Hydraulic lift has two connecting pistons?

Question

The smaller piston has a cross sectional area of 3.4 cm^2 and the larger piston has an area of 56 cm^2. What force F must be applied to the small piston to maintain the load of 82 kN at a constant elevation (in units of N)??

Answer 1

The pressure is equal throughout the hydraulic fluid - up, down, and sideways. Pressure is force/area so, at the large piston
Pl = Fl / 56 cm^2
To maintain the load of 82 kN, Fl must be 82 kN.

At the small piston
Ps = Fs / 3.4 cm^2

In my first sentence I said Pl = Ps, so set the 2 pressure expressions equal and solve for Fs.