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Facts about pressure

We want to answer the question: Given the pressure at one point in a fluid, what is the pressure at any other point? In fact, the key intuition on many problems is understanding what the pressure is everywhere.

Let us start with the simple situation shown in Figure 10-4. Point 1 is directly above point 2 in a fluid. The pressure at point 1 is P1. What is the pressure at point 2?

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Figure 10-4


Consider a vertical pipe filled with fluid and consider the body of fluid between points 1 and 2 as an object. The pipe's height is h and its cross-sectional area is A.

There are three vertical forces: the force of gravity, the force of the pressure pushing up from below, and the force of the pressure pushing down from above. The fluid is not accelerating, so the net force is zero, and we have
0 = F2F1mg
= P2AP1AρVg
= P2AP1AρAhg
We can cancel the factor A:
0 = P2P1ρhg
P2 = P1 + ρhg
This equation applies not only to this situation but to any situation involving two vertically separated points in a fluid. The pressure is greater at point 2 because more fluid is pressing down on top of point 2 than on top of point 1.

We can obtain pressures at other points in the fluid using a principle, discovered by Blaise Pascal: "Pressure applied to an enclosed fluid is transmitted to every portion of the fluid and the walls of the containing vessel in all directions." The language is a bit obscure, but it translates into principles 2 and 3 below.
Law of Hydrostatic Equilibrium
In a body of fluid,
  1. the pressures at two points separated vertically by height h are related by
  1. the pressures at two points separated only horizontally arc the same,

  2. and the pressure at a given point is the same in all directions.



We are standing on the fifty-first story of a hotel, where each story is 4 meters high. How much less is the pressure on the fifty-first story than the pressure at the ground floor? (Use 1.2 x 10–3 g/cm3 for the density of air.)



This is a straightforward application of the formula:

P2 = P1 + ρgh,

= 2400 kg/ms2

= 2400 Pa

This pressure is fairly small compared to Patm, but it is enough to make your ears pop. Notice what is going on here. The people on the ground floor have to deal with not only the air column on top of us at the fifty-first floor, but also they have the air column between us and the ground floor sitting on their head.



An underground cave is almost filled with water as shown in Figure 10-5. The air pressure above point S is 1 atm. The point Q is 20 m directly below point S, and T is at the same height as Q. R is 5 m vertically below S.

  1. What is the pressure at point Q?
  2. What is the pressure at point T against the floor?
  3. What is the pressure at point T against the walls?
  4. What is the pressure in the air chamber above R?

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Figure 10-5

a. The pressure at Q is given by the formula:

PQ = PS + (103 kg/m3)(10 m/s2)(20 m)
= 1.01 x 105 Pa + 2.0 x 105 Pa
= 3.01 x 105 Pa

b and c.

PT = 3.01 x 105 Pa


PR = PS + (103 kg/m3)(10 m/s2)(5 m)
= 1.5 x 105 Pa

The pressure inside the air chamber varies slightly with height but it is approximately equal to the pressure at R.



A pan contains a pool of mercury, with an inverted tube, as shown in Figure 10-6.
Point A is 38 cm above the surface of the mercury in the pan. Point B is 75 cm above the surface of the mercury in the pan. Point C is 76 cm above the surface of the mercury in the pan. The pressure of the air on the mercury in the pan is 1.01325 x 105 Pa, the density of mercury is 1.36 x 104 kg/m3, and the acceleration due to gravity is 9.8 m/s2.

  1. What is the pressure at point A?
  2. What is the pressure at point B?
  3. What is the pressure at point C?

..\art 10 jpg\figure 10-6.jpg
Figure 10-6


a. When we apply the equation to point A, we obtain


Patm = PA + ρgh,

PA = Patm – ρgh

= 1.01325 x 105 Pa – (1.36 x 104)(9.8)(0.38) Pa

= 5.1 x 104 Pa


b. At point B, we obtain


PB = Patm – ρgh

= 1.01325 x 105 Pa – (1.36 x 104)(9.8)(0.75) Pa

 103 Pa


c. At point C, we obtain


PC = Patm – ρgh

= 1.01325 x 105 Pa – (1.36 x 104)(9.8)(0.76) Pa

 0 Pa


Thus the pressure vanishes at the top of the column.


This is a simple barometer. Above the mercury column is a vacuum, or, more accurately, mercury vapor. The last calculation shows that the height of the mercury column is proportional to the outside pressure. For this reason, the height of a hypothetical mercury column is often given as units of pressure:


1 torr = 1 mm of mercury = pressure sufficient to lift Hg 1 mm             ...(9)


760 torr = 1 atm


These are the units used in a sphygmomanometer. But the numbers reported in blood pressure measurements are the pressures in excess of atmospheric pressure, called the gauge pressure. For instance, the systolic pressure of a woman with blood pressure 110/60 is actually (760 + 110) torr = 870 torr (assuming 760 torr atmospheric pressure). Thus:




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