[Physics] Force amplification and Newton’s third law

newtonian-mechanics

Pascal's principle in hydraulics leads to force amplification which is a common feature in hydraulic brakes and hydraulic press. A small force applied on a piston of small cross sectional area produces a larger force on another piston of larger cross sectional area when they are connected by an incompressible fluid. How does one explain the validity of Newton's Third Law in this example?

Best Answer

Lets consider a situation which is in equilibrium, which consists of two pistons with different surface area which are connected to each other by an incompressible fluid. I will also assume that gravity does not play role, so no increase in pressure inside the fluid do to gravity. In this case the situation can be represented with the following drawing:

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Here the fluid is under constant pressure, $p$, and the tubing is connected to the ground. Thus the walls of the tubing, which house the fluid, also exert a force on the fluid. In the example the bends make it a little bit more complicated, but if you would straighten out all bends, then most forces exerted by the walls would cancel each other due to symmetry. The only force contribution from the walls, which are not cancelled due to symmetry, arise from the change in cross section area. In this situation the change in cross section area generates a force to the right. In the situation, in which all bends are straighten out, the resulting force, exerted by the walls of the tubing onto the fluid, will be transferred from the connection of the ground to the tubing. In case it should become clear that in the situation in which the force $F_1$ and $F_2$ cause the system to be in equilibrium that the sum of all forces add up to zero.

It can be noted that the work done by the walls of the tubing will always be zero (assuming that they do not expand and contract due to change in pressure). When neglecting friction and shear forces, then the work done by $F_1$ and $F_2$ will always add up to zero. This is similar to a gear box, which increases torque but decreases angular velocity, or a pulley system.

External forces and stress

Your figures are mistaken for two reasons. First, you draw many small arrows directed downward, as if pressure were a vector. It isn't. Pressure has no direction.

Second, your first figure applies to a solid, not to a fluid (liquid or gas). For solids Pascal's law doesn't hold, nor you may simply speak of pressure: the useful concept is stress tensor, a much more complicated thing.

Let's see however how reasoning would work in the solid case. You have a block of solid matter, in form of a truncated cone. Suppose it is resting on a table. Then resultant force on it must vanish. Which are the forces acting on the block? The first is its weight, vertical downward, resultant $m\vg$. The second is table's "reaction" $\vR$, actually a distributed force, applied all over the contact surface. Since net force is vanishing, we know that $\vR=-m\vg$. No forces exist on the oblique lateral face.

In a solid atoms cannot move freely. They may only effect small displacements from their equilibrium positions: thermal vibrations and average net displacements if external forces are acting as in our case. That the latter exist means that no solid is absolutely rigid. Such displacements are necessary to produce forces between atoms, which globally counteract external forces.

Let me explain better this delicate point. In static conditions not only the resultant of external forces must vanish, but the same must happen for whichever portion of the body. If you mentally isolate that portion, you will see it to be subjected to two kinds of forces:

force distributed all over the portion's volume (typically it is weight)
surface force, acting between atoms residing at the two sides of the portion's boundary.

It is the second kind of forces which at a macroscopic level are represented by the stress tensor. They are generally not normal to the boundary's surface, nor have the same intensity and direction from one point to another of the body.

A solid block

Suppose now you apply a force $\vF$, vertically downward, to the upper base of block. It is easy to understand that at equilibrium $\vR$ will change, becoming $$\vR = -m \vg - \vF.$$ Note that this is true independently of where exactly the force is applied. You may also split it in several sub-forces applied in different points. Or you may think of a force continuously distributed over the entire base. In the latter case it is customary to speak of an applied "pressure" and this the reason why pressure is often thought of (wrongly) as a vector.

What can be said about $\vR$? I mean, $\vR$ is the resultant of the forces the table applies to block's lower base. But how are these forces distributed? The question cannot be answered with the given data. In mechanics jargon, this is a "statically undetermined problem".

Caution: something more could be said by keeping into account the moments of applied forces. However my last statement remains true.

Of course since we do not know the exact distribution of forces in the lower base, the same happens for internal stresses in the block, both when $\vF=0$ and when $\vF\ne0$. This shows the intrinsic complexity of our problem when a solid body is concerned.

Fluid and pressure

And now let's come back (finally! you will say) to the case of a fluid. There are two main differences between this case and that of a solid. First, a fluid needs to be contained in a vessel (a closed one for a gas). Second, what defines a fluid for our problem is that surface forces can only be orthogonal to the surface.

The first difference entails a complication: we have to take into account forces applied by the vessel's lateral wall to the fluid. And if these are not given, it could seem that the problem becomes insoluble.

The second difference instead amounts to a great simplification, because internal stress can be wholly described by a single scalar: pressure. Let's see why is it so, from a microscopic point of view. A solid can transmit a shear force, i.e. a force parallel to the surface where it is applied. The reason is that atoms are only allowed small displacements around their equilibrium positions, but there is no constraint as to the displacement's direction. Therefore atoms near boundary between two portions of the body (see above) may well displace parallel to the surface and in opposite directions on opposite sides of it.

On the contrary atoms (or molecules) in a fluid are more or less free to move around, not being constrained in the vicinity of some point. As a consequence the fluid's bulk cannot resist to a shear stress: it immediately gives in and cancels the stress.

Then in a fluid only normal stresses are allowed. But there is more: it can be shown that when this situation prevails, then in a given point the intensity of (normal) force is always the same, whichever direction you choose for the boundary between two portions of the body. Shortly, we say that stresses are isotropic. Thus we have arrived at the pressure concept.

Pressure is not a force, has no direction (is a scalar). It does not act at the surface of a body, but is present in every internal point. In fact, it is well known that we may have pressure in an unbounded gas. The omnipresent instance is our atmosphere, but think of stars too: they are gigantic gas masses with no bounds, held together only thanks their own gravity.

The case of gases

It is well to open a short parenthesis to mark an important difference between liquids and gases. In liquids atoms are very near to each other, and what we see as macroscopic forces may be correctly interpreted as the resultant of microscopical forces between them.

Not so for gases, where distances are much higher, so that macroscopic forces are better seen as the effect of a myriad of collisions in which atoms exchange momentum one with another or with the vessel's walls.

No rôle for viscosity

Another clarification is in order. I spoke of shearing stresses and of their absence in fluids. Someone could think this is not so for a viscous fluid. After all, viscosity is just defined as the ability of real fluids to transmit shear stresses!

The answer is that we strictly bounded ourselves to static situations. Viscosity only acts when a fluid is moving; it is a force arising because parts of one fluid flow one wrt to another. The classic example is a river, whose water runs faster at centre, whereas grows slower nearing the bank, where it's still.

Therefore in static problems there is no need to restrict to non-viscous fluids. Viscosity has no effect.

To begin with, let's neglect gravity

To understand Pascal's law it's well to neglect gravity, at least initially. This could appear disconcerting, as there is a common misconception that pressure is due to gravity. Many people believe that atmospheric pressure is due to the weight of air above us (which in a sense is true) and conclude (erroneously) that pressure acts "from above". I'm afraid that not all introductory physics books are free from such sin.

Curiously enough, those people forget that they are continuously using objects which are counterexamples to that idea. I'm alluding to tires: of cars, bicycles, and so on. All these are inflated with a pump pressing air within. No rôle is played by atmosphere's weight.

A more exotic example is given by ISS, where a pressure is maintained to keep astronauts' breathing comfortably. Yet there is almost no air outside!

It is easy to show that in such situations pressure is the same in the whole volume of your block, irrespective of the force you can apply to movable parts of the vessel. "Wait a moment!" - I feel like I'm hearing - "Force? Which force? Why should be a force?

It's better to begin with gases, easier to understand, I believe. If the vessel is rigid no action is required by the experimenter. Everything stands still, nothing happens. But we know (I said it before) that the gas molecules continually hit the walls and rebound, giving them some momentum. More exactly, a definite amount of momentum per unit time and surface area. Momentum per unit time equals force. Force per unit area equals pressure. Therefore this momentum exchange is a measure of the gas pressure.

You may not notice that the wall is subjected to that force since usually, if vessel is sufficiently rigid, it automatically develops internal forces that counterbalance those due to gas and keep walls to move or deform. But sometimes things go differently: a balloon inflated at too high a pressure may blow. A welding in a metal tank may leak...

In other cases the vessel is built with a movable part to make experiments (the famous cylinder-with-piston of thermodynamics). This case is obvious: the piston is steady only if force due to gas pressure (force = pressure x area) is contrasted by an equal and opposite force applied from outside.

For a liquid too things go in an analogous way. Instead of collisions exchanging momentum we have forces between neighbouring molecules. Those near a wall interact with piston's molecules and directly apply forces to them. Result is the same: to keep piston steady an opposite external force is needed. The bigger the piston's area, the stronger the force.

If there are two pistons, the same argument applies to both, and we easily conclude that the force required to keep a piston still is proportional to its area. This is what was improperly called "force amplification".

Increasing pressure

We can also see things the other way around. Gas or liquid pressure is determined by the forces applied to piston(s). If you increase external force(s) the fluid will momentarily give in. If liquid, molecules will slightly get nearer one to another; this will increase the repulsive forces between, until a new equilibrium is reached. For a gas a volume reduction will result in a greater number of molecules per unit volume, thus augmenting the number of collisions per unit time against piston, i.e. augmenting pressure. Again, compression will halt when equilibrium is attained.

Gravity comes into play

We may not always neglect gravity. Not for atmosphere only: scuba divers know very well that underwater pressure increases by one atmosphere every ten meters of depth. This contradicts what I said before about pressure being the same at all points in a (still) fluid. The increment of pressure is too easily attributed to weight of the water column above. Sometimes this works, other times doesn't: see hydrostatic paradox in the internet.

Actually what can be experimentally verified is the following law (Stevin): in a fluid at equilibrium in a uniform gravitational field $\vg$ the pressure difference between two any points is $$p_1 - p_2 = \rho\,g\,(z_1 - z_2)$$ if $z$-axis is oriented like $\vg$. Just a simple example: if fluid is water ($\rho=10^3\,\mathrm{kg/m^3}$) and $z_1-z_2=10\,\mathrm m\,$ then $p_1-p_2=98\,\mathrm{kPa}$ which is about 1 atm.

Of course Stevin's law is a consequence of fluid equilibrium under internal and external forces already discussed, with gravity added. I can't dwell on the proof, however.

What about energy?

I cant't close this extra long post without answering the above question. It would be off topic if I'd used really static arguments, but this is not so, since in several places I spoke of "displacements". And when something is displaced with a force applied to it, work is involved and therefore energy.

Here again a distinction must be made between gases and liquids, because of their very different compressibilities. Under ordinary pressures liquids may be assumed incompressible without significant error. This is far from true for gases.

Of course, even an incompressible liquid can accomplish important displacements. Incompressibility only means that overall volume does not change. If the vessel has two pistons of different surface areas it's easy to see that proportionality of force to area, derived above, together with invariable volume, entail that works of external forces on pistons are equal, save for sign. So total work done is zero and liquid's energy does not change.

We could have reasoned in the converse: since energy must be conserved, total work by external forces must be zero, then force is proportional to area. But this argument has a flaw: work is done not only by external forces. Internal forces too can do work. So we have to prove that work of internal forces vanishes. This is not too easy, and requires an inquiry at microscopic level. It is better to assume it as a characteristic property of an incompressible fluid: no work is required to displace it in any way (until kinetic energy is negligible).

Note: I hope reader did notice that my energy argument rested on a hypothesis: equality of pressures on both pistons. But we have seen that this is not true in presence of gravity, if pistons are located at different heights. Let me set aside this complication for now.

As to gases, constant volume cannot be assumed. Nothing forbids to compress or dilate a gas. Furthermore, for gases temperature too becomes important in this respect. But you are not expecting from me a treatise on mechanics and thermodynamics of fluids ... did you?

When gas volume changes, work is done on it by external forces. The relevant formula is well known: $W=-p\,\Delta V$. As I wrote it, this formula is not generally true: it holds if $p$ stays constant during displacement. Otherwise we should write an integral: $$W = -\int_A^B \!\!p\,dV.$$ And this too requires that during transformation $p$, even if not constant, is well defined and the same all over gas volume. Also remember that if volume varies and work is done, to keep $p$ constant energy must be drawn from or given to gas as heat flowing through walls.

Nothing more about gases. A short comment on what happens if pistons are placed at different heights in a vessel containing a heavy fluid. In this case pressures are different, and proportionality between force and area is not respected. Then if pistons are moved, although liquid's volume stays constant, work is done. A positive work if lower piston is moved into liquid, higher piston in the opposite direction.

Question: positive work means liquid gained energy. Where is it to be found? Easy answer: liquid was generally lifted. More precisely, its c.o.m. was lifted. Then liquid's potential energy in the gravity field has increased. For a simple geometry, e.g. a parallelepiped vessel, it is easy to prove that work equates increment of P.E. For a general shape this holds still true, but the proof is more involved.

To conclude. We may say that energy plays no relevant rôle in relation to Pascal's law. There is no "pressure energy" in a liquid, although you will easily find such expression about Bernoulli's theorem (which is out of my actual aim). Just to unravel the mystery: what is improperly called pressure energy is enthalpy density.