## Visualizing Elementary Calculus: Statics

This post assumes a little physics, specifically the relationship between work and energy.

This series:

### Equilibrium

In physics, we sometimes like to look at stuff that isn’t doing anything. This is called “statics”. It’s kind of boring after a while, which is why you would only take an entire course on it if you’re an engineer.

If something is stationary, there must be no net force on it. That means that if you move it around a little bit, the work done is zero and its energy doesn’t change. This is called the principle of virtual work. The result is that when things aren’t moving, they are generally at a minimum of potential energy. In theory they could be at a maximum or other stationary point, but these equilibrium states are unstable – the difference between a ball nestled securely at the bottom of a valley and perched precariously at the top of a mountain.

### Tubes

Here’s a picture of a disappointing jug of milk.

The water rises to the same height in the handle and in the main body. It’ll do this even if you make a hundred little tubes, all with different shapes, even if they’re curved around in strange ways. How does the water in one tube know how high the water is in all the others?

The water must be at a height such that a small movement of water from the handle to the body would cause no change in potential energy. Since the potential energy is a function of height, the water must the be same height everywhere, else we would be able to release energy by moving some water from high to low.

### Hill and Chain

There’s a lumpy hill with a chain lying on it. When will the chain be stationary, assuming the hill is frictionless?

The condition is that the potential energy of the chain shouldn’t change if you move it a little left or right. Assuming the chain is uniform, moving it a little bit to the right is identical to chopping a little bit off the left hand side and moving it all the way over to the right.

This chopping operation isn’t supposed to change the potential energy, so the left and right hand sides of the chain must be at the same height. That’s the condition for stability.

Unlike this water in the jug, this is an unstable equilibrium because if the chain is just a little bit off, it will fall towards the side that’s already further down, making that side drop down even further, and the chain will get further out of equilibrium.

### Push Ups

I have often been asked why it is much easier to do a push up than to bench press your body weight. The motion of your arms is essentially the same, so shouldn’t the tasks be about equally difficult?

A push up, unlike a bench press, is a sort of lever. We can model it like this:

The horizontal brown stick is a board – your body as you do the push up. The triangle is the fulcrum – your feet. The gray ball represents your body weight. Your true weight is actually distributed from your feet to your head, but the ball represents an average, called your center of mass. The green arrow represents the force from your arms on your body.

A force is just a force, regardless of where it comes from, so instead of your arms, we’ll imagine the same force is generated by a mass on a pulley.

The blue circle is a pulley. There’s a rope tied to the end of the board that goes around the pulley and attaches to a platform with a weight on it.

In order for the system to be in equilibrium, the pulley needs to minimize its potential energy, so it must be at a stationary point. We would like $\textrm{d}U/\textrm{d}\theta = 0$, with $U$ the potential energy of the system and $\theta$ the angle of the board with the horizontal.

$U$ changes when the weights change height. If we slant the board up by a small angle, the weight on the board will go up and the weight on the platform will go down. We need the resulting changes in potential energy to cancel each other.

Let the distance from the fulcrum to the weight on the board (the body weight) be $L_B$, and the body weight itself $w_B$. $U_B = w_B h_B$, with $h_B$ the height of the weight. Then $\textrm{d}U_B = L_B w_B \textrm{d}\theta$. (Draw a picture to see this. Also, try finding some of the assumptions being used). Similarly $\textrm{d}U_w = L_w w_w \textrm{d}\theta$. These need to be equal, so

$L_B w_B = L_w w_w$

$w_w = w_B \frac{L_B}{L_w}$

Your center of mass is maybe 70% of the way to your shoulders, so doing a push up requires a force about 70% of your body weight.

### Bridge

Sometimes it’s easy to work with energy, but other times it’s easier to work with forces. If we hang a chain between two points, we could find its shape by minimizing its energy, subject to the constraint that the length is constant. This requires the more-advanced calculus of variations.

On the other hand, we can still analyze the hanging chain in terms of force with elementary calculus.

That’s the chain, hanging between two posts. We’ll zoom in on just the red part.

$\textrm{d}x$ and $\textrm{d}y$ show the length and width of the entire segment. There are two tension forces, $T_1$ and $T_2$, acting on the segment, and their components are shown in the picture.

The segment doesn’t go left or right, so $T_{1x}$ = $T_2x$. The segment doesn’t go up or down, either, so the vertical components of the tension must balance gravity. That means $\textrm{d}T_y = \sqrt{\textrm{d}x^2 + \textrm{d}y^2}\lambda g$, with $g$ the acceleration due to gravity and $\lambda$ the mass per unit length of the segment.

Tension must also point along the direction of the rope, so $T_y/T_x = \textrm{d}y/\textrm{d}x$.

Combining this algebraically gives

$\frac{\textrm{d}T_y}{\textrm{d}x} = \sqrt{1 + (T_y/T_x)^2} g \lambda$

The trick is then to figure out what function satisfies this equation. A trig function looks good because of the relation $\sin(\theta) = \sqrt{1 - \cos^2\theta}$. In fact, the solution is

$\frac{T_y}{T_x} = \frac{\textrm{d}y}{\textrm{d}x} = \sinh \left( \frac{g\lambda}{T_x}x \right) + C_1$

$y = \frac{T_x}{g\lambda} \cosh\left( \frac{g\lambda}{T_x}x \right) + C_1 x + C_2$

This is called a catenary curve.

### Exercises

1. Rework the hanging chain problem, where the chain is a suspension bridge. Imagine the chain itself with no weight, and the bridge having constant density. This is equivalent to taking the mass of the segment to being simply $\lambda g \textrm{d}x$ rather than $\lambda g \sqrt{\textrm{d}x^2 + \textrm{d}y^2}$. (Answer: a parabola)
2. Find the shape of a hanging spring of zero rest length. In this case, the total tension is inversely proportional to the mass density. (Answer: also a parabola)