I'm trying to grab the concept of the hydraulic Lift. Someone tell me if this is correct.
1. Work does not change and to compensate this the force is going to be larger (F2 >F1) and the distance is going to be smaller on the recieving end (d1> d2).
My only real problem is if the area for the second piston is larger than the area of the first how does is the F2 larger than F1, according to the equation?
P = F/A, because pressure change is constant throughout.
Pressure change is constant throughout only at any given depth in a hydrostatic situation where there is only one fluid.
Okay, here is the pressure concept in a nutshell:
Bernoulli's equation is a conservation of energy density equation. The significant components to that equation for the MCAT are:
(1) Pressure (P)
(2) Kinetic Energy Density (1/2 * density * v^2)
(3) Gravitation Energy Density (density * g * h)
SPATIALLY at any given instant total energy density is conserved! So:
P + KE/vol + PE/vol = k = constant; I.e. there must be some transfer of energy from one component to the other if a kinetic, gravitational, or pressure term is changed (generally we see pressure terms change in Bernoulli's equation when there is friction).
Now that that issue is cleared up, I'll go onto your question.
If you have a very small area it takes a very small force increase to increase the overall pressure, correct?
P=F/A
We know that energy density will distribute equally, so if we apply a small force to a small area then that pressure at the interface of that small area will have to equal the pressure at the interface of the large area, correct (assuming the two are at about an equal height). So, if P1 = P2 but P1 has a small F and a small A, but P2 has a really big A, then for the ration to hold the F has to be huge. That is where this comes from:
F1/A1 = F2/A2
Assuming no friction between the fluid and the pipes we know that W=F*d where d is the distance the fluid has moved through the hydraulic system. Because of the different areas, the two "areas" have different forces, but the same amount of work is done at both ends:
F1d1 = F2d2
So, if point 1is the small end that requires the small force, the small force must be applied over a larger distance in order for the work to equal F2d2.
