Hydraulic press

How a hydraulic press works, the gain in force $F_2 / F_1 = S_2 / S_1$ as a consequence of Pascal's law.

Terms

Studied 0 of 2

Formulas

Hydraulic press

\dfrac{F_2}{F_1} = \dfrac{S_2}{S_1}

Quantities
Symbol	Name	Unit
$F_1$	force acting on the small piston	N
$F_2$	force acting on the large piston	N
$S_1$	area of the small piston	m²
$S_2$	area of the large piston	m²

Solved for another quantity:

F_2 = F_1\,\dfrac{S_2}{S_1}

F_1 = F_2\,\dfrac{S_1}{S_2}

S_2 = S_1\,\dfrac{F_2}{F_1}

Note. The formula follows from Pascal's law: the pressure under both pistons is the same, $\dfrac{F_1}{S_1} = \dfrac{F_2}{S_2}$ . Take the areas in the same units. Example: $F_1 = 100\ \text{N}$ , $S_1 = 10\ \text{cm}^2$ , $S_2 = 200\ \text{cm}^2$ : $F_2 = 100 \cdot \dfrac{200}{10} = 2000\ \text{N}$ . Friction is not taken into account.

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

Hydraulic press

The width of a cylinder shows the area of the piston. When the small piston goes down, the large one rises by a smaller height. The length of the arrow F₂ is limited by the frame of the picture.

Graphs

The line F₂(F₁) is drawn for the chosen areas of the pistons. Click the graph to set the force F₁.

F_2(F_1)

p

100,000 Pa

F_2

2,000 N

F_2 / F_1

Formulas with your numbers

\dfrac{F_2}{F_1} = \dfrac{S_2}{S_1} = \dfrac{200}{10} = 20

F_2 = F_1\,\dfrac{S_2}{S_1} = 100\cdot 20 = 2000\ \text{N}

How to read the graphs

The $F_2(F_1)$ graph is a straight line through the origin: the force on the large piston is directly proportional to the force on the small one.
The slope of the line is $S_2 / S_1$ — this is the gain in force. If the large piston has $20$ times the area, the force on it is $20$ times greater.
When the pistons have equal areas ( $S_2 = S_1$ ), the line rises at $45^\circ$ (with equal scales on the axes): $F_2 = F_1$ , there is no gain in force.
A gain in force is not a gain in work: the large piston gives as many times more force as it travels a shorter distance.

Loading…

Terms

Studied 0 of 2

Formulas

Hydraulic press

\dfrac{F_2}{F_1} = \dfrac{S_2}{S_1}

Quantities
Symbol	Name	Unit
$F_1$	force acting on the small piston	N
$F_2$	force acting on the large piston	N
$S_1$	area of the small piston	m²
$S_2$	area of the large piston	m²

Solved for another quantity:

F_2 = F_1\,\dfrac{S_2}{S_1}

F_1 = F_2\,\dfrac{S_1}{S_2}

S_2 = S_1\,\dfrac{F_2}{F_1}

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

Hydraulic press

The width of a cylinder shows the area of the piston. When the small piston goes down, the large one rises by a smaller height. The length of the arrow F₂ is limited by the frame of the picture.

Graphs

The line F₂(F₁) is drawn for the chosen areas of the pistons. Click the graph to set the force F₁.

F_2(F_1)

p

100,000 Pa

F_2

2,000 N

F_2 / F_1

Formulas with your numbers

\dfrac{F_2}{F_1} = \dfrac{S_2}{S_1} = \dfrac{200}{10} = 20

F_2 = F_1\,\dfrac{S_2}{S_1} = 100\cdot 20 = 2000\ \text{N}

How to read the graphs

The

F_2(F_1)

graph is a straight line through the origin: the force on the large piston is directly proportional to the force on the small one.

The slope of the line is

S_2 / S_1

— this is the gain in force. If the large piston has

20

times the area, the force on it is

20

times greater.

When the pistons have equal areas (

S_2 = S_1

), the line rises at

45^\circ

(with equal scales on the axes):

F_2 = F_1

, there is no gain in force.

A gain in force is not a gain in work: the large piston gives as many times more force as it travels a shorter distance.