Fisica

CHAPTER 2 - FLUIDS

2.1 Mass Density
• Fluids are materials that can flow which include both gases and liquids.
← Example:
i) gas: air
ii) liquid: water
• Mass density of liquid or gas is important factor that determine its behaviour as a fluid.
← Example:
Air (gas) has smallest densities compared to water (liquid) because gas molecules are relatively far apartand contains large fraction of empty space. While for liquids, the molecules are much more tightly packed that leads to larger densities.
• Equation of mass density, (




• A convenient way to compare densities is to use the concept of specific gravity.





EXERCISE 1
What is the approximate mass of air in a living room 4.8m x 3.8m x 2.8m?

SOLUTION

The mass is found fromthe density of air and the volume of air.
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EXERCISE 2
Accomplished silver workers in India can pound silver into incredibly thin sheets as thin as 3.00 x 10-7 m. Find the area of such sheet that can be formed from 1.00 kg of silver.

SOLUTION
[pic]

The area of the silver, is, therefore, [pic]
EXERCISE 3
A bottle has a mass of 35.00 g when empty and 98.44 g when filledwith water. When filled with another fluid, the mass is 88.78 g. What is the specific gravity of this other fluid?

SOLUTION
To find the specific gravity of the fluid, take the ratio of the density of the fluid to
that of water, noting that the same volume is used for both liquids.
[pic]
2.2 Pressure
• The pressure P exerted by a fluid is the magnitude F of the force actingperpendicular to a surface in the fluid divided by the area A over which the force acts:



SI Unit : N/m2 or Pascals ( Pa ) where: 1 Pa = 1 Nm 2
Other units : atmosphere (atm) where 1 atm = 1.013 x 105 Pa

EXERCISE 4
High heeled shoes can cause tremendous pressure to be applied to a floor. Suppose the radius of a heel is 6.00x10-3m. Find the pressure that is applied to the floor under theheel because of the weight of a 50 kg woman.

SOLUTION
[pic]

EXERCISE 5
A person who weighs 625 N is riding a 98 N mountain bike. Suppose the entire weight of the rider and bike is supported equally by the two tires. If the gauge pressure in each tire is 7.6 x 105 pa, what is the area of contact between each tire and the ground?

SOLUTION
Since the weight is distributed uniformly, eachtire exerts one-half of the weight of the rider and bike on the ground. According to the definition of pressure,
Equation 11.3, the force that each tire exerts on the ground is equal to the pressure P inside the tire times the area A of contact between the tire and the ground. From this relation, the area of contact can be found.

The area of contact that each tire makes with the ground is
[pic]2.3 Pressure and Depth in a Static Fluid

← How the pressure in a fluid varies with depth?

• Figure 1 shows a container of fluid (assumed to be liquid) in which one column of the liquid is outlined.

• The liquid is at rest. (If a fluid is at rest in a container, all portions of the fluid must be in static equilibrium).

• This column has a cross sectional area, A and extends to adepth, h.

• 3 external forces act on this volume of liquid:
i) Force of gravity or weight, mg
ii) Downward force, P1A
iii) Upward force, P2A exerted by the liquid below it.


• Since the column is in equilibrium, the sum of vertical forces equal to zero.
(Fy = 0
P2A (P1A – mg = 0 OR
P2A = P1A + mg

• But mass, m = ( V where V = A x htherefore, P2A = P1A + (Ahg OR
where P1 = pressure at higher level
P2 = pressure at deeper level
h = vertical distance between 2 points

• according to the equation, pressure P2 at a depth h is greater than P1


← Conclusion
Pressure in a fluid increases with depth due to the weight of the fluid above
the point of interest.

EXERCISE 6
At a given instant...