6.15°}{2}+\frac{122}{72}k{L}^2\left(\cos 2\times 6.15°\right)\\ {\left(\frac{d^{2}V}{d{\theta }^2}\right)}_{\theta =6.15°}=-0.05356k{L}^2+1.655k{L}^2\\ {\left(\frac{d^{2}V}{d{\theta }^2}\right)}_{\theta =6.15°}=1.60k{L}^2\left(+ve\right)\end{array}$

Thus, system is at stable equilibrium at $\theta =6.15°$.

Conclusion:

Therefore, the value ofangle at the stable equilibrium positions is $\theta =6.15°$