Sumário

Pressure

You have no doubt heard the word pressure being used in relation to blood (high or low blood pressure) and in relation to the weather (high- and low-pressure weather systems). These are only two of many examples of pressures in fluids. Pressure $P P size 12{P} {}$ is defined as

P = \frac{F}{A} P = \frac{F}{A} size 12{P= {  {F}  over  {A} } } {}

11.6

where $F F size 12{P} {}$ is a force applied to an area $A A size 12{P} {}$ that is perpendicular to the force.

Pressure

Pressure is defined as the force divided by the area perpendicular to the force over which the force is applied, or

P = \frac{F}{A} . P = \frac{F}{A} . size 12{P= {  {F}  over  {A} } } {}

11.7

A given force can have a significantly different effect depending on the area over which the force is exerted, as shown in Figure 11.6. The SI unit for pressure is the pascal, where

1 Pa = 1 {N/m}^{2} . 1 Pa = 1 size 12{1`"Pa"=1`"Nm" rSup { size 8{2} } } {}

11.8

In addition to the pascal, there are many other units for pressure that are in common use. In meteorology, atmospheric pressure is often described in units of millibar (mb), where

100 mb = 1 \times 10^{4} Pa . 100 mb = 1 \times 10^{4} . size 12{"100"`"mb"=1 times "10" rSup { size 8{5} } `"Pa"} {}

11.9

Pounds per square inch $({lb/in}^{2} or psi) ({lb/in}^{2}) size 12{ left ("lb/in" rSup { size 8{2} } `"or"``"psi" right )} {}$ is still sometimes used as a measure of tire pressure, and millimeters of mercury (mm Hg) is still often used in the measurement of blood pressure. Pressure is defined for all states of matter but is particularly important when discussing fluids.

In figure a, the person is poked with a finger exerting a small pressure due to the large area of contact and, in b, he is poked with a syringe exerting a large pressure due to the small area of contact.

Figure 11.6 (a) While the person being poked with the finger might be irritated, the force has little lasting effect. (b) In contrast, the same force applied to an area the size of the sharp end of a needle is great enough to break the skin.

Example 11.2

Calculating Force Exerted by the Air: What Force Does a Pressure Exert?

An astronaut is working outside the International Space Station where the atmospheric pressure is essentially zero. The pressure gauge on her air tank reads $6 . 90 \times 10^{6} Pa 6 . 90 \times 10^{6} size 12{6 "." "90" times "10" rSup { size 8{6} } `"Pa"} {}$ . What force does the air inside the tank exert on the flat end of the cylindrical tank, a disk 0.150 m in diameter?

Strategy

We can find the force exerted from the definition of pressure given in $P = \frac{F}{A} P = \frac{F}{A} size 12{P= { {F} over {A} } } {}$ , provided we can find the area $A A size 12{A} {}$ acted upon.

Solution

By rearranging the definition of pressure to solve for force, we see that

F = PA . F = PA . size 12{F= ital "PA"} {}

11.10

Here, the pressure $P P size 12{P} {}$ is given, as is the area of the end of the cylinder $A A size 12{A} {}$ , given by $A = {πr}^{2} A = {πr}^{2} size 12{A=πr rSup { size 8{2} } } {}$ . Thus,

\begin{array}{l} F & = & (6.90 \times 10^{6} {N/m}^{2}) (3.14) {(0.0750 m)}^{2} \\ = & 1.22 \times 10^{5} N. \end{array} \begin{array}{l} F & = & (6.90 \times 10^{6}) (3.14) {(0.0750 m)}^{2} \\ = 1.22 \times 10^{5} \end{array} alignl { stack {
 size 12{F= left (6 "." "90" times "10" rSup { size 8{6} } `"N/m" rSup { size 8{2} }  right ) left (3 "." "14" right ) left (0 "." "0750"`m right ) rSup { size 8{2} } }  {} #
=1 "." "22" times "10" rSup { size 8{5} } `N "."  {}
} } {}

11.11

Discussion

Wow! No wonder the tank must be strong. Since we found $F = PA F = PA size 12{F= ital "PA"} {}$ , we see that the force exerted by a pressure is directly proportional to the area acted upon as well as the pressure itself.

The force exerted on the end of the tank is perpendicular to its inside surface. This direction is because the force is exerted by a static or stationary fluid. We have already seen that fluids cannot withstand shearing (sideways) forces; they cannot exert shearing forces, either. Fluid pressure has no direction, being a scalar quantity. The forces due to pressure have well-defined directions: they are always exerted perpendicular to any surface. (See the tire in Figure 11.7, for example.) Finally, note that pressure is exerted on all surfaces. Swimmers, as well as the tire, feel pressure on all sides. (See Figure 11.8.)

The forces inside a tire are shown by arrow lines. An inset shows an enlarged view of the valve in the tire. Air pressure in the tire keeps the valve closed.

Figure 11.7 Pressure inside this tire exerts forces perpendicular to all surfaces it contacts. The arrows give representative directions and magnitudes of the forces exerted at various points. Note that static fluids do not exert shearing forces.

A man swimming underwater has many arrows pointing toward him to represent the directions and magnitudes of the forces exerted on him at various points.

Figure 11.8 Pressure is exerted on all sides of this swimmer, since the water would flow into the space he occupies if he were not there. The arrows represent the directions and magnitudes of the forces exerted at various points on the swimmer. Note that the forces are larger underneath, due to greater depth, giving a net upward or buoyant force that is balanced by the weight of the swimmer.

PhET Explorations

Gas Properties

Pump gas molecules to a box and see what happens as you change the volume, add or remove heat, change gravity, and more. Measure the temperature and pressure, and discover how the properties of the gas vary in relation to each other.

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