Question 1013256
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The area of master piston of a hydraulic press is 10 cm^2 and the area of the slave piston is 1000 cm^2. 
When an effort of 50 n is exerted on the master piston calculate: the pressure in the hydraulic fluid, and the maximum weight the press can lift
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The pressure (in Pascals) in the hydraulic fluid is an effort (in newtons) divided by the area of master piston ({{{m^2}}}):

P = {{{F[1]/S[1]}}} = {{{50/0.001}}} = 50000 Pa.

This pressure is exerted on the slave piston surface area too.

Hence, the maximum weight the press can lift is the product of this pressure P by the area on the slave piston {{{S[2]}}} in {{{m^2}}}:

{{{F[2]}}} = {{{P*S[2]}}} = 50000*0.1 = 5000 newtons.

The general rule for a hydraulic press : the forces are proportional to areas: {{{F[2]/F[1]}}} = {{{S[2]/S[1]}}},  or  {{{F[1]/S[1]}}} = {{{F[2]/S[2]}}}

Each side of the last proportion is the pressure of hydraulic liquid.

On hydrailc press see <A HREF=https://en.wikipedia.org/wiki/Hydraulic_press>this article</A> in Wikipedia.
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