Chapter 10.2, Problem 10.100P

Solve Prob. 10.99, assuming that the vertical force applied at B is increased to 5P.

*10.99 Bars AB and CD, each of length l and of negligible weight, are attached to a spring of constant k. The spring is 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. P10.99

To determine

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

Answer to Problem 10.100P

The range of values of P for which the equilibrium position is stable is $\underset{\_}{P<\frac{4kl}{5}}$.

Explanation of Solution

Given information:

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

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.

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

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

$y_B=-l\cos {\theta }_1$

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

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

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.

$\begin{array}{c}V=V_g+V_s\\ =5P{y}_B+P{y}_C+\frac{1}{2}k{\left(x_C-x_B\right)}^2\end{array}$

Here, the spring constant is k.

Substitute $-l\cos {\theta }_1$ for $y_B$, $l\cos {\theta }_2$ for $y_C$, $l\sin {\theta }_2$ for $x_C$, and $l\sin {\theta }_1$ for $x_B$.

$\begin{array}{c}V=5P\left(-l\cos {\theta }_1\right)+P\left(l\cos {\theta }_2\right)+\frac{1}{2}k{\left(l\sin {\theta }_2-l\sin {\theta }_1\right)}^2\\ =-5Pl\cos {\theta }_1+Pl\cos {\theta }_2+\frac{k{l}^2}{2}{\left(\sin {\theta }_2-\sin {\theta }_1\right)}^2\end{array}$

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 $\cos {\theta }_2$.

$V=-5Pl\left(1-\frac{{\theta }_1^2}{2}\right)+Pl\left(1-\frac{{\theta }_2^2}{2}\right)+\frac{k{l}^2}{2}{\left({\theta }_2-{\theta }_1\right)}^2$ (1)

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

$\begin{array}{c}\frac{\partial V}{\partial {\theta }_1}=-5Pl\left(-\frac{2{\theta }_1}{2}\right)+Pl\left(0\right)-\frac{2k{l}^2}{2}\left({\theta }_2-{\theta }_1\right)\\ =5Pl{\theta }_1-k{l}^2\left({\theta }_2-{\theta }_1\right)\end{array}$ (2)

Differentiate the Equation (2).

$\frac{{\partial }^{2}V}{\partial {\theta }_1^2}=5Pl+k{l}^2$

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 ${\theta }_2$.

$\begin{array}{c}\frac{\partial V}{\partial {\theta }_2}=-5Pl\left(0\right)+Pl\left(-\frac{2{\theta }_2}{2}\right)+\frac{2k{l}^2}{2}\left({\theta }_2-{\theta }_1\right)\\ =-Pl{\theta }_2+k{l}^2\left({\theta }_2-{\theta }_1\right)\end{array}$ (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).

$\begin{array}{c}\frac{\partial V}{\partial {\theta }_1}=5Pl\left(0\right)-k{l}^2\left(0-0\right)\\ =0\end{array}$

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

$\begin{array}{l}{\left(-k{l}^2\right)}^2-\left(5Pl+k{l}^2\right)\left(-Pl+k{l}^2\right)<0\\ k^2{l}^4+5P^2{l}^2-5Pk{l}^3+Pk{l}^3-k^2{l}^4<0\\ 5P-4kl<0\\ P<\frac{4kl}{5}\end{array}$

Condition 3:

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

Refer to all the conditions,

The minimum value of P is 0.

The maximum value of P is $P_{\max }<\frac{4kl}{5}$.

Therefore, the range of values of P for which the equilibrium position is stable is $\underset{\_}{P<\frac{4kl}{5}}$.