No need for gravity.
F1/Area1 = F2/AREA2
Solver for F1
In a hydraulic system a 20.0-N force is applied to the small piston with cross sectional
area 25.0 cm2. What weight can be lifted by the large piston with cross sectional
area 50.0 cm2?
Would I need to divide both pistons areas then mutiply the force? and or have gravity in there somewhere since its weight?
4 answers
So it would be 50.0/25.0*20.0=40
20/25 = F/50
F = (20/25)*50
F= 40 N
F = (20/25)*50
F= 40 N
thank you