10.2: How Force Is Multiplied Using Pascal’s Law
- Page ID
- 117046
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)One of the most powerful and useful aspects of hydraulics is its ability to multiply force, allowing a small input force to generate a much larger output force. This is known as hydraulic leverage or force multiplication, and it works thanks to Pascal’s Law.
How It Works
Let’s look at a simple example:
Suppose you apply 10 pounds of force on a small piston with a surface area of 1 in². This creates 10 psi of pressure in the fluid.
P = F x A =10 lbs x 1 in² = 10 psi
Now, imagine this pressure is transmitted through the fluid to a larger piston with a surface area of 10 in². According to Pascal’s Law, the pressure is the same, 10 psi, everywhere in the system.
So on the large piston:
F = P × A
=10 psi × 10 in² = 100 lbsF
Result
Your small 10-pound input force on the small piston has resulted in a 100-pound output force at the large piston.
Real-World Example: The Brake System
Automotive brake systems provide a practical application of hydraulic leverage. When you press the brake pedal, a small master cylinder sends pressurized fluid through the brake lines to larger slave cylinders at the wheels. This small input force from your foot can be multiplied to generate enough force to stop a moving vehicle.
Industrial Applications
In industrial hydraulic systems, we usually don’t use pistons to push fluid. Instead, a hydraulic pump (driven by an electric motor) generates flow. The pressure produced by this flow acts on a cylinder's large piston area to create a powerful output force.
This means that even a small motor can produce a large mechanical output when paired with the right hydraulic components —another example of force multiplication in action.
The Tradeoff: Speed vs. Force
Force multiplication comes at a cost: distance (or speed).
In the earlier example, while we increased the force from 10 to 100 pounds, the displacement of fluid must also be considered. To move the large piston 1 inch, the small piston must be pushed 10 inches, assuming incompressible fluid and perfect efficiency.
This is an important concept:
As the output force increases, the distance moved by the piston on the output side decreases. The same occurs with speed as the volume is larger to fill.
In practical terms, a larger cylinder area will provide more force, but it will also require more fluid volume to move, making the cylinder slower if the pump flow rate remains the same.