Is the weight suspended from the piston or pushed up by it? If the former, what is the diameter of the suspending shaft?

it is piston dia of direct lift.

OK. Back to the Physics class: "Show that the minimum pressure needed on the underside of the piston is a little under 8e5N/m² (10 marks)."

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I need technical help in order to decide which product I need. Where can I get help?

you'll need need more information than whats currently asked, is it a direct lift, or is the cylinder on a scissors or some type of linkage lift?

First, learn to use the internet.

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Then you can discover all the information you need.

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Hydraulic Cylinder Calculations

Isn't using CR4 a way of using the internet?

O well, I gave you a GA anyway because this shouldn't have made it to CR4...and it certainly didn't merit so much argument.

Force = Pressure X Area

Pressure = Force / Area

You need to account for the rod diameter.

123 is the diameter, not the area

Sorry; you are wrong both times.

While you doing this homework please also think about if it would be different with Water, Butan, Methan and Air.

It is 24 deg C outside and the rest is up to you!

Dear Mr. dharam ver9,

What is the diametre of the Shaft on which the Ram or Piston is mounted.?

The area of the shaft will NOT BE AVAILABLE for providing the lifting Force.

125 MM Dia of Cylinder is equal to 12.5 CM.

Tthe Pressure required will be

P x A = 1000, where P=oil pressure in Kg/CM^2,

A= NETT area or Effective Area calculated from (3.14/4)x (12.5^2 - d^2) and once you substitute d and get the NETT area, you can calculate the oil Pressure required.

DHAYANANDHAN.S

Well, first off, you need to include the weight of the piston and connecting hardware to the 1000kg weight to determine the actual force you will need:

m_T = m_W + m_P

The force required to lift must be greater than the mass of what you are lifting:

F > m_T

The pressure required is the force required divided by the area it is acting on, in this case the face of the piston:

P = F/A

Remember to keep consistent units when doing the final calculation.

Determine the extra weight involved (piston, etc.), do the calculations and come back with an answer. We'll check it for you.

force(1000kg) / eff area (3.14159 x 6.25cm(r) x 6.25cm(r) = 1000kg / 122.71835cm2

=8.1487405kg/cm2.

we manufacture pressure gauges and re-range them to suit tons/tonne/kn every day this is the calculation that would be used, tornado is correct.

Try with 8.2 kg/ cm2 (theoretical value)

Because friction, specially at the start the pressure must be hihger,

10 kg/cm2 should be enough, when there are no special mechanical

systems that request again more force

edo

Tornado is absolutely right. However, the pressure mentioned by him is just enough to overcome the dead weight of 1000Kgs. Any addition to pressure beyond 8.15Kg/cm² will start moving it. The speed of lift will depend on how quickly (LPM) you are able to pump.

But My dear, this is all elementary.