 The hydraulic jack in the figure below is filled with oil with a specific weight of 56 pounds of force per cubic foot Neglecting the weight of the two pistons what force on the handle would be needed to support the 2,000 pound force weight for this design Pascal's law allows us to Say the pressure on both sides of the oil channel are the same. I'm gonna call this a I'm gonna call this B And What we can say as a result of Pascal's law is that PA is equal to PB Because the pressure increases on a constrained fluid everywhere at the same height the same Furthermore these pistons which are shown in white are assumed to be weightless, which means that effectively The weight of the 2,000 pound weight is applied directly at point A and the force exerted on this lever or by this lever is Exerted at point B So we're going to use Pascal's law to relate the force applied at A to the force applied at B By recognizing that the pressure A can be written as The force at A divided by the area at A and the pressure at point B can be described as the force at B divided by the area B These are presumably circular pistons, which means that their area can be written as power four times diameter squared FA is known so I can write FB as being Pi over four times Diameter B squared divided by pi over four times diameter A squared times F B Pi over four cancels So I can write this as Diameter B over diameter A Quantity squared times FB Excuse me Time FA I could calculate that numerically if I wanted to or I could leave it symbolically B is a one inch piston is one inch in diameter A is three inches so I can write this as One divided by three quantity squared times two thousand And I get two thousand divided by nine Thank you, cacutter two hundred and twenty two point two two two pounds of force and Handling the lever is just going to be a statics problem so I have a fulcrum I have a force at a distance of one inch and then at a distance of 16 inches Because it's 15 plus one I have a force down which is F So I'm going to call this point over here C for fulcrum C and then some of the moments Call this positive must equal zero therefore the Clockwise moment must equal the counterclockwise moment Which is going to be F times 16 inches Is equal to FB times one inch So F is equal to FB times the quantity one over sixteen And if I had wanted to I could leave FB as Diameter B over diameter a squared times F a and handle all of this in a single calculation That's fun, right? So times one over 16 yields 125 divided by nine Thank you calculator thirteen point eight eight nine pounds of force So we have two mechanical advantages here We have the lever of the handle itself and then we have the mechanical advantage as the result of different diameter pistons in a hydraulic system So it takes only thirteen point nine pounds to support this two thousand pound weight We are offsetting literally a ton Cool, huh