Chapter 10.2, Problem 10.100P

(a)

To determine

Find the range of values of P for which the equilibrium is stable.

Answer to Problem 10.100P

The range of values of P for which the equilibrium position is stable is $\underset{\_}{P<10\text{lb}}$.

Explanation of Solution

Given information:

The system is in equilibrium when ${\theta }_1={\theta }_2=0$.

The value of spring constant is $k=20\text{lb/in}\text{.}$.

The radius of the drums is $r=3\text{in}\text{.}$.

The length of the rods AB and CD is $l=6\text{in}\text{.}$.

The weight acting at point A is $W=15\text{lb}$.

Calculation:

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

Consider the movement of spring at left end is from a to b, and the right end is from $a^{\prime }$ to $b^{\prime }$.

Find the elongation of the spring (s) using the relation.

$\begin{array}{c}s=a^{\prime }{b}^{\prime }-ab\\ =r{\theta }_1-r{\theta }_2\\ =r\left({\theta }_1-{\theta }_2\right)\end{array}$

Find the potential energy (V) using the relation.

$\begin{array}{c}V=V_g+V_s\\ =\frac{1}{2}k{s}^2+Pl\cos {\theta }_1-Wl\cos {\theta }_2\\ =\frac{1}{2}k{r}^2{\left({\theta }_1-{\theta }_2\right)}^2+Pl\cos {\theta }_1-Wl\cos {\theta }_2\end{array}$ (1)

Here, the spring constant is k.

Differentiate the Equation (1) with respect to ${\theta }_1$.

$\begin{array}{c}\frac{\partial V}{\partial {\theta }_1}=2\times \frac{1}{2}k{r}^2\left({\theta }_1-{\theta }_2\right)-Pl\sin {\theta }_1-0\\ =k{r}^2\left({\theta }_1-{\theta }_2\right)-Pl\sin {\theta }_1\end{array}$ (2)

Differentiate the Equation (2) with respect to ${\theta }_1$.

$\frac{{\partial }^{2}V}{\partial {\theta }_1^2}=k{r}^2-Pl\cos {\theta }_1$

Differentiate the equation (2) with ${\theta }_2$ to find the derivative of $\frac{{\partial }^{2}V}{\partial {\theta }_1\partial {\theta }_2}$.

$\frac{{\partial }^{2}V}{\partial {\theta }_1\partial {\theta }_2}=-k{r}^2$

Differentiate the Equation (1) with respect to ${\theta }_2$.

$\begin{array}{c}\frac{\partial V}{\partial {\theta }_2}=-2\times \frac{1}{2}k{r}^2\left({\theta }_1-{\theta }_2\right)+0-Wl\sin {\theta }_2\\ =-k{r}^2\left({\theta }_1-{\theta }_2\right)-Wl\sin {\theta }_2\end{array}$ (3)

Differentiate the Equation (3) with respect to $\partial {\theta }_2$.

$\frac{{\partial }^{2}V}{\partial {\theta }_2^2}=k{r}^2+Wl\cos {\theta }_2$

Condition 1:

When the equilibrium is stable, ${\theta }_1={\theta }_2=0$.

Substitute 0 for ${\theta }_1$ and 0 for ${\theta }_2$ in Equation (2).

$\begin{array}{c}\frac{\partial V}{\partial {\theta }_1}=k{r}^2\left(0-0\right)-Pl\sin 0\\ =0\end{array}$

Substitute 0 for ${\theta }_1$ and 0 for ${\theta }_2$ in Equation (3).

$\begin{array}{c}\frac{\partial V}{\partial {\theta }_2}=-k{r}^2\left(0-0\right)-Wl\sin 0\\ =0\end{array}$

$\frac{\partial V}{\partial {\theta }_1}=\frac{\partial V}{\partial {\theta }_2}=0$

The condition is satisfied. The equilibrium is stable.

Condition 2:

Check the condition,

${\left(\frac{{\partial }^{2}V}{\partial {\theta }_1\partial {\theta }_2}\right)}^2-\frac{{\partial }^{2}V}{\partial {\theta }_1^2}\frac{{\partial }^{2}V}{\partial {\theta }_2^2}<0$

Substitute $-k{r}^2$ for $\frac{{\partial }^{2}V}{\partial {\theta }_1\partial {\theta }_2}$, $\left(k{r}^2-Pl\cos {\theta }_1\right)$ for $\frac{{\partial }^{2}V}{\partial {\theta }_1^2}$, and $\left(k{r}^2+Wl\cos {\theta }_2\right)$ for $\frac{{\partial }^{2}V}{\partial {\theta }_2^2}$.

${\left(-k{r}^2\right)}^2-\left(k{r}^2-Pl\cos {\theta }_1\right)\left(k{r}^2+Wl\cos {\theta }_2\right)<0$

Substitute 0 for ${\theta }_1$ and 0 for ${\theta }_2$.

$\begin{array}{l}{\left(-k{r}^2\right)}^2-\left(k{r}^2-Pl\cos 0°\right)\left(k{r}^2+Wl\cos 0°\right)<0\\ {\left(-k{r}^2\right)}^2-\left(k{r}^2-Pl\right)\left(k{r}^2+Wl\right)<0\\ k^2{r}^4-k^2{r}^4-k{r}^{2}Wl+k{r}^{2}Pl+PW{l}^2<0\\ -k{r}^{2}W+P\left(k{r}^2+Wl\right)<0\end{array}$

$\begin{array}{l}P<\frac{k{r}^{2}W}{\left(k{r}^2+Wl\right)}\\ P<\frac{k{r}^2}{l}\left(\frac{W}{\left(\frac{k{r}^2}{l}+W\right)}\right)\end{array}$ (4)

Condition 3:

$\begin{array}{l}\frac{{\partial }^{2}V}{\partial {\theta }_1^2}>0\\ k{r}^2-Pl\cos {\theta }_1>0\end{array}$

Substitute 0 for ${\theta }_1$.

$\begin{array}{l}k{r}^2-Pl\cos 0°>0\\ P<\frac{k{r}^2}{l}\end{array}$

Refer to all the conditions,

The minimum value of P is 0.

The maximum value of P is $P_{\max }<\frac{k{r}^2}{l}\left(\frac{W}{\left(\frac{k{r}^2}{l}+W\right)}\right)$.

Substitute $20\text{lb/in}\text{.}$ for k, 6 in. for l, 15 lb for W, and 3 in. for r in Equation (4).

$\begin{array}{c}P<\frac{20\times 3^2}{6}\left(\frac{15}{\left(\frac{20\times 3^2}{6}+15\right)}\right)\\ <\frac{20\times 9}{6}\left(\frac{15}{45}\right)\\ <10\text{lb}\end{array}$

Therefore, the range of values of P for which the equilibrium position is stable is $\underset{\_}{P<10\text{lb}}$.

(b)

To determine

Find the range of values of P for which the equilibrium is stable.

Answer to Problem 10.100P

The range of values of P for which the equilibrium position is stable is $\underset{\_}{P<20\text{lb}}$.

Explanation of Solution

Given information:

The system is in equilibrium when ${\theta }_1={\theta }_2=0$.

The value of spring constant is $k=20\text{lb/in}\text{.}$.

The radius of the drums is $r=3\text{in}\text{.}$.

The length of the rods AB and CD is $l=6\text{in}\text{.}$.

The weight acting at point A is $W=60\text{lb}$.

Calculation:

Substitute $20\text{lb/in}\text{.}$ for k, 6 in. for l, 60 lb for W, and 3 in. for r in Equation (4).

$\begin{array}{c}P<\frac{20\times 3^2}{6}\left(\frac{60}{\left(\frac{20\times 3^2}{6}+60\right)}\right)\\ <\frac{20\times 9}{6}\left(\frac{60}{90}\right)\\ <20\text{lb}\end{array}$

Therefore, the range of values of P for which the equilibrium position is stable is $\underset{\_}{P<20\text{lb}}$.