(8.1)A. Density
B. Specific Gravity
C. Pressure
D. Pressure in a Confined Fluid
E. Pascal’s Principle
F. Archimedes’ Principle
Problems
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A simple way to classify matter is as either a solid or a fluid. Solids are objects with a definite size and shape, and over the first seven chapters we developed laws that describe the equilibrium conditions or motion of solid objects. Fluids, which include liquids and gases, are distinct from solids because of their abilities to flow and change shape. In this chapter we will use Newton’s laws to develop techniques for describing fluids in equilibrium. These techniques will be somewhat similar to methods used for solid bodies, but because the nature of a fluid id so much different that a solid we will need to introduce variables which are appropriate for describing fluids.
A substance can be identified by properties such as color, odor, or crystalline structure. An important distinguishing attribute of any substance is its density. Density tells us how much of a substance occupies a given volume, and it can be used to identify solids as well as liquids. For example, we know that 1 kg of lead will occupy less volume than 1 kg of water, which would occupy less volume than 1 kg of air under standard conditions. Furthermore, we know that 1 m3 of lead would contain more mass that 1 m3 of water, which would contain more mass than 1 m3 of air under standard conditions.
The density of an object is defined to be mass per unit volume. Or in symbols
(8.1)
The two most common units for mass density are grams per cubic centimeter (g/cm3) and kilograms per cubic meter (kg/m3).
It is common in the British system to make use of the weight density rather than the mass density. The weight density is defined to be the weight of a unit volume of material, or, in equation form
(8.2)
Weight density is commonly expressed in units of pounds per cubic foot.
There are a vast array of units in the field of fluids. The following relations are very useful for converting units.
Volume:
1 cm3 = 1 millimeter (ml)
1000 ml = 1000 cm3 = 1 liter
1 m3 = 1,000,000 cm3
1 m3 = 35.3 ft3
1 ft3 = 1728 in3
1 gallon = 231 in3
Density: 1 gram/cm3 = 1000 kg/m3 = “62.4 lb/ft3”
Note: The quotation marks surrounding 62.4 lb/ft3 are to emphasize that we have equated a mass density and a weight density. This simply means that an object at the surface of the earth with a mass density of 1000 kg/m3 would have a weight density of 62.4 lb/ft3.
Examples:
| Substance | Density (gm/cm3) |
| Water | 1.0 |
| Air | 0.0013 |
| Styrofoam | 0.1 |
| Bone | 1.8 |
| Blood | 1.06 |
| Iron | 7.8 |
| Mercury | 13.6 |
| Silicon | 2.4 |
| Earth | 5.6 |
| Jupiter | 1.3 |
The specific gravity, or relative density, of a substance is defined to be the ratio of the density of the substance to the density of water.
(8.3)
Since it is defined as a ratio of densities, specific gravity
is a unitless quantity. Objects with a specific gravity less than one will
float in water. This simply means that it is less dense than water.
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Consider the simple example of pushing your thumb against your nose with a force F. Such an action may be uncomfortable but it is doubtful that you will seriously injure yourself. Now try pushing against your nose with the same force F, but this time use a sharp nail instead of you thumb. In each case described the applied force is the same. However, much more pain is experienced when you use a nail. The reason why the nail hurts more than your thumb is because the force is exerted over a large area when your thumb is used. When the nail is employed this same force is exerted over a more restricted area. Pressure is a measure of the force exerted per unit area. It is defined as
Or in symbols 
(8.4)
The units of pressure are a force divided by an area. As was the case with volume, there are many units of pressure that are commonly used today. The following list of conversions illustrates some of the more popular units of pressure.
1 N/m2 = 1 Pascal = 0.021 lb/ft2
1 lb/in2 = 1 psi = 144 lb/ft2
1 Atmosphere = 76 cm of mercury = 100,000 N/m2 = 14.7 lb/in2
Example:
Let us examine a student with a mass of 70 kg (weight = 700 N = 154 lb). Calculate the pressure exerted by the student if he
Note: In Chapter 3 we discussed coefficients of kinetic friction.
In that chapter you were told that the coefficient of kinetic friction between
ice and steel was 0.9. However, if there is a layer of water between the ice
and steel, then the water acts as a lubricant and reduces the coefficient of
kinetic friction to 0.01. It is clear that this lubricant is necessary for
ice skating. The melting point of ice can be lowered if it is subjected to
high pressure. It is known that every 1200 N/cm2 of pressure will
lower the melting point of ice by 1˚C. In the previous example we calculated
the pressure exerted by the ice skate to be 175 N/cm2. This pressure
will lower the melting point of ice only a fraction of a degree. Thus it seems
impossible to ice skate! Can you think of any mistake or incorrect assumption
we have made in this problem? Think about how the blades on an ice skate are
shaped.
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Shown below is a block of wood that weighs 100 N (22.5 lb) and an aquarium which holds water weighing 100 N also. The surface areas on the bottom of the block and the bottom of the aquarium are both 0.1 m2.
Because of its weight the block exerts a pressure on the surface on which it rests. Using Eq. (8.4) we can readily determine this pressure to be 1,000 N/m2. Clearly, the weight of the water in the aquarium also exerts a pressure of 1,000 N/m2 on the bottom of the aquarium.
There are two important differences between the pressure exerted by the block and the water in the aquarium. The first is that the water exerts a pressure on the sides of the aquarium and this pressure increases as the depth below the surface increases. Because of its nature, however, the block exerts no sideways pressure.
The second difference is a matter of how we measure these pressures. For a solid it is a simple task to measure its weight and contact area. This is not the case for a fluid. Suppose you were 20 meters under water. How would you determine the weight of water above you? How would you determine the contact area? Since both of these measurements would be very difficult we need to find an alternative method for determining pressure in a fluid.
The container below holds a liquid of height h and cross sectional area A. We will attempt to find an expression for the pressure on the bottom of this container due to weight of the fluid above it. In addition, we want this expression in terms of variables that are easily measured when dealing with fluids.
The definition of pressure is
Where the force on the bottom of the container is just equal to the weight of the fluid above it, F = mg. Remember now the definition of density, d = m/V, which allows use to write m = dV. Substitute these into the pressure equation
Now the volume can be expressed as V = height x length x width = hA. This allows us to write h = V/A. Substituting we then find
Or
(8.5)
This tells us the pressure due to a liquid of density d of height h.
Examples:
Notes:
Examine the two blocks in contact shown below. If a force is exerted on block 1 it will exert a force on block 2.
It is easy to see that the force on block 2 is in the same direction as the applied force on block 1.
Now examine what happens when we exert a force on an incompressible fluid. Shown below are two such examples.
In both cases the pressure is transmitted to every point within the fluid. They are examples of Pascal’s principle.
Pressure applied to an incompressible confined fluid
is transmitted to every point in the fluid.
Example: The best known application of Pascal’s principle is the hydraulic lift, shown schematically below.
If force F1 is applied to the piston of area A1, the pressure in the fluid is increased throughout by F1/A1. To maintain the liquid equilibrium, the upward pressure of the liquid on the piston of area A2, must be balanced by an equal downward pressure. We have by Pascal’s principle that
Solving this for F2 we find
If A2 is much larger that A1 then F2
is much larger than F1. Thus a relatively small force F1
can support (or lift) a very large weight resting on the piston of area A2.
The walls of the container must, of course, also be strong enough to withstand
the large pressure without rupturing. The same hydraulic principle is used
in the brake system of a car.
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Let us examine an empty plastic milk container, a baseball, and a rock. You know that if you try to completely submerge the milk container you must push it under the water. You also realize that the baseball will float, and that the rock, even though it sinks, appears to weigh less than when it is not submerged. These observations reflect the fact that an upward force helps to support a submerged object. We call this a buoyant force.
The buoyancy principle , first discovered by Archimedes is as follows:
A body partially or wholly immersed in a fluid is buoyed up
by a force equal to the weight if the fluid when it displaces.
Buoyant Force = Weight if Fluid Displaced
BF = mFg (8.6)
where mF is the mass of the displaced fluid.
Note: When solving buoyancy problems it is helpful to remember the definition of density
D = m/V, m = dV, or V = m/d
Examples: We will consider three different examples where buoyant forces play an important role.
| Mass Density | Pressure in a Confined Fluid |
| Weight Density | Pascal’s Principle |
| Specific Gravity | Achimedes’ Principle |
| Pressure |
