Question

A hydraulic lift consists of two pistons that connect to each other by an incompressible fluid. If one piston has an area of 0.15 m^2 and the other an area of 6.0 m^2, how large a mass can be raised by a force of 130 N exerted on the smaller piston?

Answer 1

You have:
 
 A1= 0.15 m²
 A2=6.0 m²
 F1=130.0 N
 
 To solve this problem you must use the pressure formula, as below:
 
 P1=P2
 F1/A1=F2/A2
 
 Then, you have:
 
 m2=A2xF1/A1xg
 
 g=9.81 m/s² (This is the acceleration of gravity)
 
 When you substitute the values in m2=A2xF1/A1xg, you obtain:
 
 m2=(0.60 m² x 130.0 N)/(0.15 m² x 9.81 m/s²)
 
 The answer is:
 
 m2=530 Kg