How to calculate distance in a hydraulic press when you have 2 pistons

[arhzz]

Homework Statement

In a hydraulic press, a small piston with the cross-sectional area a exerts a small force f on the enclosed (incompressible) liquid. A connecting tube leads to a larger piston with cross-sectional area A.

a) Which force F can the large piston counter without moving?
b) The small flask has a diameter of 3.80 cm, the large flask a diameter of 53 cm. What force on the small piston compensates for a force of 20 kN on the large piston?
c) How far does the large piston have to be moved so that the small piston is lifted 0.85 cm?

Relevant Equations

p = F/a

Hello!a) Now here I've simply though "logically" and drew a litle sketch and I think the answer should be 0N.No calculation needed I'd pressume (if needed I don't know how to do it)
b) Here I've used the fact that $$ p1 = \frac{F1}{A1} = \frac{F2}{A2} $$ For the surface I just went with ## A = r^2 * \pi ## (I found online that this is how you calculate the volume of a piston if not please let me know). Than I tried to get F1 and I did that like this $$ F1 = \frac{F2*A1}{A2} $$ Where F2=20000 N, and I get that F1 = 102,81 N.

Now for c) I am kind of stuck.I know it isn't hard but I just can't get a decent idea.What would you suggest for a start?Thank you!

 

[hmmm27]

'a' is pretty obvious, agreed.

'b' you don't really need to mention ##P_1## Also, there's no real need to explicitly invoke '3.14159...' : the ##\pi##'s and a few other things in the equations can be canceled out before you plug in the actual numbers.

'c' ... also, pretty straightforwards. How much volume is being moved by displacing piston_1 ? Where does that volume go ?

(assumed is that the pistons and fluid are treated as massless. etc., just in case that's what was making your brain itch in 'c')

Suggested is to optimize the equations first, before plugging in the numbers.

 

[arhzz]

'a' is pretty obvious, agreed.

'b' you don't really need to mention ##P_1## Also, there's no real need to explicitly invoke '3.14159...' : the ##\pi##'s and a few other things in the equations can be canceled out before you plug in the actual numbers.

'c' ... also, pretty straightforwards. How much volume is being moved by displacing piston_1 ? Where does that volume go ?

(assumed is that the pistons and fluid are treated as massless. etc., just in case that's what was making your brain itch in 'c')

Suggested is to optimize the equations first, before plugging in the numbers.

for b) that would make things easier with simply ##\pi##, I'll keep that in mind.

Now for c; Well I think I get what you are aiming for. We can use this ## V1 = V2 ## and the volume equals to ## V = A * s ## Now if we use this property in this sence

$$ A*s = A*D $$ we can get an equation for D $$ D = \frac {A*s}{A}$$ that should give out that D = 0,0437 mm

 

[haruspex]

for b) that would make things easier with simply ##\pi##, I'll keep that in mind.

Now for c; Well I think I get what you are aiming for. We can use this ## V1 = V2 ## and the volume equals to ## V = A * s ## Now if we use this property in this sence

$$ A*s = A*D $$ we can get an equation for D $$ D = \frac {A*s}{A}$$ that should give out that D = 0,0437 mm

Looks right, but do avoid using the same symbol (A here) with two meanings, particularly in the one equation.

 

[arhzz]

Looks right, but do avoid using the same symbol (A here) with two meanings, particularly in the one equation.

Yea that does make ti a bit confusing,sadly I cannot edit it now but I'm glad my calculation is correct

Thank you!