53.1: Hydraulics- The Hydraulic Press
- Page ID
- 92338
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The properties of liquids may be exploited to make it possible to lift large, heavy objects using a machine called a hydraulic press (Fig. \(\PageIndex{1}\)). Referring to the figure, we know by Pascal's law that pressure \(P_{1}\) must be equal to pressure \(P_{2}\) :
\[P_{1}=P_{2} \text {. }\]
Therefore
\[\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}\]
where \(A_{1}\) and \(A_{2}\) are the cross-sectional areas of the pistons on the left and right. Solving for \(F_{1}\), we find
\[F_{1}=F_{2} \frac{A_{1}}{A_{2}}\]
Since \(A_{2}>A_{1}\), the force \(F_{1}\) is multiplied by the factor \(A_{2} / A_{1}\). One may place a heavy object like an automobile on the right, and lift it by applying a relatively small force on the left. The price for gaining this multiplication of force is that the piston on the left must be moved through a greater distance than the object on the right will be raised. To find the distance \(d_{1}\) through which the piston on the left must be moved in order to lift the object on the right a distance \(d_{2}\), we note that the liquid is essentially incompressible; therefore the volume change on the left must equal the volume change on the right:
\[A_{1} d_{1}=A_{2} d_{2}\]
Therefore the distance \(d_{1}\) is
\[d_{1}=d_{2} \frac{A_{2}}{A_{1}}\]
We can find \(d_{1}\) in terms of the ratio of forces using Eq. \(\PageIndex{3}\) to substitute for \(A_{2} / A_{1}\); we get\[d_{1}=d_{2} \frac{F_{2}}{F_{1}}\]
The force in the small cylinder must be exerted over a much larger distance. A small force exerted over a large distance is traded for a large force over a small distance.
Suppose the piston on the left has a diameter of \(10 \mathrm{~cm}\), and the piston on the right has a diameter of \(1 \mathrm{~m}\). What force must be applied on the left to lift a 1000-kg automobile on the right? (See Fig. \(\PageIndex{2}\.)
Solution
The automobile has a weight \(F_{2}=m g=(1000 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)=9.8 \times 10^{3} \mathrm{~N}\). The area \(A_{1}=\pi r^{2} / 4=(\pi / 4)(0.1 \mathrm{~m})^{2}=(\pi / 4) \times 10^{-2} \mathrm{~m}^{2}\). The area \(A_{2}=\pi r^{2} / 4=(\pi / 4)(1 \mathrm{~m})^{2}=\pi / 4 \mathrm{~m}^{2}\). The force \(F_{1}\) is then
\[F_{1} =F_{2} \frac{A_{1}}{A_{2}} \]
\[ =\left(9.8 \times 10^{3} \mathrm{~N}\right) \frac{\pi / 4 \times 10^{-2} \mathrm{~m}^{2}}{\pi / 4 \mathrm{~m}^{2}} \]
\[ =98 \mathrm{~N}\]
In this case, the piston on the left must be pushed in \(1 \mathrm{~m}\) to lift the car by \(1 \mathrm{~cm}\).