# Force amplification and Newton's third law

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?

• In equilibrium, the net force on the lever is zero. Whereas, in this example the net force on the fluid at equilibrium isn't zero. – Ajit May 14 '15 at 13:46
• I missundertood your question – user66432 May 14 '15 at 13:50
• No prob, @brucesmitherson. I totally agree with your earlier answer. I was thinking about a chain of fluid molecules starting from the smaller piston to the larger piston and how one explains Newton's Third Law at each molecular level. – Ajit May 14 '15 at 13:52
• It feels absurd to say the system is in equilibrium, yet the net force isn't zero. I offer an explanation. The cylinder containing the fluid and the pistons needs to be attached to another body to prevent it from accelerating under the unbalanced forces on the two pistons. This external body neutralises the unbalanced force by providing an equal and opposite force. This force provided by the body accounts for the force amplification. Yet, I still want to see the picture at the molecular level. – Ajit May 14 '15 at 14:31

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:

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.

• Appreciate your comment. Let us consider the cylinder with the fluid and pistons as the system. In the situation in which all bends are straighten out, the forces F1 and F2 do not cancel each other. There is a net force acting on the system towards the left. This unbalanced force can be compensated only by attaching the system to another body. Yes, the net work by the two pistons is zero. – Ajit May 14 '15 at 14:44
• @Ajit I forgot to mention that the tubing is assumed to be connected to the ground, which in turn will provide the reaction force. – fibonatic May 14 '15 at 14:58

Let's say the small piston has inner surface 1 centimeter^2, and the large piston has inner surface 5 centimeter^2. Let's say the pressure in the system is 1 Pascal (1 Newton/meter^2). The pressure per cubic centimeter over ALL surfaces inside the system is the same. The pressure of 1 Pascal at the small piston is also 1 Pascal at the large piston.

BUT, the pressure at the large piston is applied over a larger surface area, so the total force exerted by pressure on the large piston is 5 times the total force generated by pressure at the small piston. Both pistons move, and the overall pressure inside the system remains 1 Pascal.

Newton's Third law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body. As the pressure inside the system remains the same on all surfaces, the large piston does indeed exert an equal pressure on the small piston PER UNIT OF SURFACE AREA.

Pascal's principle describes a force multiplier, like a lever. But Newton's Third Law says there is no free lunch. In order to move the large piston 1 centimeter, the small piston must be moved 5 centimeters. The amount of work done by both pistons is the same, but the large piston moves a lesser distance than the small piston.

Here is a graphic explanation: http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html. Scroll down to the second box.