Chapter 10.2, Problem 10.98P

Solve Prob. 10.97 knowing that l = 30 in. and k = 50 lb/in.

*10.97 Bars AB and BC, each with a length l and of negligible weight, are attached to two springs, each of constant k. The springs are undeformed, and the system is in equilibrium when θ₁ = θ₂ = 0.

Determine the range of values of P for which the equilibrium position is stable.

Fig. P10.97

To determine

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

Answer to Problem 10.98P

The range of values of P for which the equilibrium position is stable is $\underset{\_}{P<573\text{lb}}$.

Explanation of Solution

Given information:

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

The length of the bars AB and BC is $l=30\text{in}\text{.}$

The spring constant is $k=50\text{lb/in}\text{.}$

Calculation:

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

Find the horizontal distance $\left(x_B\right)$ using the relation.

$x_B=l\sin {\theta }_1$

Find the horizontal distance $\left(x_C\right)$ using the relation.

$x_C=l\sin {\theta }_1+l\sin {\theta }_2$

Find the vertical distance $\left(y_C\right)$ using the relation.

$y_C=l\cos {\theta }_1+l\cos {\theta }_2$

When the values are small,

$\begin{array}{l}\sin {\theta }_1\approx {\theta }_1;\sin {\theta }_2\approx {\theta }_2\\ \cos {\theta }_1\approx 1-\frac{{\theta }_1^2}{2};\cos {\theta }_2\approx 1-\frac{{\theta }_2^2}{2}\end{array}$

Find the potential energy (V) using the relation.

$\begin{array}{c}V=V_g+V_s\\ =P{y}_C+\frac{1}{2}k{x}_B^2+\frac{1}{2}k{x}_C^2\end{array}$

Here, the magnitude of the force applied at C is P and the spring constant is k.

Substitute $\left(l\cos {\theta }_1+l\cos {\theta }_2\right)$ for $y_C$, $l\sin {\theta }_1$ for $x_B$, and $\left(l\sin {\theta }_1+l\sin {\theta }_2\right)$ for $x_C$.

$V=P\left(l\cos {\theta }_1+l\cos {\theta }_2\right)+\frac{1}{2}k{\left(l\sin {\theta }_1\right)}^2+\frac{1}{2}k{\left(l\sin {\theta }_1+l\sin {\theta }_2\right)}^2$

Substitute ${\theta }_1$ for $\sin {\theta }_1$, ${\theta }_2$ for $\sin {\theta }_2$, $\left(1-\frac{{\theta }_1^2}{2}\right)$ for $\cos {\theta }_1$, and $\left(1-\frac{{\theta }_2^2}{2}\right)$ for $\cos {\theta }_2$.

$\begin{array}{c}V=P\left(l\left(1-\frac{{\theta }_1^2}{2}\right)+l\left(1-\frac{{\theta }_2^2}{2}\right)\right)+\frac{1}{2}k{\left(l{\theta }_1\right)}^2+\frac{1}{2}k{\left(l{\theta }_1+l{\theta }_2\right)}^2\\ =Pl\left(1-\frac{{\theta }_1^2}{2}+1-\frac{{\theta }_2^2}{2}\right)+\frac{1}{2}k{l}^2\left({\theta }_1^2+{\left({\theta }_1+{\theta }_2\right)}^2\right)\end{array}$ (1)

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

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

Differentiate the Equation (2).

$\frac{{\partial }^{2}V}{\partial {\theta }_1^2}=-Pl+2k{l}^2$

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

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

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

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

Differentiate the Equation;

$\frac{{\partial }^{2}V}{\partial {\theta }_2^2}=-Pl+k{l}^2$

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).

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

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

$\begin{array}{c}\frac{\partial V}{\partial {\theta }_2}=-Pl\left(0\right)+k{l}^2\left(0+0\right)\\ =0\end{array}$

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

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

$\begin{array}{l}{\left(k{l}^2\right)}^2-\left(-Pl+2k{l}^2\right)\left(-Pl+k{l}^2\right)<0\\ k^2{l}^4-P^2{l}^2+Pk{l}^3+2Pk{l}^3-2k^2{l}^4<0\\ -P^2+3Pkl-k^2{l}^2<0\\ P^2-3Pkl+k^2{l}^2>0\end{array}$

Solve the equation using the mathematical equation.

$\begin{array}{c}P<\frac{3-\sqrt{5}}{2}kl\\ <0.382kl\end{array}$

Condition 3:

Check the condition;

$\begin{array}{l}\frac{{\partial }^{2}V}{\partial {\theta }_1^2}>0\\ -Pl+2k{l}^2>0\\ -P+2kl>0\\ P<2kl\end{array}$

Condition 4:

$\begin{array}{l}\frac{{\partial }^{2}V}{\partial {\theta }_2^2}>0\\ -Pl+k{l}^2>0\\ -P+kl>0\\ P<kl\end{array}$

Refer to all the conditions,

The minimum value of P is 0.

The maximum value of P is $P_{\max }<0.382kl$.

Substitute $50\text{lb/in}\text{.}$ for k and 30 in. for l.

$\begin{array}{c}{P}_{\max }<0.382\times 50\times 30\\ <573\text{lb}\end{array}$

Therefore, the range of values of P for which the equilibrium position is stable is $\underset{\_}{P<573\text{lb}}$.