Hydraulic Physics (How Hydraulic Jacks Work)

I started looking at a hydraulic trolley jack and got to thinking it would make a good physics video. In this video we will be going over how the hydraulic trolley jack works and the physics behind hydraulics.
So lets begin by going over how a hydraulic trolley jack works.The way the hydraulic trolley jack works is you pump the lever arm which is a second class lever. This lever drives down the small piston which pumps hydraulic fluid through a one way valve also known as a check valve. After the lever is raised again more fluid is allowed to go into small piston cylinder. The hydraulic fluid pressure builds up in a chamber with a larger piston until the pressure is great enough to push up on the load arm of the jack
Now we talked about levers in a previous video so I am only going to concentrate on the hydraulic portion of this trolley jack.
So lets say we have a small piston with diameter of 12.7 mm (half inch) and a 10 mm stroke. This piston is connected to a larger piston which has a diameter of 25.4 mm (one inch). How far will the large piston move after the small piston strokes down 10 mm.
To solve this problem we can take the volume of the small pistons stroke and divide by the area of the large piston. Doing unit analysis we see that dividing volume by area gives us our desired units of mm length.
So after pluging in our numbers we notice that we really are taking the stroke length times the diameter squared of the small piston over the diameter squared of the large piston. All other numbers cancel out. In this example it results in a stroke distance at the large piston of 2.5 mm. So the displacement is reduced to a quarter of the small pistons displacement. What have we learned from previous simple machine videos about what happens when the displacement is reduced.
That right the force is increased. In this example if we take the diameter of the large piston squared over the diameter of the small piston squared we get the mechanical advantage. Note we really are taking area over area but all other units get canceled out.
In this case we have a mechanical advantage of 4. This means that the force applied at the small piston is multiplied times 4 at the large piston
The way this works is that the fluid is evenly distributing a pressure. Being there is a larger area on the large piston more area is affected by the pressure. You can think of it as each circle has so many squares of force on them and each square equals one unit of force
Completing unit analysis we see that taking the area times the pressure we are left with force. This greater force on the pistoncomes at the cost of having less displacement
That concludes this video hopefully I have earned a share like or subscription

Disclaimer
These videos are intended for educational purposes only (students trying to pass a class) If you design or build something based off of these videos you do so at your own risk. I am not a professional engineer and this should not be considered engineering advice. Consult an engineer if you feel you may put someone at risk.