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Imagine that there's a force that is given to you in $psi$, say $1450\ psi$. And there's some body of weight say, $100\ kg$.

What I want to know is that, how much $psi$ will I need to lift the "body" above the ground? Is there any formula to calculate the same?

Moreover, I'd like to know that how much $kg$ can $1450\ psi$ lift above the ground?

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closed as off-topic by Chris, stafusa, Kyle Kanos, tom, M. Enns Apr 10 '18 at 15:58

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    $\begingroup$ PSI is a unit of pressure not of force. As the name says it is "pounds of force per square inch", if you have a piston of the area of 1 square inch, with a pressure of one PSI a force of 1 pound will be exerted on the piston. $\endgroup$ – Sebastian Riese Jul 19 '15 at 16:57
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To know the force, you'd know the area. Since Psi is a unit of pressure, you could have almost have no pressure and lift thousands of tons. On the contrary, you could have just a tiny fraction of a millionth of a gram but a very high pressure.

Since pressure is defined as force per unit area, $p = \frac{F}{A} \Leftrightarrow pA = F$, the pressure needed to make a body with the mass $m$ "hover" can be described as $\frac{mg}{A}$, where $g$ is the acceleration due to gravity and $A$ the area of which the pressure is applied on.

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  • $\begingroup$ the units of pressure, mass and area would be? $\endgroup$ – Anoneemus Jul 19 '15 at 18:03
  • $\begingroup$ @user86243 Whatever units you like. $\endgroup$ – Madde Anerson Jul 19 '15 at 18:06
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    $\begingroup$ @user86243 - You do understand that "psi" stands for "Pounds (of force) per Square Inch", right? Does that give you a clue to the units used? $\endgroup$ – WhatRoughBeast Jul 19 '15 at 18:30
  • $\begingroup$ @user86243 Yeah, if you choose to use psi as a unit of pressure, then you have also determined your units for force, length, area and volume. The unit of force would then be pounds and the units of area would be square inch. $\endgroup$ – Madde Anerson Jul 19 '15 at 18:32
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To lift 100 kg (220 lbs) using a pressure of 1450 psi (about 100 atm), the minimum area needed to apply this force is: $$area=\frac {220\ lbs}{1450\ psi}$$ $$area=0.152\ in^2$$

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