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.
1 answer:
answer:
<h2>F2= 5kN</h2>
step by step explanation
<em>Note the second diameter was used as 50cm and not 500cm</em>
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
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