Pressure

Section 17.2 Pressure

You have studied forces between bodies. They can be broadly classified as “action-at-a-distance” forces, such as gravity, electric and magnetic, and “contact” forces. Contact forces, such as normal and frictional forces, either between two bodies or between one part of a body to another, act through a common area of contact.

We define pressure exerted by a contact force \(\vec F\) that acts over the area \(A\) to be equal to the magnitude of the normal component (\(F_\perp\)) of the force per unit contact area.

\begin{equation} \text{Pressure exerted by force, }\vec F = \frac{|F_\perp|}{A}.\tag{17.2} \end{equation}

Thus, if the normal force by table on a box is \(100\,\text{N}\) and the area of contact is \(20\,\text{cm}^2\text{,}\) then pressure exerted by this normal force on the box will be

\begin{equation*} \frac{100\,\text{N}}{20\,\text{cm}^2} = 5\,\text{N/cm}^2 = 5\times 10^{-4}\,\text{N/m}^2. \end{equation*}

If the same normal acted over the same box, but, say the contact surface was made smaller and was only \(2\,\text{cm}^2\text{.}\) Then, the pressure wil now be

\begin{equation*} \frac{100\,\text{N}}{2\,\text{cm}^2} = 50\,\text{N/cm}^2 = 5\times 10^{-3}\,\text{N/m}^2. \end{equation*}

Subsection 17.2.1 Pressure in a Fluid

An important characteristics of fluids is that there is no significant reaction force from the fluid when you apply a force horizontal, i.e., tangentially, to the free surface of the fluid as in Figure 17.1. The fluid simply flows under such a force. We say that in fluids you do not have a shear strain.

On the other hand, if you push on a fluid such that the fluid cannot flow due to it being confined as in Figure 17.2. Such a force will compresses the volume of the fluid, which changes the volume (\(\Delta V\)) ever so slighltly causing a volumetric strain (\(\Delta V/V\)) in the fluid. By Hooke’s law this will lead to develop a stress inside the fluid. The stress in fluid is called pressure.

\begin{equation} p_\text{fluid} = B \frac{\Delta V}{V},\tag{17.3} \end{equation}

where \(B\) is the Bulk modulus of the fluid.

If the fluid is not flowing, i.e., in a static fluid, the force at each point of the fluid must balance out. Therefore, when we look at the boundary surface between the piston and the fluid, the force due to the applied force from above and force by internal stress, i.e., pressure times area, must be equal. That is,

\begin{equation*} F_\perp = p_\text{fluid}\, A. \end{equation*}

From this, we get the following interesting equality.

\begin{equation*} F_\perp = B \frac{A}{V} \Delta V. \end{equation*}

Although we have attached a subscript to pressure \(p_\text{fluid}\text{,}\) in the following, we will omit this and denote pressure in the fluid by just \(p\text{.}\)

Let \(L\) be the length of the cylinder that contains the fluid. If the fluid is compressed along the normal direction by a distance \(\Delta L\text{,}\) then

\begin{equation*} \Delta V = A \Delta L,\ \ \ V = AL. \end{equation*}

Using this in the prior equation we get the following for the normal component of the applied force

\begin{equation*} F_\perp = \left( B \frac{A}{L}\right)\, \Delta L. \end{equation*}

That is the fluid acts as spring of spring constant \(k\) with

\begin{equation*} k = \frac{BA}{L}, \end{equation*}

and pressure in the fluid is like the restoring force of the spring.

Example 17.4. Pressure of Same Magnitude Force Acting on a Pin Versus on a Nail.

Compare the pressures by a \(1\text{ N}\) force on a pin and a nail. The tip of the pin has a diameter of \(10\ \mu\text{m}\) and the nail \(100\ \mu\text{m}\text{.}\)

Answer.

pin: \(1.27\times 10^{10}\text{ Pa}\text{,}\) nail: \(1.27\times 10^8\text{ Pa}\text{.}\)

Solution.

First we work out the areas in \(\text{m}^2\text{.}\)

\begin{align*} A_\text{pin} \amp = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4}\times 10^{-10}\text{ m}^2, \\ A_\text{nail} \amp = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4}\times 10^{-8}\text{ m}^2, \end{align*}

Dividing the force by the area will give pressure in each case.

\begin{align*} p_\text{pin} \amp = 1.27\times 10^{10}\text{ Pa},\\ p_\text{nail} \amp = 1.27\times 10^{8}\text{ Pa}. \end{align*}

Subsection 17.2.2 Units of Pressure

The dimension of pressure is

\begin{equation*} [p] = [M][L]^{-1}[T]^{-2}, \end{equation*}

and therefore the unit in SI system is

\begin{equation*} \text{kg/m.s}^2\text{ called Pascal (Pa)}, \end{equation*}

in honor of the French mathematician and physicist Blaise Pascal (1623-1662).

Actually, other units of pressure are in more common use in physics, engineering, and other sciences. For instance, the ambient pressure due to the atmosphere is called one atmosphere (atm) which is

\begin{equation*} 1\text{ atm} \approx 1.013\times10^5\text{ Pa}. \end{equation*}

The pressure at the bottom of a 760-mm high column of mercury at \(0^{\circ}\text{ C}\) in a container where the top part is evacuated is equal to the atmospheric pressure. Thus, 760-mmHg is also used in place of 1 atmospheric pressure.

\begin{equation*} 1\text{ atm} = 760\text{ mmHg}. \end{equation*}

In vacuum physics labs, one often uses another unit called Torr, named after the Italian inventor Evangelista Torricelli (1608-1647), who invented the mercury manometer for measuring pressure. One Torr is equal to a pressure of \(1\text{ mmHg}\text{.}\) Low pressures in atmospheric physics and high vacuum physics are often expressed in yet another unit called millibar or mbar, which is equal to 100 Pa, with a bar being \(10^5\text{ Pa}\text{.}\) The multitude of units for pressure is certainly very confusing to students. Perhaps, when doing problems, you should stick to \(\text{Pa}\text{.}\)

\begin{gather*} 1\text{ Torr} = 1\text{ mmHg},\\ 1\text{ bar} = 10^5\text{ Pa},\\ 1\text{ mbar} = 100\text{ Pa}. \end{gather*}

Not to be left out is the pounds per square inch (\(\text{psi} \)) will be the English unit of pressure - this is used, for instance in tire pressure in USA:

\begin{equation*} 1\text{ psi} = 7.015\times10^3\text{ Pa} = \dfrac{1}{14.7}\text{ atm}. \end{equation*}

Subsection 17.2.3 Range of Pressure

Table 17.5 displays pressure under various conditions in nature.

Table 17.5. Pressures in universe (Source: various)

Place	Pressure (Pa)	Place	Pressure (Pa)
Center of the Sun	\(\approx 3 \times 10^{16} \)	Mechanical Vacuum Pump	\(100 \text{ to } 10^{-4} \)
Center of the Earth	\(\approx 4 \times 10^{11} \)	Cryopump molecular beam epitaxy	\(10^{-9} \)
Atmosphere	\(1.013\times10^5 \)	On the Moon	\(10^{-9} \)
Vacuum cleaner	\(\approx8\times10^4\)	Interstellar space	\(10^{-15} \)

Exercises 17.2.4 Exercises

1. Balancing Force by Gas Pressure.

In a zero gravity and vacuum environment, nitrogen gas is kept at a pressure of \(50,000\text{ Pa}\) in a rigid cylinder with a movable piston. The area of the piston is \(20\text{ cm}^2\text{.}\) What force must be applied on the piston to keep the gas from expanding?

Hint.

Balance the forces on the piston.

Answer.

\(100\text{ N}\text{.}\)

Solution.

Let \(F\) be the magnitude of the force required to act normally to the piston. Then, this force must balance the force by the pressure of the gas.

\begin{equation*} F = pA. \end{equation*}

Now, before we put in the numerical values, we need to convert area in \(\text{m}^2\text{.}\)

\begin{equation*} A = 20\text{ cm}^2 = 20\times 10^{-4}\text{ m}^2, \end{equation*}

where note that we need a factor of \((1/100)^2\text{.}\) Therefore, we have

\begin{equation*} F = 50,000 \times 20\times 10^{-4} = 100\text{ N}. \end{equation*}

2. Conversion Among Units of Pressure.

In a low pressure laboratory, partial vacuum of \(1.0 \times 10^{-6}\text{ Torr}\) is achieved. (a) What is the corresponding reading in millibar? (b) How much is that in psi? (c) How much is that in Pa?

Hint.

Look up the unit conversion factors.

Answer.

(a) \(1.33\times 10^{-6}\text{ mbar}\text{.}\) (b) \(1.9\times 10^{-8}\text{ psi}\text{.}\) (c) \(1.33\times 10^{-4}\text{ Pa}\text{.}\)

Solution.

Using the conversion factors gives the results.

(a) \(1.33\times 10^{-6}\text{ mbar}\text{.}\)

(b) \(1.9\times 10^{-8}\text{ psi}\text{.}\)

3. Opening a Door on a Windy Day - Handling Pressure Difference.

On a windy day, the air pressure outside a 90 cm x 200 cm door is 943 mbar, while the air pressure inside the building is 1013 mbar. How much force you must apply at the handle of the door to open it if the door is hinged on the long side?

Hint.

(1) Place the force by pressures at the center. (2) Balance the torque by your force.

Answer.

\(630\text{ N} \text{.}\)

Solution.

The force by outside air and inside air could both be placed at the center of the door. Say, we apply a force \(F\) at the handle. Then, the torque balancing will give us the following equation.

\begin{equation*} F w = \left[ (p_\text{in} - p_\text{out})wh \right]\, \dfrac{w}{2}. \end{equation*}

Solving for \(F\) we get

\begin{equation*} F = \dfrac{1}{2} (p_\text{in} - p_\text{out})wh. \end{equation*}

Putting in the numerical values requires us to convert \(\text{mbar}\) to \(\text{Pa}\text{.}\)

\begin{equation*} \dfrac{1}{2} \times (1013 - 943)\times \times 10^{2} \times 0.9\times 2.0 = 630\text{ N}. \end{equation*}

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