Pressure is defined as the physical force exerted on an object. The force applied is perpendicular to the surface of objects per unit area. The smaller the area on which the force acts, the greater the impact. This impact is known as pressure. 

For example, If an elephant were to step on a man’s chest, his ribs would crack. A circus performer across whose chest a large, light but strong wooden plank is placed first, is saved from this accident. Such everyday experiences convince us that both the force and its coverage area are important.

When an object is submerged in a fluid at rest, the fluid exerts a force on its surface. This force is always normal to the object’s surface. This is so because if there were a component of force parallel to the surface, the object will also exert a force on the fluid parallel to it; as a consequence of Newton’s third law. This force will cause the fluid to flow parallel to the surface. Since the fluid is at rest, this cannot happen. Hence, the force exerted by the fluid at rest has to be perpendicular to the surface in contact with it. This is shown in Fig.10.1(a).

The normal force exerted by the fluid at a point may be measured. An idealised form of one such pressure-measuring device is shown in Fig. 10.1(b). It consists of an evacuated chamber with a spring that is calibrated to measure the force acting on the piston. This device is placed at a point inside the fluid. The inward force exerted by the fluid on the piston is balanced by the outward spring force and is thereby measured.

If \(F\) is the magnitude of this normal force on the piston of area \(A\) then the average pressure \(P_{a v}\) is defined as the normal force acting per unit area.

\(P_{a v}=\frac{F}{A} \dots(10.1)\)

In principle, the piston area can be made arbitrarily small. The pressure is then defined in a limiting sense as

\(
P=\lim _{\Delta A \rightarrow 0} \frac{\Delta F}{\Delta A} \dots(10.2)
\)

Pressure is a scalar quantity. Its dimensions are \(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\). The SI unit of pressure is \(\mathrm{N} \mathrm{m}^{-2}\). It has been named as pascal \((\mathrm{Pa})\) in honour of the French scientist Blaise Pascal (1623-1662) who carried out pioneering studies on fluid pressure. A common unit of pressure is the atmosphere (atm), 1.e. the pressure exerted by the atmosphere at sea level ( \(1 \mathrm{~atm}=1.013 \times 10^5 \mathrm{~Pa}\) ).

Density of a Fluid

For a fluid of mass \(m\) occupying volume \(V\), the density

\(\rho=\frac{m}{V} \dots(10.3)\)

The dimensions of density are \(\left[\mathrm{ML}^{-3}\right]\). Its SI unit is \(\mathrm{kg} \mathrm{m}^{-3}\). It is a positive scalar quantity. A liquid is largely incompressible and its density is therefore, nearly constant at all pressures. Gases, on the other hand exhibit a large variation in densities with pressure.

The density of water at \(4^{\circ} \mathrm{C}(277 \mathrm{~K})\) is \(1.0 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\). The relative density of a substance is the ratio of its density to the density of water at \(4^{\circ} \mathrm{C}\). It is a dimensionless positive scalar quantity. For example the relative density of aluminium is \(2.7\). Its density is \(2.7 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\). The densities of some common fluids are displayed in Table \(10.1\).

Example 1: The two thigh bones (femurs), each of cross-sectional area \(10 \mathrm{~cm}^2\) support the upper part of a human body of mass \(40 \mathrm{~kg}\). Estimate the average pressure sustained by the femurs.

Solution:

Total cross-sectional area of the femurs is \(A=2 \times 10 \mathrm{~cm}^2=20 \times 10^{-4} \mathrm{~m}^2\). The force acting on them is \(F=40 \mathrm{~kg}\) wt \(=400 \mathrm{~N}\) (taking \(g=10 \mathrm{~m} \mathrm{~s} \mathrm{~s}^{-2}\) ). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
\(
P_{a v}=\frac{F}{A}=2 \times 10^5 \mathrm{~N} \mathrm{~m}^{-2}
\)

Pascal’s Law

The French scientist Blaise Pascal observed that the pressure in a fluid at rest is the same at all points if they are at the same height. This fact may be demonstrated in a simple way.

Fig. 10.2 shows an element in the interior of a fluid at rest. This element ABC-DEF is in the form of a right-angled prism. In principle, this prismatic element is very small so that every part of it can be considered at the same depth from the liquid surface and therefore, the effect of the gravity is the same at all these points. But for clarity we have enlarged this element. The forces on this element are those exerted by the rest of the fluid and they must be normal to the surfaces of the element as discussed above. Thus, the fluid exerts pressures \(P_{\mathrm{a}}, P_{\mathrm{b}} \text { and } P_{\mathrm{c}} \text { on }\) this element of area corresponding to the normal forces \(F_{\mathrm{a}}, F_{\mathrm{b}}\) and \(F_{\mathrm{c}}\) as shown in Fig. \(10.2\) on the faces BEFC, ADFC and ADEB denoted by \(A_a, A_b\) and \(A_c\) respectively. Then

\(
\begin{aligned}
& F_{\mathrm{b}} \sin \theta=F_{\mathrm{c}}, \quad F_{\mathrm{b}} \cos \theta=F_{\mathrm{a}} \quad \text { (by equilibrium) } \\
& A_{\mathrm{b}} \sin \theta=A_{\mathrm{c}}, \quad A_{\mathrm{b}} \cos \theta=A_{\mathrm{a}} \text { (by geometry) } \\
&
\end{aligned}
\)
Thus,
\(
\frac{F_b}{A_b}=\frac{F_c}{A_c}=\frac{F_a}{A_a} ; \quad P_b=P_c=P_a \dots(10.4)
\)

Hence, the pressure exerted is the same in all directions in a fluid at rest. The force against any area within (or bounding) a fluid at rest and under pressure is normal to the area, regardless of the orientation of the area.

Variation of Pressure with Depth

Consider a fluid at rest in a container. In Fig. \(10.3\) point 1 is at height \(h\) above a point 2. The pressures at points 1 and 2 are \(P_1\) and \(P_2\) respectively. Consider a cylindrical element of fluid having area of base \(A\) and height \(h\). As the fluid is at rest the resultant horizontal forces should be zero and the resultant vertical forces should balance the weight of the element. The forces acting in the vertical direction are due to the fluid pressure at the top \(\left(P_1 A\right)\) acting downward, at the bottom \(\left(P_2 A\right)\) acting upward. If \(m g\) is weight of the fluid in the cylinder we have
\(
\left(P_2-P_1\right) A=m g \dots(10.5)
\)
Now, if \(\rho\) is the mass density of the fluid, we have the mass of fluid to be \(m=\rho V=\rho h A\) so that
\(
P_2-P_1=\rho g h \dots(10.6)
\)

Pressure difference depends on the vertical distance \(h\) between the points ( 1 and 2), mass density of the fluid \(\rho\) and acceleration due to gravity \(g\). If the point 1 under discussion is shifted to the top of the fluid (say, water), which is open to the atmosphere, \(\mathrm{P}_1\) may be replaced by atmospheric pressure \(\left(\mathrm{P}_{\mathrm{a}}\right)\) and we replace \(\mathrm{P}_2\) by \(P\). Then Eq. \((10.6)\) gives
\(
P=P_{\mathrm{a}}+\rho g h \dots(10.7)
\)
Thus, the pressure \(P\), at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount \(\rho g h\). The excess of pressure, \(P-P_a\), at depth \(h\) is called a gauge pressure at that point.

The area of the cylinder is not appearing in the expression of absolute pressure in Equation (10.7). Thus, the height of the fluid column is important and not cross-sectional or base area or the shape of the container. The liquid pressure is the same at all points at the same horizontal level (same depth).

The result is appreciated through the example of hydrostatic paradox. Consider three vessels A, B, and C [Fig.10.4] of different shapes. They are connected at the bottom by a horizontal pipe. On filling with water, the level in the three vessels is the same, though they hold different amounts of water. This is so because water at the bottom has the same pressure below each section of the vessel.

Example 2: What is the pressure on a swimmer \(10 \mathrm{~m}\) below the surface of a lake?

Solution:

Here
\(h=10 \mathrm{~m}\) and \(\rho=1000 \mathrm{~kg} \mathrm{~m}^{-3}\). Take \(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\)
From Eq. (10.7)
\(P=P_{\mathrm{a}}+\rho g h\)
\(=1.01 \times 10^5 \mathrm{~Pa}+1000 \mathrm{~kg} \mathrm{~m}^{-3} \times 10 \mathrm{~m} \mathrm{~s}^{-2} \times 10 \mathrm{~m}\)
\(=2.01 \times 10^5 \mathrm{~Pa}\)
\(\approx 2 \mathrm{~atm}\)

This is a \(100 \%\) increase in pressure from surface level. At a depth of \(1 \mathrm{~km}\), the increase in pressure is \(100 \mathrm{~atm}\) ! Submarines are designed to withstand such enormous pressures.

Atmospheric Pressure and Gauge Pressure

The pressure of the atmosphere at any point is equal to the weight of a column of air of unit cross-sectional area extending from that point to the top of the atmosphere. At sea level, it is \(1.013 \times 10^5 \mathrm{~Pa}(1 \mathrm{~atm})\). Italian scientist Evangelista Torricelli (1608-1647) devised for the first time a method for measuring atmospheric pressure. A long glass tube closed at one end and filled with mercury is inverted into a trough of mercury as shown in Fig. 10.5 (a). This device is known as ‘mercury barometer’. The space above the mercury column in the tube contains only mercury vapour whose pressure \(P\) is so small that it may be neglected.

Thus, the pressure at Point \(\mathrm{A}=0\). The pressure inside the column at Point B must be the same as the pressure at Point C, which is atmospheric pressure, \(\mathrm{P}_a\).
\(
P_{\mathrm{a}}=\rho g h \dots(10.8)
\)
where \(\rho\) is the density of mercury and \(h\) is the height of the mercury column in the tube, and \(g\) is the acceleration due to gravity.

In the experiment, it is found that the mercury column in the barometer has a height of about \(76 \mathrm{~cm}\) at sea level equivalent to one atmosphere ( \(1 \mathrm{~atm})\). This can also be obtained using the value of \(\rho\) in Eq. (10.8). A common way of stating pressure is in terms of \(\mathrm{cm}\) or \(\mathrm{mm}\) of mercury (Hg). A pressure equivalent of \(1 \mathrm{~mm}\) is called a torr (after Torricelli).

1 torr \(=133 \mathrm{~Pa}\).

The \(\mathrm{mm}\) of \(\mathrm{Hg}\) and torr are used in medicine and physiology. In meteorology, a common unit is the bar and millibar.

\(1 \mathrm{bar}=10^5 \mathrm{~Pa}\)

Gauge Pressure and Absolute Pressure

Gauge pressure is the pressure relative to atmospheric pressure. Gauge pressure is positive for pressures above atmospheric pressure, and negative for pressures below it.

The total pressure, or absolute pressure, is thus the sum of gauge pressure and atmospheric pressure:

\(P=P_g + P_a\), where \(P\) is absolute pressure, \(P_g\) is gauge pressure, and \(P_a\) is atmospheric pressure.

Absolute pressure is the sum of gauge pressure and atmospheric pressure.

For example, if your tire gauge reads 34 psi (pounds per square inch), then the absolute pressure is 34 psi plus \(14.7 \mathrm{psi}\) \(\left(P_{\mathrm{a}} \text { in psi), or } 48.7 \mathrm{psi} \text { (equivalent to } 336 \mathrm{kPa}\right) \text {. }\)

In most cases, the absolute pressure in fluids cannot be negative. Fluids push rather than pull, so the smallest absolute pressure is zero. (A negative absolute pressure is a pull.) Thus the smallest possible gauge pressure is \(P_g=-P_a\) (this makes \(P=\) zero). There is no theoretical limit to how large a gauge pressure can be.

Pressure Measurement using a Manometer

Let us examine how a manometer is used to measure pressure. An open-tube manometer is a useful instrument for measuring pressure differences. It consists of a U-tube containing a suitable liquid i.e., a low-density liquid (such as oil) for measuring small pressure differences and a high-density liquid (such as mercury) for large pressure differences. One end of the tube is open to the atmosphere and the other end is connected to the system whose pressure we want to measure [see Fig. 10.5 (b)]. The pressure \(P\) at \(A\) is equal to pressure at point \(B\). What we normally measure is the gauge pressure, which is \(P_g=P-P_{\mathrm{a}}\), given by Eq. (10.8) and is proportional to manometer height \(h\).

Pressure is the same at the same level on both sides of the U-tube containing a fluid. For liquids, the density varies very little over wide ranges in pressure and temperature and we can treat it safely as a constant for our present purposes. Gases on the other hand, exhibits large variations of densities with changes in pressure and temperature. Unlike gases, liquids are, therefore, largely treated as incompressible.

Example 3: The density of the atmosphere at sea level is \(1.29 \mathrm{~kg} / \mathrm{m}^3\). Assume that it does not change with altitude. Then how high would the atmosphere extend?

Solution:

We use Eq. (10.7)
\(\rho g h=1.29 \mathrm{~kg} \mathrm{~m}^{-3} \times 9.8 \mathrm{~m} \mathrm{~s}^2 \times h \mathrm{~m}=1.01 \times 10^5 \mathrm{~Pa}\)
\(\therefore h=7989 \mathrm{~m} \approx 8 \mathrm{~km}\)
In reality, the density of air decreases with height. So does the value of \(g\). The atmospheric cover extends with decreasing pressure over \(100 \mathrm{~km}\). We should also note that the sea level atmospheric pressure is not always \(760 \mathrm{~mm}\) of Hg. A drop in the \(\mathrm{Hg}\) level by \(10 \mathrm{~mm}\) or more is a sign of an approaching storm.

Example 4: At a depth of \(1000 \mathrm{~m}\) in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? (c) Find the force acting on the window of area \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\) of a submarine at this depth, the interior of which is maintained at sea level atmospheric pressure. (The density of sea water is \(1.03 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\), \(g=10 \mathrm{~ms}^{-2}\).)

Solution:

Here \(h=1000 \mathrm{~m}\) and \(\rho=1.03 \times 10^3 \mathrm{~kg} \mathrm{~m}^3\).
(a) From Eq. (10.6), absolute pressure
\(
\begin{aligned}
P & =P_{\mathrm{a}}+\rho g h \\
= & 1.01 \times 10^5 \mathrm{~Pa} \\
& +1.03 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} \times 10 \mathrm{~m} \mathrm{~s}^{-2} \times 1000 \mathrm{~m} \\
= & 104.01 \times 10^5 \mathrm{~Pa} \\
& \approx 104 \mathrm{~atm}
\end{aligned}
\)

(b) Gauge pressure is \(P-P_{\mathrm{a}}=\rho g h=P_{\mathrm{g}}\)
\(
\begin{aligned}
P_{\mathrm{g}} & =1.03 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} \times 10 \mathrm{~ms}^2 \times 1000 \mathrm{~m} \\
& \approx 103 \times 10^5 \mathrm{~Pa} \\
& \approx 103 \mathrm{~atm}
\end{aligned}
\)

(c) The pressure outside the submarine is \(P=P_{\mathrm{a}}+\rho g h\) and the pressure inside it is \(P_{\mathrm{a}}\). Hence, the net pressure acting on the window is gauge pressure, \(P_{\mathrm{g}}=\rho g h\). Since the area of the window is \(A \stackrel{g}{=} 0.04 \mathrm{~m}^2\), the force acting on it is
\(
F=P_{\mathrm{g}} A=103 \times 10^5 \mathrm{~Pa} \times 0.04 \mathrm{~m}^2=4.12 \times 10^5 \mathrm{~N}
\)

Hydraulic Machines

Let us now consider what happens when we change the pressure on a fluid contained in a vessel. Consider a horizontal cylinder with a piston and three vertical tubes at different points [Fig. 10.6 (a)]. The pressure in the horizontal cylinder is indicated by the height of the liquid column in the vertical tubes. It is necessarily the same in all. If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them.

This indicates that when the pressure on the cylinder was increased, it was distributed uniformly throughout. We can say whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions. This is another form of the Pascal’s law and it has many applications in daily life.

A number of devices, such as hydraulic lift and hydraulic brakes, are based on the Pascal’s law. In these devices, fluids are used for transmitting pressure. In a hydraulic lift, as shown in Fig. \(10.6\) (b), two pistons are separated by the space filled with a liquid. A piston of small cross-section \(A_1\) is used to exert a force \(F_1\) directly on the liquid. The pressure \(P=\frac{F_1}{A_1}\) is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area \(A_2\), which results in an upward force of \(P \times A_2\). Therefore, the piston is capable of supporting a large force (large weight of, say a car, or a truck, placed on the platform) \(F_2=P A_2=\frac{F_1 A_2}{A_1}\). By changing the force at \(A_1\), the platform can be moved up or down. Thus, the applied force has been increased by a factor of \(\frac{A_2}{A_1}\) and this factor is the mechanical advantage of the device. The example below clarifies it.

Example 5: Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are \(1.0\) \(\mathrm{cm}\) and \(3.0 \mathrm{~cm}\) respectively. (a) Find the force exerted on the larger piston when a force of \(10 \mathrm{~N}\) is applied to the smaller piston. (b) If the smaller piston is pushed in through \(6.0 \mathrm{~cm}\), how much does the larger piston move out?

Solution:

(a) Since pressure is transmitted undiminished throughout the fluid,
\(
\begin{aligned}
F_2=\frac{A_2}{A_1} F_1= & \frac{\pi\left(3 / 2 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(1 / 2 \times 10^{-2} \mathrm{~m}\right)^2} \times 10 \mathrm{~N} \\
& =90 \mathrm{~N}
\end{aligned}
\)

(b) Water is considered to be perfectly incompressible. Volume covered by the movement of smaller piston inwards is equal to volume moved outwards due to the larger piston.
\(
\begin{aligned}
& L_1 A_1=L_2 A_2 \\
& L_2=\frac{A_1}{A_2} L_1=\frac{\pi\left(1 / 2 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(3 / 2 \times 10^{-2} \mathrm{~m}\right)^2} \times 6 \times 10^{-2} \mathrm{~m} \\
& \simeq 0.67 \times 10^{-2} \mathrm{~m}=0.67 \mathrm{~cm} \\
&
\end{aligned}
\)
Note, atmospheric pressure is common to both pistons and has been ignored.

Example 6: In a car, lift compressed air exerts a force \(F_1\) on a small piston having a radius of \(5.0 \mathrm{~cm}\). This pressure is transmitted to a second piston of radius \(15 \mathrm{~cm}\) (Fig 10.7). If the mass of the car to be lifted is \(1350 \mathrm{~kg}\), calculate \(F_1\). What is the pressure necessary to accomplish this task? \(\left(g=9.8 \mathrm{~ms}^2\right)\).

Solution:

Since pressure is transmitted undiminished throughout the fluid,
\(
\begin{aligned}
& F_1=\frac{A_1}{A_2} F_2=\frac{\pi\left(5 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(15 \times 10^{-2} \mathrm{~m}\right)^2}\left(1350 \mathrm{~kg} \times 9.8 \mathrm{~m} \mathrm{~s}^{-2}\right) \\
& =1470 \mathrm{~N} \\
& \approx 1.5 \times 10^3 \mathrm{~N} \\
&
\end{aligned}
\)
The air pressure that will produce this force is
\(
P=\frac{F_1}{A_1}=\frac{1.5 \times 10^3 \mathrm{~N}}{\pi\left(5 \times 10^{-2}\right)^2 \mathrm{~m}}=1.9 \times 10^5 \mathrm{~Pa}
\)
This is almost double the atmospheric pressure.