Question

a.  A force of 50N is applied to the input piston of a hydraulic system the area of the input piston’s 60 cm2, and that of the output piston’s 15 cm2. (a) How much force does the output piston expert? (b) If the input piston moves 2.0cm per stroke, how far does the output piston move?

Answer 1

a) By the hydraulic press formula, we have:

$$\dfrac{F_{input}}{A_{input}}=\dfrac{F_{output}}{A_{output}},$$ $$F_{output}=\dfrac{F_{input}A_{output}}{A_{input}}=\dfrac{50\ N\cdot15\ cm^2}{60\ cm^2}=12.5\ N.$$

b) We can find how far does the output piston move as follows:

$$F_{input}d_{input}=F_{output}d_{output},$$ $$d_{output}=\dfrac{F_{input}d_{input}}{F_{output}}=\dfrac{50\ N\cdot0.02\ m}{12.5\ N}=0.08\ m=8\ cm.$$