# Interweaving the Principle of Least Potential Energy in School and Introductory University Physics Courses

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## Abstract

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## 1. Introduction

## 2. Description of the Principle of Least Potential Energy

## 3. Examples Demonstrating the Use of the Least Potential Energy Principle

#### 3.1. Example a: Fluid Static in a Piston of Small Cross-Sectional Area

#### 3.1.1. Solution Based on Newton’s Laws

#### 3.1.2. Solution Based on the Least Potential Energy Principle

#### 3.2. Example b: A System of Point Charges

#### 3.2.1. Solution Based on Newton’s Laws

#### 3.2.2. Solution Based on the Least Potential Energy Principle

#### 3.3. Example c: The Catenary Problem

#### 3.3.1. Solution Based on Newton’s Laws

#### 3.3.2. Solution Based on the Least Potential Energy Principle

## 4. Conclusions

## Author Contributions

## Conflicts of Interest

## References

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**Figure 1.**Schematic illustration of the hydrostatic problem. We have to divide ${A}_{0}$ volume of water between two vessels with basis areas ${A}_{1}$ and ${A}_{2}$ such that the total potential energy of the water will be minimum.

**Figure 2.**Four positive electric charges are fixed in the corner of a square. We have to find the place where a fifth positive charge will be in equilibrium. The repulsive forces acting on the charge are illustrated schematically in the diagram.

**Figure 3.**Contour plots showing the energy function $U(x,y)-{U}_{0}$ for $k=1$ and $a=1$. In the first row (from left to right), we plot the symmetric case ${q}_{i}=1,\text{}\mathrm{for}\text{}i=1,\dots ,5$, and then the asymmetric case $({q}_{1}=1.1)$. In the second row, we zoom to the relevant minimum area. The values of the minima points can be found using Mathematica (or similar mathematical package) and are 5.657 and 5.933 for the symmetric and asymmetric cases, respectively.

**Figure 4.**A chain hanging in a catenary form against the background of the Tzin Valley in central part of the Israeli Negev.

**Figure 5.**The forces acting on a small chain element $ds$ spans a horizontal distance of $dx$. The gravity force is given by $ds\rho g$ where $\rho $ is chain’s constant mass per unit length and $g$ is the gravity acceleration. The tensions forces are ${T}_{A}$ and ${T}_{B}$, the direction of which is along the chain.

**Figure 6.**The solution of the two-segment catenary problem, which can be obtained from geometric constraints and symmetry considerations. The rhomboid sign marks the center of mass of the chain.

**Figure 7.**Schematic illustration of the four-segment chain. Consider a uniform chain ($l=1.4\text{}\mathrm{m}$) which is suspended between two same level points $A(0,1)$and $B(1,1)$. Find points $C({x}_{1},{y}_{1})$and $D(0.5,{y}_{2})$ that minimize the potential energy of the chain (Figure 8).

**Figure 8.**Numerical solution of the four-segments chain. The rhomboid sign marks the center of mass of the chain. It can be seen that compared to the two-segment solution its position is lower.

**Figure 9.**Successive solutions of the chain, which consists of 16 segments, show how the solutions become closer to the catenary form.

**Figure 10.**The exact solution of the problem (in black) which is a hyperbolic cosine function given by $y=a\mathrm{cosh}((x-0.5)/a)+b$ where $a=0.340564$ and $b=0.221546$ and the fourth order approximation (in red). Galileo’s solution to the problem given by the parabola $y(x)=1.788{(x-0.5)}^{2}+0.553$ with the same length $l=1.4\text{}\mathrm{m}$ is plotted in blue. It can be seen that fourth order approximation is very close to the exact solution (there is a slight difference at the bottom of the curve).

**Figure 11.**The real catenary solution of the problem ($\rho =0.132\text{}\mathrm{kg}/\mathrm{m}$). The measured sag is 42 cm. whereas the computed sag at the fourth order is 43.9 cm.

**Figure 12.**(

**a**) The total potential energy of the chain (in Joules) for a chain with $\rho =0.132\text{}\mathrm{kg}/\mathrm{m}$ and for $g=9.8{\text{}\mathrm{m}/\mathrm{s}}^{2}$ for the different stages of the approximations; and (

**b**) the 16-line chain. The cross mark indicates the center of mass, whereas the plus mark is the location of the center of mass for $n=2$.

**Figure 13.**An asymmetrical catenary formed by two equal length (0.7 m) chains with different mass densities ${\rho}_{1}=0.132$ kg/m and ${\rho}_{2}=0.589$. Note that the chain shifts to the heavier side.

© 2017 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( http://creativecommons.org/licenses/by/4.0/).

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**MDPI and ACS Style**

Ben-Abu, Y.; Eshach, H.; Yizhaq, H.
Interweaving the Principle of Least Potential Energy in School and Introductory University Physics Courses. *Symmetry* **2017**, *9*, 45.
https://doi.org/10.3390/sym9030045

**AMA Style**

Ben-Abu Y, Eshach H, Yizhaq H.
Interweaving the Principle of Least Potential Energy in School and Introductory University Physics Courses. *Symmetry*. 2017; 9(3):45.
https://doi.org/10.3390/sym9030045

**Chicago/Turabian Style**

Ben-Abu, Yuval, Haim Eshach, and Hezi Yizhaq.
2017. "Interweaving the Principle of Least Potential Energy in School and Introductory University Physics Courses" *Symmetry* 9, no. 3: 45.
https://doi.org/10.3390/sym9030045