Worked exercise — Fruit press

Problem

A hydraulic press has a small piston of cross-section $A_1 = 5\;\text{cm}^2$ and an upper plate of cross-section $A_2 = 200\;\text{cm}^2$. Pushing with $F_1 = 80\;\text{N}$ on the small piston, what force is applied to the fruit? If we want the plate to descend by $1\;\text{cm}$, by how much must the small piston move down?

Solution

Step 1 — Pascal’s principle. The pressure transmitted by the fluid is the same on both pistons, $P = F_1/A_1 = F_2/A_2$. The force on the fruit is therefore amplified by the ratio of the areas:

$F_2 = F_1\,\frac{A_2}{A_1} = 80\cdot\frac{200}{5} = 80\cdot 40 = 3200\;\text{N}$

$\ev{F_2 = 3200\;\text{N} \approx 326\;\text{kgf}}$

Step 2 — Conservation of volume. The fluid is incompressible: the volume displaced by the small piston rises beneath the plate, $A_1 h_1 = A_2 h_2$. To make the plate descend by $h_2 = 1\;\text{cm}$:

$h_1 = h_2\,\frac{A_2}{A_1} = 1\cdot 40 = 40\;\text{cm}$

Step 3 — Work is conserved. The force is multiplied $40$ times, but the stroke is $40$ times longer: work is not created from nothing.

$L = F_1 h_1 = 80\cdot 0{,}40 = 32\;\text{J} = 3200\cdot 0{,}01 = F_2 h_2$

Linked atoms

Argomenti: Fluid statics and dynamics Concetti: Pressure · Pascal’s principle Competenze: Unit conversion