Chapter 10.2, Problem 10.80P

A slender rod AB with a weight W is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when AB is horizontal, determine three values of θ corresponding to equilibrium when W = 300 lb, l = 16 in., and k = 75 lb/in. State in each case whether the equilibrium is stable, unstable, or neutral.

Fig. P10.79 and P10.80

To determine

Find three values of $\theta$ corresponding to equilibrium and check whether the equilibrium is stable, unstable, or neutral.

Answer to Problem 10.80P

The value of $\theta$ for equilibrium is $\underset{\_}{\theta =9.39°;\theta =34.2°;\theta =90°}$.

The equilibrium is $\underset{\_}{\text{stable}\text{for}\theta =9.39°}$.

The equilibrium is $\underset{\_}{\text{unstable}\text{for}\theta =34.2°}$.

The equilibrium is $\underset{\_}{\text{stable}\text{for}\theta =90°}$.

Explanation of Solution

Given information:

The weight of the slender rod AB is $W=300\text{lb}$.

The length of the slender rod AB is $l=16\text{in}\text{.}$.

The spring constant is $k=75\text{lb/in}\text{.}$.

Calculation:

The spring is unstretched when AB is horizontal.

Show the free-body diagram of the arrangement as in Figure 1.

Find the elongation of the spring s using the relation.

$\begin{array}{c}s=\left(l\sin \theta +l\cos \theta \right)-l\\ =l\left(\sin \theta +\cos \theta -1\right)\end{array}$

Here, the length of the slender rod AB is l.

Find the potential energy (V) using the relation;

$\begin{array}{c}V=V_s+V_g\\ =\frac{1}{2}k{s}^2+W\left(-\frac{l}{2}\sin \theta \right)\\ =\frac{1}{2}k{\left(l\left(\sin \theta +\cos \theta -1\right)\right)}^2-\frac{Wl}{2}\sin \theta \\ =\frac{1}{2}k{l}^2{\left(\sin \theta +\cos \theta -1\right)}^2-\frac{Wl}{2}\sin \theta \end{array}$

Differentiate the equation;

$\frac{dV}{d\theta }=k{l}^2\left(\sin \theta +\cos \theta -1\right)\left(\cos \theta -\sin \theta \right)-\frac{Wl}{2}\cos \theta$ (1)

Here, the weight of the slender rod AB and the spring constant is k.

Differentiate the Equation (1).

$\begin{array}{c}\frac{d^{2}V}{d{\theta }^2}=k{l}^2\left[\begin{array}{l}\left(\cos \theta -\sin \theta \right)\left(\cos \theta -\sin \theta \right)\\ +\left(\sin \theta +\cos \theta -1\right)\left(-\sin \theta -\cos \theta \right)\end{array}\right]+\frac{Wl}{2}\sin \theta \\ =k{l}^2\left[\begin{array}{l}{\cos }^2\theta -\sin \theta \cos \theta -\sin \theta \cos \theta +{\sin }^2\theta -{\sin }^2\theta \\ -\sin \theta \cos \theta +\sin \theta -\sin \theta \cos \theta -{\cos }^2\theta +\cos \theta \end{array}\right]+\frac{Wl}{2}\sin \theta \\ =k{l}^2\left[-4\sin \theta \cos \theta +\sin \theta +\cos \theta \right]+\frac{Wl}{2}\sin \theta \\ =k{l}^2\left[-2\sin 2\theta +\sin \theta +\cos \theta \right]+\frac{Wl}{2}\sin \theta \end{array}$ (2)

Equate the Equation (1) to zero for equilibrium.

$\begin{array}{l}\frac{dV}{d\theta }=0\\ k{l}^2\left(\sin \theta +\cos \theta -1\right)\left(\cos \theta -\sin \theta \right)-\frac{Wl}{2}\cos \theta =0\\ \cos \theta \left[kl\left(\sin \theta +\cos \theta -1\right)\left(1-\tan \theta \right)-\frac{W}{2}\right]=0\\ kl\left(\sin \theta +\cos \theta -1\right)\left(1-\tan \theta \right)=\frac{W}{2};\cos \theta =0\end{array}$

Substitute 300 lb for W, $75\text{lb/in}\text{.}$ for k, and 16 in. for l.

$\begin{array}{l}75\times 16\times \left(\sin \theta +\cos \theta -1\right)\left(1-\tan \theta \right)=\frac{300}{2}\\ \left(\sin \theta +\cos \theta -1\right)\left(1-\tan \theta \right)=0.125\end{array}$

Solve the equation by trial and error procedure.

$\begin{array}{l}\theta =9.39°;\theta =34.2°;\text{for}kl\left(\sin \theta +\cos \theta -1\right)\left(1-\tan \theta \right)=\frac{W}{2}\\ \theta =90°;\text{for}\cos \theta =0\end{array}$

Therefore, the value of $\theta$ for equilibrium is $\underset{\_}{\theta =9.39°;\theta =34.2°;\theta =90°}$.

For $\theta =9.39°;$

Substitute 300 lb for W, $75\text{lb/in}\text{.}$ for k, 16 in. for l, and $9.39°$ for $\theta$ in Equation (2).

$\begin{array}{c}\frac{d^{2}V}{d{\theta }^2}=75\times {16}^2\times \left[-2\sin 2\left(9.39°\right)+\sin 9.39°+\cos 9.39°\right]+\frac{300\times 16}{2}\sin 9.39°\\ =19,200\left[-2\sin 2\left(9.39°\right)+\sin 9.39°+\cos 9.39°\right]+2,400\sin 9.39°\\ =10,104.54\text{lb}\cdot \text{in}\text{.}>\text{0,stable}\end{array}$

Therefore, the equilibrium is $\underset{\_}{\text{stable}\text{for}\theta =9.39°}$.

For $\theta =34.2°;$

Substitute 300 lb for W, $75\text{lb/in}\text{.}$ for k, 16 in. for l, and $34.2°$ for $\theta$ in Equation (2).

$\begin{array}{c}\frac{d^{2}V}{d{\theta }^2}=75\times {16}^2\times \left[-2\sin 2\left(34.2°\right)+\sin 34.2°+\cos 34.2°\right]+\frac{300\times 16}{2}\sin 34.2°\\ =19,200\left[-2\sin 2\left(34.2°\right)+\sin 34.2°+\cos 34.2°\right]+2,400\sin 34.2°\\ =-7,682.5\text{lb}\cdot \text{in}\text{.}<\text{0,unstable}\end{array}$

Therefore, the equilibrium is $\underset{\_}{\text{unstable}\text{for}\theta =34.2°}$.

For $\theta =90°;$

Substitute 300 lb for W, $75\text{lb/in}\text{.}$ for k, 16 in. for l, and $90°$ for $\theta$ in Equation (2).

$\begin{array}{c}\frac{d^{2}V}{d{\theta }^2}=75\times {16}^2\times \left[-2\sin 2\left(90°\right)+\sin 90°+\cos 90°\right]+\frac{300\times 16}{2}\sin 90°\\ =19,200\left[-2\sin 2\left(90°\right)+\sin 90°+\cos 90°\right]+2,400\sin 90°\\ =21,600\text{lb}\cdot \text{in}\text{.}>\text{0,stable}\end{array}$

Therefore, the equilibrium is $\underset{\_}{\text{stable}\text{for}\theta =90°}$.