Chapter 10, Problem 10.60P

The weightless bars AB and CE, together with the 5-lb weight BE, form a parallelogram linkage. The ideal spring attached to D has a free length of 2 in. and a stiffness of 7.5 lb/in. Find the two equilibrium positions that are in the range $0<\theta <\pi /2,$ and determine their stability. Neglect the weight of slider F.

To determine

The equilibrium positions.

Answer to Problem 10.60P

The system is at unstable equilibrium at 29.6^o and at 53.1^o the system comes in stable equilibrium.

Explanation of Solution

Given Information:

The weight of the load at BE = 5-lb

The stiffness of spring k = 7.5 lb/in

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 = $V=V_g+V_e$

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

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

  $\begin{array}{l}\text{Here,}\\ Y_g=10\sin \theta +2.5\\ s=5\cos \theta \end{array}$

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

  $V=\left(W\times \left(10\sin \theta +2.5\right)\right)+\left(\frac{1}{2}\times k\times {\left(5\cos \theta \right)}^2\right)$

The potential energy of the system comes out to be

  $\begin{array}{l}V=\left(5\times \left(10\sin \theta +2.5\right)\right)+\left(\frac{1}{2}\times 7.5\times {\left(5\cos \theta \right)}^2\right)\\ V=50\sin \theta +12.5+93.75{\cos }^2\theta \end{array}$ ... (2)

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

  $\frac{dV}{d\theta }=50\cos \theta +2\times 93.75\cos \theta \times \left(-\sin \theta \right)$ ... (3)

  $\frac{dV}{d\theta }=50\cos \theta -187.5\sin \theta \cos \theta$

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.

  $\begin{array}{l}\frac{dV}{d\theta }=0\\ 50\cos \theta -187.5\sin \theta \cos \theta =0\\ \cos \theta \left(50-187.5\sin \theta \right)=0\\ \text{The roots of this equation are}\\ \cos \theta =0and\sin \theta =0.267\\ \text{The equlibrium positions are}\\ \cos \theta =0,\theta =\frac{\pi }{2}\\ \sin \theta =0.267,\theta ={\sin }^{-1}\left(0.267\right),\\ \theta =29.6°and\theta =53.1°\end{array}$

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

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

  $\begin{array}{l}\frac{d^{2}V}{d{\theta }^2}=-50\sin \theta -187.5\left(\cos 2\theta \right)\\ \text{Now}\text{for}\theta =\frac{\pi }{2}\\ \frac{d^{2}V}{d{\theta }^2}=-50\sin \frac{\pi }{2}-187.5\left(\cos 2\times \frac{\pi }{2}\right)\\ {\left(\frac{d^{2}V}{d{\theta }^2}\right)}_{\theta =\frac{\pi }{2}}=-50+187.5\\ {\left(\frac{d^{2}V}{d{\theta }^2}\right)}_{\theta =\frac{\pi }{2}}=137.5\left(+ve\right)\end{array}$

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

  $\begin{array}{l}\frac{d^{2}V}{d{\theta }^2}=-50\sin \theta -187.5\left(\cos 2\theta \right)\\ \text{Now}\text{for}\theta =29.6°\\ \frac{d^{2}V}{d{\theta }^2}=-50\sin 29.6°-187.5\left(\cos 2\times 29.6°\right)\\ {\left(\frac{d^{2}V}{d{\theta }^2}\right)}_{\theta =\frac{\pi }{2}}=-24.69-96\\ {\left(\frac{d^{2}V}{d{\theta }^2}\right)}_{\theta =\frac{\pi }{2}}=120.7\left(-ve\right)\end{array}$

Thus, system is at unstable equilibrium at $\theta =29.6°$.

  $\begin{array}{l}\frac{d^{2}V}{d{\theta }^2}=-50\sin \theta -187.5\left(\cos 2\theta \right)\\ \text{Now}\text{for}\theta =53.1°\\ \frac{d^{2}V}{d{\theta }^2}=-50\sin 53.1°-187.5\left(\cos 2\times 53.1°\right)\\ {\left(\frac{d^{2}V}{d{\theta }^2}\right)}_{\theta =\frac{\pi }{2}}=-39.98+52.31\\ {\left(\frac{d^{2}V}{d{\theta }^2}\right)}_{\theta =\frac{\pi }{2}}=12.33\left(+ve\right)\end{array}$

Thus, system is at stable equilibrium at $\theta =53.1°$.

Conclusion:

Therefore, thesystem is at unstable equilibrium at 29.6^o and at 53.1^o the system comes in stable equilibrium.