Calculate the force output by the larger piston of a hydraulic lift when a force of 800 N is exerted on the smaller piston. The areas of the two pistons are 20 cm2 and 500 cm2.

step by step explanation

Note the second diameter was used as 50cm and not 500cm

given

F1=800N

D1=20cm

A1=πD1^2/4

=3.142*20^2/4

=314.2cm^2

D2= 50cm

A2=πD2^2/4

=3.142*50^2/4

=1963.75cm^2

The pressure applied on the smaller relative to the bigger end is expressed as

F1/A1=F2/A2

substitute we have

800/314.2=F2/1963.75

cross multiply we have

F2=800*1963.75/314.2

F2=1571000/314.2

F2=5000N

F2=5kN

The larger Force is 5kN