# Homework Help: Help please f= PxA

1. Oct 13, 2011

### jezman

hey everyone, can someone please help me out with this one, im a bit stuck with this
using the law F = PXA and F = mxa

calculate the pressure in a hydraulic cylinder, when the load of 375kg, is applied to the piston
piston diameter it 95mm

so far what i have worked out is as follows
force on the piston is:
F = M x 9.81
= 250x9.81 = 2452.5N

now for the area
pi x rxr
= 3.1416 x 37.5x37.5
=4417.86 mm2

now if im correct i need to change the mm2 to m2?
so...
4417.86mm2 /1,000,000
=0.00441786m2

therefore....
f=2452.5N
A=0.00441786m2
P = F/A ?
555,133.028p
or 555.133kp

is this correct guys or am i way out?
if anyone could help me out with this it would be greatly appreciated
thansk

2. Relevant equations

3. The attempt at a solution

2. Oct 13, 2011

### Staff: Mentor

You might want to check the value that you used for the radius of the cylinder.

Other than that your method looks fine.

3. Oct 13, 2011

### SteamKing

Staff Emeritus
If the load is 375 kg, why did you use 250 kg in your calculations?

4. Oct 13, 2011

### Staff: Mentor

Good catch SteamKing. I missed that second error!

5. Oct 17, 2011

### Nucengable

Actually , all your work is messed up , the pressure inside the cylinder would be the summation of the external force and the weight of the piston divided by the area of the piston..

So:

P= ( Fex + W ) / A

A= 3.14 * d^2 / 4 = 3.14 * (0.095)^2 / 4 = 0.0071 m^2

Fex = 375 * 9.81 = 3.44 k N

..and since you didn't mention the mass of the piston I'll assume its negligible

so now:

P = 3.44 / 0.0071 = 484.5 kPa