Force and Pressure on Hydraulic System

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BlackSideburns
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Rank in order from largest to smallest, the forces required to balance the the masses (in kgs)
You can find the diagram and answer at this link on the third slide http://www.gwu.edu/~phy21bio/Presentations/PHYS1021-15a.pdf
Equation
P=F/A

The Attempt at a Solution


I'm really having trouble picturing the FBDs and seeing how this works. It's easy when comparing the two with a single weight but the one with two is what trips me up.
 
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The area which the force is applied to is the same between the two systems (the one with two weights and the one with a single weight) therefore the forces will be equal if (and only if) the pressure is equal in the two systems.

Even though one of the systems has twice the weight being applied to it, it is also being applied to (what appears to be) twice the area, and so the pressure inside the two systems is the same.

And so the force must be the same.It can be tricky because of a tendency to look at just the forces involved, but the forces only interact with each other through the fluid (and not directly) so you must think of it in terms of pressure.
 
ohhh that makes sense. I totally overlooked the fact that there is double the area, thanks. My only concern is how would I show that in terms of formulas/numbers? I would be required to show my work on a test
 
BlackSideburns said:
ohhh that makes sense. I totally overlooked the fact that there is double the area, thanks. My only concern is how would I show that in terms of formulas/numbers? I would be required to show my work on a test

If I was required to show my work I might put something like this:

A[itex]_{f}[/itex](600g/A[itex]_{w}[/itex]) = A[itex]_{f}[/itex](1200g/2A[itex]_{w}[/itex])

Where A[itex]_{f}[/itex] is the area the force is applied to and A[itex]_{w}[/itex] is the area the weights are applied to (and g is the acceleration of gravity)