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 104. Point 1 is directly above point 2 in a fluid. The pressure at point 1 is P_{1}. What is the pressure at point 2?
Figure 104
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
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,

We are standing on the fiftyfirst story of a hotel, where each story is 4 meters high. How much less is the pressure on the fiftyfirst story than the pressure at the ground floor? (Use 1.2 x 10^{–3 }g/cm^{3 }for the density of air.)
This is a straightforward application of the formula:
P_{2} = P_{1} + ρgh,
= 2400 kg/ms^{2}
= 2400 Pa
This pressure is fairly small compared to P_{atm}, 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 fiftyfirst 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 105. 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.
 What is the pressure at point Q?
 What is the pressure at point T against the floor?
 What is the pressure at point T against the walls?
 What is the pressure in the air chamber above R?
Figure 105
P_{Q} = P_{S} + (10^{3} kg/m^{3})(10 m/s^{2})(20 m)
= 1.01 x 10^{5} Pa + 2.0 x 10^{5} Pa
= 3.01 x 10^{5} Pa
b and c.
P_{T} = 3.01 x 10^{5} Pa
d.
P_{R} = P_{S} + (10^{3} kg/m^{3})(10 m/s^{2})(5 m)
= 1.5 x 10^{5} 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 106.
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 10^{5} Pa, the density of mercury is 1.36 x 10^{4} kg/m^{3}, and the acceleration due to gravity is 9.8 m/s^{2}.
 What is the pressure at point A?
 What is the pressure at point B?
 What is the pressure at point C?
Figure 106
a. When we apply the equation to point A, we obtain
P_{atm} = P_{A} + ρgh,
P_{A} = P_{atm} – ρgh
= 1.01325 x 10^{5} Pa – (1.36 x 10^{4})(9.8)(0.38) Pa
= 5.1 x 10^{4} Pa
b. At point B, we obtain
P_{B} = P_{atm} – ρgh
= 1.01325 x 10^{5} Pa – (1.36 x 10^{4})(9.8)(0.75) Pa
≈ 10^{3} Pa
c. At point C, we obtain
P_{C} = P_{atm} – ρgh
= 1.01325 x 10^{5} Pa – (1.36 x 10^{4})(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:
...(10) 