# A simple Fluid Dynamics/Pressure HW question

1. Apr 21, 2015

### SnakeDoc

1. The problem statement, all variables and given/known data
A piston of cross-sectional area a is used in a hydraulic press to exert a small force of magnitude f on the enclosed liquid. A connecting pipe leads to a larger piston of cross-sectional area A (the figure). If the piston diameters are 3.63 cm and 50.7 cm, what force magnitude on the small piston will balance a 41.8 kN force on the large piston?

2. Relevant equations
P=F/A

3. The attempt at a solution
So I'm pretty sure I'm doing this right especially since we did a similar one in class so I think it might just be a math error here's what I did.

For the smaller side Ps = Fs/as

For the larger side Pb = Fb/Ab

Ps = Pb

Fs/as = Fb/Ab

Solve for Fs
Fs = Fb/Ab * as

Plug in values (Fb 41.8kN = 41800N) (Ab 50.7cm =.507m) (as 3.63cm = .0363m)

Fs = (41800/.507)*.0363 = 2992.78N WileyPlus says its wrong though, so did I make a mistake?

#### Attached Files:

• ###### fig14_35.gif
File size:
4.3 KB
Views:
71
Last edited: Apr 21, 2015
2. Apr 21, 2015

### paisiello2

For some reason you plugged in the diameters instead of the areas.

3. Apr 21, 2015

### SnakeDoc

I guess I did. I didn't even notice it said diameters thanks. I must remember to read more thoroughly next time.