# Pressing hydraulic pressure questions

## Main Question or Discussion Point

Hi,
It's been a long time since I've been to school, and never did specialize in physics. But I now have a series of many questions that will build on each other I am hoping someone can help me with. I can only ask so man questions in each session as my next question set will depend on previous answers. And I'm looking for general understandings more than actual formulas.
Ok, Here it starts:
Suppose I have a hydraulic jack with a one-inch square piston to keep things simple. And suspended above this jack is a fixed plate.
-1- If I pump up this jack to 100 psi then there will be 100 pounds pressing against this top plate, correct?
-2- And if I place a 1 inch cube on top of the jack's piston, then pump the jack (I will always pump it to 100 psi and keep pumping so it stays at 100 psi unless I state otherwise) even if the cube is crushed slightly, there will still be 100 pounds of pressure squishing the cube and 100 pounds of pressure pressing against the top plate, correct?
-3- And if I put two cubes, one on top of the other, on top of the jack and pump it, each cube will be being squished by 100 pounds of pressure, and 100 pounds of pressure will be pressing against the top plate, correct?
-4- And if I put a 1 in. by 3 in. plate on top of the jack, (and assume the plate will not bend or fall off) and place the 1 in cube on top of this 3 in plate, pump the jack, then regardless where I place the cube on this plate, there will be 100 pounds of pressure squishing this cube against the top plate, is this correct?
-5- But if this is correct, then every sq in of that 3 in plate has a "squishing" potential of 100 pounds, correct?
-6- But if that's true, then without the cube, will that 100 psi jack be exerting 300 pounds of pressure against that top plate simply because I put a 3 in plate directly on top of the jack's piston? If true, it seems like we're getting more pressure from no-where with no additional jack pressure "work".
-7- And if I put two 1 in cubes directly on top of the 3 in plate, next to each other, will each cube be squished by 100 pounds or will the pressure spread and each cube be squished by 50 pounds? And the same question if I put 3 one-inch cubes on top of the plate - 100 pounds squishing each cube or 33.3 pounds? And the same question with one cube 3 in long.
-8- And will evenly stacking more cubes on top of each other change the answers?
I have lots more variations of questions on this "pressing" situation but I need to see the answers before I know which way to go with my next set of questions.
And thank you much for your help and education.
steve

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1.- yes
2.- yes
3.- yes
4.- theoretcilly correct, I guess....100 pounds, regardless of where you put the cube, BUT you put only 1 cube.
5.- no...the total force coming out of the jack is 100 pounds...if you extend the plate and actually use all of its surface, then the 100 pounds will distribute itself into a new pressure of 100/3.
6.- no
7.- the force re-distributes
8.- no

Thank you so much! Now I understand. Based on your answers I can skip a lot of my questions I had concerning pyramid shaped plates, etc.

-9- Based on my understanding, then a 1/2 in cube in the example above would be squished by 200 pounds of pressure, correct?

Let me jump to what may be the next-to final set of questions.
I have two very old presses I'm restoring for my own use. The only info I could find on one of the presses is an ad in a 1964 Analytical Chemistry magazine, otherwise no info is available from or about the manufacturer. I can replace the piston seals etc but the pressure gauge must also be replaced in each press.
The jack in the first press has a piston with an o.d. of 2.25 in and is stamped "20 Ton". The gauge it has with broken glass and a bent pointer states "Total Load in 1000 pounds" and a scale of 0 to 40. The press has a 8 in by 8 in plate, if that matters.

-10- What range psi gauge would I use and what formula would I use in reading the gauge to convert from psi to "Total Load"? I would like to mark the replacement gauge with the same scale.
-11- And am I correct in understanding that this press, at full gauge deflection, is putting 40000 / (8 x 8) pounds of pressure on each square inch of its plate, assuming I'm using the plate's entire surface?

And I have similar questions for another press with a jack stamped "12 Ton". Its broken gauge has two scales - "Metric Tons" 0 to 11 and "Pounds" 0 to 24000. The o.d. of its piston is 1.75 in diameter and it uses a 6 by 6 in plate.

-12- What range psi gauge would I use and what formula would I use in reading the gauge to convert from psi to "Pounds" and "Metric Tons"? I would like to mark the replacement psi gauge with the corresponding two scales.

Thank you again for your help and education, it is appreciated.
Steve

9.- Yes, the 1/2 inch cube will be subjected to a pressure of 200 psi...but that is only because a force of 100 pounds is being used within 0.5 square inches, and so...the pressure is 200 psi, but the total force still only is 100 pounds.
10.- I think you need to simply multiply by the cross-section area of the piston.
11.- yes
12.- from psi to pounds, multiply by the cross-sectional area of the piston; from pounds to metric tons...google it.

For the first part of answer to -10- is the following correct?
Diameter is 2.25 in. therefore Radius is 1.125 in. Therefore area is 3.14159 x 1.125 x 1.125 = 3.976 sq in.
Therefore to find the proper range psi gauge needed for this press is: 40000 / 3.976 = 10060
-13- Therefore I need a 0 to 10000 psi gauge for this press, correct?
-14- And to mark the gauge, I simply multiply the gauge psi reading by 3.976, correct? For example, next to the 2012 psi gauge marking I would write 8 for the "Total Load" reading of 8000 pounds, correct? (Total Load in 1000 Pounds) (2012 x 3.976 = 7999.7)

And for the first part of the answer to question -12- is the following correct?
Diameter is 1.75 in. therefore Radius is 0.875 in. Therefore area is 3.14159 x 0.875 x 0.875 = 2.4052 sq in.
Therefore to find the proper range psi gauge needed for this press is: 24000 / 2.4052 = 9978
-13- Therefore I also need a 0 to 10000 psi gauge for this press, correct?
-14- And to mark the gauge, I simply multiply the gauge psi reading by 2.4052, correct? For example, next to the 1663 psi gauge marking I would write 4 for the "Total Load" reading of 4000 pounds, correct? (Total Load in 1000 Pounds) (1663 x 2.4052 = 3999.8)

-15- And with 2204 pounds in a metric ton, I would use the following formula to mark the gauge in metric tons, correct?
psi reading x 2.4052 / 2204. For example, 6 metric tons would be marked at the 5498 psi gauge reading (6 tons = 5498 x 2.4052 / 2204)

Thanks.
Steve

looks good

Hi,
I mostly understand your answers, but for my application is there a difference between force and pressure and are they both measured in psi?
Thanks.
Steve

aaahhh! So, that's the problem...I thought that might be the case due to the way you spoke of them but manage to stay borderline correct.

I might not have too much time, so you might have to simply google the two terms.

But, basically, in the english system of units, force has units of pounds-force (lbf)...typically, people simply drop the 'force' part and just call them pounds, but this is incorrect BECAUSE pounds alone is meant to me pounds of mass (lbm)

Force is a force and is typically thought of being applied on a single point.

Pressure has units of pounds-per-square-inch...again, these are pounds-force.

Pressure is the application of a force over a given area, and so, the force gets diluted to a force per units area kind of thing.

Hello,
Do you have any idea of what the significance of seven additional markings on the 24000 pound gauge might be? They are labeled 1 to 7 and appear to be at the following readings: 3100, 6500, 9700, 12900, 16100, 19400, and 22600.
Thanks.
Steve