88

The anthropomorphic formulation "tries to" is misleading. Under the effect of ambient noise, matter explores the possible configurations around its current state: e.g., two single hydrogen atoms wiggle around and meet. If they happen to bind, this releases energy which goes away, and we say that the energetic state of this new $H_2$ molecule is lower than ...


77

Take a look at this picture of a cup slightly out-of-balance : In case (A), generated torque is directed out of your reference axis and in case (B) - towards your reference axis. So in case A), you need to compensate out of balance movement with your finger contra-movement. But in case B), torque assists you and makes balancing for yourself, so that you ...


45

Consolidating some of the points made in the answers to the question you linked, and comments: When constructing a chair, 4 legs is easy when you use traditional (wooden) construction - 90 degree angles, and easy to make stackable. A little bit harder than three legs because you have to make sure they are all the same length (or the chair will wobble). Once ...


43

You exchange heat with the objects around you. If the objects around you are hotter than you, you'll heat up. If the objects around you are cooler than you (neglecting the heat you're generating due to metabolic processes), you'll cool off. In space, the objects around you (mostly interstellar medium) is cooler than you so you radiate more heat away from ...


26

The question doesn't make sense, as not every system has to be in thermal equilibrium. The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, since by statistical mechanics, the position of the ball must obey the Boltzmann distribution."


26

In the centre of a bowl there is equilibrium. Put a ping pong ball in it. If this ball is ever shaken slightly away from equilibrium, it will immediately roll back. A "shake-proof" equilibrium is called stable. Now, turn the bowl around and put the ball on the top. This is an equilibrium. But at the slightest shake, the ball rolls down. This "non-shake-...


25

It entirely depends on what you think "thermodynamics" is. The traditional idea of thermodynamics dealing with systems whose macrostate can be fully described by e.g. temperature, pressure and volume indeed only applies to systems in equilibrium. Of course, as an approximation it also applies to systems "not far" from equilibrium, for some suitable notion ...


24

"A state of rest" is a relative term. Relative means - measured in comparison to the things around it. When you sit in a train and sip from a cup of coffee, you can do so because the cup is still relative to you even though both of you might be hurtling through the countryside at 200 km/h. For most experiments, objects can be considered "at rest" if they ...


22

This is a consequence of the second law of thermodynamics, which states that In a closed system with fixed internal energy (i.e. an isolated system), entropy is maximized at equilibrium. It can be shown that this statement is equivalent to the following: In a closed system with fixed entropy, the energy is minimized at equilibrium. Callen in his ...


21

It is wrong to think potential energy is stored in the object. The earth pulls the object down, but the object pulls the earth up. They share the potential energy. The object fails to fall down because the tabletop pushes it up. The earth fails to fall up because the bottom of the table legs push the earth down. The table pushes up and down because it is ...


21

Indeed, you need friction. Otherwise this object will slide to equator just as you expect. In fact, all the objects will slide there to form equatorial bulge and eventually the sphere will be turned into new shape with surface orthogonal to $\vec{g}+\omega^2 \vec{r}$ at each point. That is exactly what happens to planets in reality while they are still ...


18

This is really a statistical effect, as pretty much all of thermodynamics. You have two free hydrogen atoms. They tend to move around the space they have, and when conditions are favourable (there's enough energy, the atoms come "close enough" together), they might interact - chemically or otherwise. Now, "enough energy" is the important bit here. When a ...


15

Maybe because when the cup is the right way up, it’s centre-of-mass is above the point on your finger meaning that as your finger tries to balance the cup any small motion will generate a torque about this c.o.m making it harder to balance. When the cup is upside down, you have your finger on or going through the c.o.m and so any small motion by your finger ...


14

I'm going to take a slightly different approach and say it's because we defined energy to make it so. In other words, systems "try" to find the lowest energy state because energy is a concept humans invented in order to describe what we observe. This is the reason that for any given set of constraints, you might need a different "energy" to describe the ...


14

You have to be careful to distinguish between microstates and macrostates. Thermodynamic equilibrium is a macrostate which consists of a mixture of all possible microstates of energy $E$ weighted by a Boltzmann weight $e^{- \beta E} / Z$. A state in macroscopic thermal equilibrium can be thought of as "moving through phase space" ergodically (i.e. the ...


12

The reason is, quite simply, that the walls of the container exert a force on the fluid given by the pressure at each point, and orthogonal to the container's surface. In your second container, the wall is slanted and therefore provides an upward force which helps stabilize the fluid elements next to it. If you draw a free-body diagram of the bit of fluid ...


12

"Equilibrium" means thermal equilibrium. The solid has one well defined temperature, and a constant Fermi energy. The Fermi energy is an energy value against which energy levels are compared to determine how fully occupied (or not) an energy level is. Generally when the Fermi level is constant throughout a solid electrons diffuse equally in all directions....


12

Sketched proof of all possible Lagrange points: Consider first the 2-body problem. Deduce that possible Lagrange points must lie in the orbital plane (because a probe will always be gravitationally attracted towards the orbital plane). So from now on we restrict attention to the orbital plane, which we identify with the complex plane $\mathbb{C}$. ...


11

To answer your question, you should first understand when is a system most stable. Firstly it shouldn't have a tendency to move or change state, thus it should be under equilibrium conditions, i.e. the net Force should be zero. We know that $$F = - \frac{dU}{dx}$$ Putting $F=0$, we get $$\frac{dU}{dx}=0 \tag{1}$$ Secondly, it should be able to maintain ...


11

This is a very interesting, important and at the same time subtle matter which is useful not only in thermodynamics, but other areas as well, and not everyone quite understands it. First off, two statements that you'd better just memorize for now: In a reversible process, equilibrium is never left. Yes, that is a paradox. That's why all real processes are ...


11

Let us look at a one-dimensional example: Recall that $f(x) = \dot{x}$, so $f$ encodes the time evolution of $x$. If $f < 0$, then $x$ will move to the left. If $f > 0$, then $x$ will move to the right. If $f = 0$, $x$ will not move at all, this is why $f(x_0) = 0$ is the equilibrium condition. Now, look what happens if you perturb the equilibria $...


11

does it also leads to ... No, it doesn't. A simple Counterexample: Consider the figure below (the bar $\textrm {AB}$ and forces $F$ are on a plane parallel to $\textrm {xy}$ plane) We have $\Sigma \vec F=\vec 0$, but, if we calculate vector sum of torques about point $\textrm A$ we will obtain $\Sigma \vec M_A=F (\overline{AB})\vec k\neq \vec 0$ ($\vec k$...


11

Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.


11

If the centre of gravity of the object is vertically above the edge of the table then the object is in equilibrium. However, this equilibrium position is unstable (like a pencil balanced on its point) because a small tilt of the object will lower the centre of gravity, which will then cause the tilt to increase. This is a positive feedback loop. However, if ...


10

If you made the most perfect cone possible, so that its tip was a single atom, and stood it on the most perfect surface possible (a perfectly smooth, perfectly hard sheet of atoms), and completely removed all forces other than gravity, it would still topple. This is because those atoms are all jiggling around due to thermal motion. This effect fundamentally ...


10

A system in perfect equilibrium is not perfectly stationary in space or time. But, any change in the system is offset by another change such that there is no net change. Let's look at a simple example. Take a perfectly insulated box isolated from the environment and fill it with a gas. The gas molecules are all bouncing around, hitting each other and the ...


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