SHS

General Physics 1
Quarter 2 – Week 6
Module 6 - Pascal’s Principle in Analyzing Fluids Versus
Archimedes’ Principle of Buoyancy
General Physics 1
Grade 11/12 Quarter 2 - Module 6 - Pascal’s Principle in Analyzing Fluids
Versus Archimedes’ Principle of Buoyancy
First Edition, 2020

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La Union Schools Division
Region I

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General Physics 1
Quarter 2 – Week 6
Module 6 - Pascal’s Principle in
Analyzing Fluids Versus
Archimedes’ Principle of
Buoyancy
 Target

Fluids play a vital role in many aspects of everyday life. We drink them, breathe
them, swim in them. They circulate through our bodies and control our weather.
Airplanes fly through them; ships float in them. A fluid is any substance that can flow;
we use the term for both liquids and gases. We usually think of gas as easily
compressed and a liquid as nearly incompressible, although there are exceptional
cases.

In the preceding discussion, you have learned about density, specific gravity,
mass, and volume (and their relationships with each other). In this section, we would
be going to deal with pressure in a fluid, fluid density, depth, Pascal’s principle, and
Archimedes’ principle or Buoyancy.

After going through this module, you are expected to:

1. relate pressure to area and force (STEM_GP12FM-IIf41);
2. relate pressure to fluid density and depth (STEM_GP12FM-IIf42);
3. apply Pascal’s principle in analyzing fluids in various systems (STEM_GP12FM-
IIf43);
4. apply the concept of buoyancy and Archimedes’ principle (STEM_GP12FM-
IIf44)

Before going on, check how much you know about this topic. Answer the
pretest on the next page in a separate sheet of paper.
 Jumpstart

For you to understand the lesson well, do the following activities.

Have fun and good luck!

Direction: Write the letter of the term or phrase that best completes the statement
or answers the question.

Activity: Press on me!

_____1. Related to the word press.

A. Pressure B. Friction C. Force D. Power

_____2. The formula for pressure is ___________.

A. Pressure = Force divided by Area C. Pressure = Force times Area
B. Pressure = Area divided by Force D. Pressure = Area times Force

_____3. The unit for pressure is ________________.

A. Gram B. Pascal C. Newton D. Miles

_____4. When force increase, ________________.

A. pressure decreases C. pressure stays the same
B. pressure increases D. pressure increases then decreases

_____5. When area increases, ______________________.

A. pressure increases C. pressure decreases
B. pressure stays the same D. pressure increases then decreases

_____6. Any substance that can easily flows and therefore can easily change shape is ____.
A. Gas B. Liquid C. Fluid D. Solid

_____7. As elevation increases, ______________________.

A. pressure increases C. pressure increases then decreases
B. pressure stays the same D. pressure decreases

_____8. As elevation decreases, ___________________.

A. pressure decreases C. pressure increases then decreases
B. pressure stays the same D. pressure increases

_____9. As depth decreases, ___________________.

A. pressure decreases C. pressure increases then decreases
B. pressure stays the same D. pressure increases

_____10. As depth increases, ____________________.

A. pressure decreases C. pressure increases then decreases
B. pressure increases D. pressure stays the same
 Discover

Whenever we visit the hospital for a check-up and some kind of diagnosis, we
commonly hear about the word pressure, which is related to blood (high or low blood
pressure) and concerning the weather (high and low-pressure weather systems). These
are only two of many examples of pressure in fluids. Other than that we have the
drivers that are so used to fixing flat tires by reinflating them to the proper pressure.
The idea of pressure involves a force acting on an area.

If your ears have ever popped on a plane flight or ached during a deep dive in
a swimming pool, you have experienced the effect of depth on pressure in a fluid. At
the Earth’s surface, the air pressure exerted on you is a result of the weight of air
above you. This pressure is reduced as you climb up in altitude and the weight of air
above you decreases. Underwater, the pressure exerted on you increases with
increasing depth. In this case, the pressure being exerted upon you is a result of both
the weight of water above you and that of the atmosphere above you. You may notice
an air pressure change on an elevator ride that transports you many stories, but you
need only dive a meter or so below the surface of a pool to feel a pressure increase.
The difference is that water is much denser than air, about 775 times as dense.

Pressure

The pressure is defined as the magnitude of the force acting perpendicular to a

surface divided by the area of the surface. Pressure (P) is defined as
𝑭
𝑷=
𝑨
where F is the force applied to an area A that is perpendicular to the force.

A given force can have a significantly different effect depending on the area over
which the force is exerted, as shown in Figure 1.
Figure 1. (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.

Photo credit: Lumen Learning

 Pressure is a scalar quantity. So, it is expressed in magnitude only. The SI unit
of pressure is the pascal (Pa), named after the French scientist Blaise Pascal, where
𝑁
1 𝑃𝑎 = 1 2
𝑚

In addition to the pascal, there are many other units for the pressure that is in
common use. In meteorology, atmospheric pressure is often described in units of a
millibar (mb), where
100 𝑚𝑏 = 1𝑥105 𝑃𝑎

The atmosphere or atmospheric pressure (atm) is, as the name suggests, the
approximate average pressure of the atmosphere at sea level. The torr (named for
Evangelista Torricelli, who invented the mercury barometer in 1674) was formerly
called the millimeter of mercury (mmHg).

𝑙𝑏
Pounds per square inch ( 2 𝑜𝑟 𝑝𝑠𝑖) is still sometimes used to measure tire
𝑖𝑛
pressure, and millimeters of mercury (mmHg) 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.

There are older units for pressure UNITS OF PRESSURE

that are still in use. Conversion factors 1 atm = 1.013𝑥105 𝑃𝑎
between the atmosphere and other units of 1 atm = 1.013 𝑏𝑎𝑟
pressure are given in Table 1. 1 atm = 1.013𝑥106 𝑑𝑦𝑛𝑒
𝑐𝑚2
1 atm = 14.7 𝑙𝑏
𝑖𝑛2
1 atm = 76 𝑐𝑚 𝐻𝑔
1 atm = 760 𝑚𝑚 𝐻𝑔
Table 1. Units of Pressure
1 atm = 760 𝑡𝑜𝑟𝑟

In simpler terms, pressure is defined as force per unit area (P=F/A). It is

convenient to use pressure rather than force to describe the influence upon fluid
behavior. The standard unit for pressure is the Pascal, which is a newton per square
meter.

For an object sitting on a surface, the force pressing on the surface is the
weight of the object, but in different orientations, it might have a different area in
contact with the surface and therefore exert a different pressure.

Figure 2. Pressure difference

of a block in different
orientation.

Photo credit: Hyperphysics

 PHYS Bit (FYI)

Do you use stiletto-heeled shoes? Do you know exactly

how much pressure is under your foot? You would be
surprised! Pressure exerted by a lady wearing these shoes
is much greater than an elephant in barefoot. Ladies wear
these shoes for fashion’s sake, but with the immense
pressure it can produce, this may also be used as a
weapon!

Figure 3. Foot of a lady and an elephant

Photo credit: Science Photo Library

Sample Problem 1.
We walk with one foot on the ground, while an elephant walks with two feet.
Find the pressure exerted on the ground by a 700 N lady wearing stiletto-heeled
shoes assuming her entire weight is supported by one heel. The area of the heel is
2.00 cm2. Compare this to the pressure exerted by a 30 000 N elephant whose two
feet has a total area of 560 cm2.

Given: weight of the lady = 700 N

area of heel = 2.00 cm2 = 2.0x10-4 m2
weight of the elephant = 30 000 N
area of the feet (elephant)= 560 cm 2 = 5.60x10-2 m2

Unknown: Pressure exerted by the lady = ?

Pressure exerted by the elephant = ?

𝐹
Formula: 𝑃 =
𝐴

Solution and Answer:

𝐹 7.00 𝑥 102 𝑁
For the lady 𝑃𝐿 = = PL = 3.50 x 106 Pa
𝐴 2.00 𝑥 10−4 𝑚2

𝐹 3.00 𝑥 104 𝑁
For the elephant 𝑃𝐸 = = 5.60 𝑥 10−2 𝑚2 PE = 5.36 x 105 Pa
𝐴

The lady exerted more pressure on the ground as compared to the elephant.
The elephant’s feet has a larger area as compared to the heel of the lady.

Sample Problem 2.
The two feet of a 60 kg person covers an area of 500 cm2. (a) Determine the
pressure exerted by the two feet on the ground. (b) If the person stands on one foot,
what will the pressure be under that foot?

Approach - Assume the person is at rest. Then the ground pushes up on her with a
force equal to her weight mg, and she exerts a force mg on the ground where her feet
(or foot) contact it. Because 1 cm2 = (10-2 m)2 = 10-4 m2, then 500 cm2 = 0.050 m2.
Given: mass = 60 kg
agravity = g = 9.8 m/s2
Area = 500 cm2 = 0.050 m2

Unknown: (a) Pressure by two feet = ?

(b) Pressure by one foot = ?

𝐹
Formula: 𝑃 =
𝐴

Solution and Answer:

(a) The pressure on the ground exerted by the two feet is
𝑚
𝐹 𝑚𝑔 (60𝑘𝑔)(9.8 2 ) 𝑵
𝑃=𝐴= = 𝑠
= 𝟏𝟐𝒙𝟏𝟎𝟑
𝐴 0.050𝑚2 𝒎𝟐

(b) If the person stands on one foot, the force is still equal to the person’s weight, but
𝑵
the area will be half as much, so the pressure will be twice as much: 𝟐𝟒𝒙𝟏𝟎𝟑 .
𝒎𝟐

A fluid is a substance that flows easily. Gases and liquids are fluids, although
sometimes the dividing line between liquids and solids is not always clear. Because of
their ability to flow, fluids can exert buoyant forces, multiply forces in hydraulic
systems, allow aircraft to fly and ships to float.

A fluid remains at rest when the net force acting everywhere on the fluid is
equal to zero. We say that the fluid is on hydrostatic equilibrium when this
condition is met. If external forces are absent, hydrostatic equilibrium requires that
the pressure in the fluid is constant throughout the fluid. A pressure difference
within the fluid results in forces acting on the fluid which makes the fluid move.

Static Fluid Pressure

The pressure exerted by a static fluid depends only upon the depth of the fluid,
the density of the fluid, and the acceleration of gravity.

The pressure in a static fluid arises from the weight of the fluid and is given by
the expression
𝑷 = 𝝆𝒈𝒉

where 𝜌 = (rho) the density of the fluid (m/V),

g = the acceleration of gravity, and
h = the height of the fluid above the object (depth of fluid)

The pressure from the weight of a column of liquid of area A and height h is
(see figure 3)
 Figure 4. Static fluid pressure does not depend on the shape, total mass, or surface area
of the liquid.

Photo credit: Hyperphysics

The most remarkable thing about this expression is what it does not include.
The fluid pressure at a given depth does not depend upon the total mass or total
volume of the liquid. The above pressure expression is easy to see for the straight,
unobstructed column, but not obvious for the cases of different geometry which are
shown.

Variation of pressure with depth in a fluid of constant density

Pressure is defined for all states of matter, but it is particularly important when
discussing fluids. An important characteristic of fluids is that there is no significant
resistance to the component of a force applied parallel to the surface of a fluid. The
molecules of the fluid simply flow to accommodate the horizontal force. A force applied
perpendicular to the surface compresses or expands the fluid. If you try to compress
a fluid, you find that a reaction force develops at each point inside the fluid in the
outward direction, balancing the force applied on the molecules at the boundary.

Consider a fluid of constant density as shown in Figure 5. The pressure at the

bottom of the container is due to the pressure of the atmosphere (𝑝0 ) plus the pressure
due to the weight of the fluid. The pressure due to the fluid is equal to the weight of
the fluid divided by the area. The weight of the fluid is equal to its mass times the
acceleration due to gravity.
 Figure 5. The bottom of this container
supports the entire weight of the fluid
in it. The vertical sides cannot exert an
upward force on the fluid (since it
cannot withstand a shearing force), so
the bottom must support it all.

Photo credit: Lumen Learning

Since the density is constant, the weight can be calculated using the density:

𝑭 = 𝒘 = 𝒎𝒈 = 𝝆𝑽𝒈 = 𝝆𝑨𝒉𝒈

The pressure at the bottom of the container is therefore equal to atmospheric

pressure added to the weight of the fluid divided by the area:

𝝆𝑨𝒉𝒈
𝒑 = 𝒑𝟎 + = 𝒑𝟎 + 𝝆𝒉𝒈
𝑨
This equation is only good for pressure at a depth for a fluid of constant density.

PRESSURE AT A DEPTH FOR A FLUID OF CONSTANT DENSITY

The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere
plus the pressure due to the weight of the fluid, or
𝑝 = 𝑝0 + 𝜌ℎ𝑔
where 𝑝 is the pressure at particular depth, 𝑝0 is the pressure of the atmosphere, 𝜌 is the
density of the fluid, g is the acceleration due to gravity, and h is the depth.
Sample Problem 1.
Find the pressure on a scuba diver when she is 12 meters below the surface of
the ocean. Assume standard atmospheric conditions.

Given: d (depth) = 12 m
Density of sea water = 1.03 x103 kg/m3
Atmospheric pressure = 1.01 x 105 N/m2

Unknown: Pressure on a scuba diver = ?

Formula: 𝑷 = 𝝆𝒈𝒉

𝑘𝑔 𝑚 𝑵 𝑵
Solution: 𝑃𝑓𝑙𝑢𝑖𝑑 = 𝜌𝑔ℎ = (1.03𝑥103 ) (9.8 2 ) (12𝑚) = 𝟏𝟐𝟏𝟏𝟐𝟖 𝑜𝑟 𝟏. 𝟐𝟏𝒙𝟏𝟎𝟓
𝑚3 𝑠 𝒎𝟐 𝒎𝟐

𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑝0 + 𝑝𝑓𝑙𝑢𝑖𝑑 = (1.01𝑥105 ) + (1.21𝑥105 ) 𝑃𝑎 = 𝟐. 𝟐𝟐𝒙𝟏𝟎𝟐 𝒌𝑷𝒂

Sample Problem 2.
A cylindrical tank with a cross-sectional area of 2m2 contains water 4 high.
Find (a) the pressure of the water at the bottom of the tank and (b) the total force
acting on the bottom of the tank due to the water.

Given: cross-sectional area of the tank = 2m2

height of water in the tank = 4 m

Unknown: (a) P at the bottom of the tank = ?

(b) Total force acting on the bottom of the tank due to the water = ?

Formula: (a) P = F/A (b) F = PA

Solution: (a) The mass density of water is 1x103 kg/m3.

Weight of water = 𝜌𝑉𝑔

𝑘𝑔 𝑚
= (1000 ) (2𝑚2 )(4𝑚) (9.8 2) = 78400 𝑁
𝑚3 𝑠
𝐹 78400 𝑁 𝑁
𝑃= = 2 = 39 200 2 𝑜𝑟 𝟑𝟗 𝟐𝟎𝟎 𝑷𝒂
𝐴 2𝑚 𝑚

(b) Total force F due to the water = PA

𝑁
𝐹 = (39 000 ) (2𝑚2 ) = 𝟕𝟖 𝟒𝟎𝟎 𝑵
𝑚2
 The equation 𝑝 = 𝑝0 + 𝜌𝑔ℎ shows that an increase in surface, 𝑝0 , results in the
same pressure increase throughout the fluid. In general, a pressure increase
anywhere in the fluid is felt throughout the fluid. So, if an external force is applied to
a confined fluid, the fluid will be subjected to additional pressure. This fact was
recognized in 1653 by the French scientist Blaise Pascal (1623–1662) and is called
Pascal’s law.

Hydrostatic Paradox
Simon Stevin discovered the so-called “hydrostatic paradox” that is the
downward pressure of a liquid does not depend on the shape of the container but only
on the height of the liquid. Blaise Pascal built an apparatus, now known as Pascal
vases, consisting several interconnected containers of different shapes, heights, and
volumes. If liquid is poured into them, the liquid will stand at the same level in each
container.

Figure 6. Pascal vases

Photo credit: TEX StackExchange

Pascal’s principle states that a change in the pressure applied to an enclosed

fluid is transmitted undiminished to every portion of the fluid and the walls of the
containing vessel.

Demonstrating Pascal’s Principle

Consider the case in which the incompressible fluid is a liquid contained in a
tall cylinder, as in Fig. 7. The cylinder is fitted with a piston on which a container of
lead shot rests. The atmosphere, container, and shot exert pressure 𝑝𝑒𝑥𝑡 on the piston
and thus on the liquid. The pressure p at any point P in the liquid is then
𝑝 = 𝑝𝑒𝑥𝑡 + 𝜌𝑔ℎ

Let us add a little more lead shot to the container to increase 𝑝𝑒𝑥𝑡 by an amount
∆𝑝𝑒𝑥𝑡 . The quantities 𝜌, g, and h are unchanged, so the pressure change at P is
∆𝑝 = ∆𝑝𝑒𝑥𝑡 .
 Figure 7. Lead shot (small balls of lead) loaded
onto the piston create a pressure 𝑝𝑒𝑥𝑡 at the top
of the enclosed (incompressible) liquid. If 𝑝𝑒𝑥𝑡 is
increased, by adding more lead shot, the
pressure increases by the same amount at all
points within the liquid.

Photo credit: Fundamentals of Physics Halliday

& Resnick

This pressure change is independent of h, so it must hold for all points within
the liquid, as Pascal’s principle states.

Pascal’s Principle and the Hydraulic Lever

Figure 6 shows how Pascal’s principle can be made the basis of a hydraulic
lever. In operation, let an external force of magnitude Fi be directed downward on the
left hand (or input) piston, whose surface area is Ai. An incompressible liquid in the
device then produces an upward force of magnitude Fo on the right-hand (or output)
piston, whose surface area is Ao. To keep the system in equilibrium, there must be a
downward force of magnitude Fo on the output piston from an external load (not
shown). The force ⃗⃗𝐹𝑖 applied on the left and the downward force 𝐹 ⃗⃗⃗𝑜 from the load on
the right produce a change ∆𝑝 in the pressure of the liquid that is given by

𝐹𝑖 𝐹𝑜
∆𝑝 = =
𝐴𝑖 𝐴𝑜

so
𝐴𝑜
𝐹𝑜 = 𝐹𝑖
𝐴𝑖

𝑭𝒊𝒏 𝑭 𝐹𝑖𝑛 𝐴
= 𝑨𝒐𝒖𝒕 or = 𝐴 𝑖𝑛 or 𝑷𝒊 = 𝑷𝒐
𝑨𝒊𝒏 𝒐𝒖𝒕 𝐹𝑜𝑢𝑡 𝑜𝑢𝑡

For elaboration. Since A2 is greater than A1, F2 is greater than

F1.
 Figure 8. A hydraulic arrangement
that can be used to magnify a force.
The work done is, however, not
magnified and is the same for both
the input and output forces.

Photo credit: Fundamentals of Physics

Halliday & Resnick

The equation above shows that the output force Fo on the load must be greater
than the input force Fi if 𝐴𝑜 > 𝐴𝑖 , as is the case in Fig. 8. If we move the input piston
downward a distance di, the output piston moves upward a distance do, such that the
same volume V of the incompressible liquid is displaced at both pistons. Then

𝑉 = 𝐴𝑖 𝑑 𝑖 = 𝐴𝑜 𝑑 𝑜
which we can write
𝐴𝑖
𝑑 𝑜 = 𝑑𝑖 .
𝐴𝑜

This shows that, if 𝐴𝑜 > 𝐴𝑖 (as in Fig. 6), the output piston moves a smaller
distance than the input piston moves.

From the previous equations, we can write the output work as

𝐴𝑜 𝐴𝑖
𝑊 = 𝐹𝑜 𝑑𝑜 = (𝐹𝑖 ) (𝑑𝑖 ) = 𝐹𝑖 𝑑𝑖 ,
𝐴𝑖 𝐴𝑜

which shows that the work W done on the input piston by the applied force is equal
to the work W done by the output piston in lifting the load placed on it.

The advantage of a hydraulic lever is this: “With a hydraulic lever, a given force
applied over a given distance can be transformed to a greater force applied over a
smaller distance.”

The product of force and distance remains unchanged so that the same work
is done. However, there is often a tremendous advantage in being able to exert the
larger force. Most of us, for example, cannot lift an automobile directly but can with a
hydraulic jack, even though we have to pump the handle farther than the automobile
rises and in a series of small strokes.
Applications of Pascal’s Principle
The hydraulic lift is a force-multiplying device with a multiplication factor
equal to the ratio of the areas of the two pistons. Dentist’s chairs, car lifts and jacks,
many elevators, and hydraulic brakes all use this principle.

Figure 9. Some applications of Pascal’s principle: (a) hydraulic lift; (b) hydraulic brakes
in a car

Figure 9-a. A small input force is Figure 9-b. The brake system of the car.
used to exert a large output force by When the driver presses the brake pedal, the
making the area of the output piston pressure in the master cylinder increases.
larger than the area of the input This pressure increase occurs throughout the
piston. brake fluid, thus pushing the brake pads
against the disk attached to the car’s wheel.
Photo credit: Giancoli Physics (6th)
Photo credit: Giancoli Physics (6th)

Sample Problem 1.
The ratio of the diameters of the small piston to the large piston of a hydraulic
lift is 1:10. What weight can the large piston support when a force of 150 N is applied
to the small piston?

𝑑𝑖 1
Given: = 𝐹𝑖 = 150 𝑁
𝑑𝑜 10

(𝜋𝑑 2 ) 𝐴𝑖 𝑑 2 1 𝐹𝑖 𝐴𝑖
Approach- Since 𝑎𝑟𝑒𝑎 = , = ( 𝑖) = . But =
4 𝐴𝑜 𝑑𝑜 100 𝐹𝑜 𝐴𝑜

Unknown: Weight can the large piston support = ?

𝐴𝑜
Formula: 𝐹𝑜 = 𝐹𝑖
𝐴𝑖

Solution: The weight that can be lifted is equal to Fo,

𝐴𝑜 100
𝐹𝑜 = 𝐹𝑖 = (150 𝑁) ( )
𝐴𝑖 1
Answer: 𝑭𝒐 = 𝟏𝟓 𝟎𝟎𝟎 𝑵
Sample Problem 2.
A hydraulic system is used to lift a 2000-kg vehicle in an auto garage. If the vehicle
sits on a piston of area 0.5 square meter, and a force is applied to a piston of area
0.03 square meters, what is the minimum force that must be applied to lift the vehicle?

Given: m = 2 000 kg, g = 9.8 m/s2, Ao= 0.5 m2, Ai= 0.03 m2
Unknown: Fi = ?
𝐹 𝐴
Formula: 𝐹𝑖 = 𝐴𝑜 𝑖
𝑜
𝐹𝑜 𝐴𝑖 (196 000 𝑁)(0.03 𝑚2 )
Solution: 𝐹𝑖 = =
𝐴𝑜 0.5 𝑚2
Answer: 𝑭𝒊 = 𝟏𝟏 𝟕𝟔𝟎 𝑵

We see different materials floating and sinking in water. All of us know that a
stone thrown into water will sink, and a cork stopper will float. Also, we often observe
that our body feels lighter while taking a swim in the pool.

Archimedes’ principle states that when a body is fully or partially submerged

in a fluid, the fluid pushes upward with a buoyant force with magnitude
𝑭𝑩 = 𝒎𝒇 𝒈,
where 𝑚𝑓 is the mass of the fluid that has been pushed out of the way by the body.

KEYPOINT
“Archimedes’ principle: When a body is completely or partially
immersed in a fluid, the fluid exerts an upward force on the body equal
to the weight of the fluid displaced by the body.”

Determination of the Buoyant Force

The buoyant force occurs because the
pressure in a fluid increases with depth. Thus,
the upward pressure on the bottom surface of
a submerged object is greater than the
downward pressure on its top surface. To see
this effect, consider a cylinder of height ∆ℎ
whose top and bottom ends have an area A and
which is completely submerged in a fluid of
density 𝜌𝐹 , as shown in Fig. 10. The fluid
exerts a pressure 𝑃1 = 𝜌𝐹 𝑔 ℎ1 at the top surface

Figure 10. Cylinder submerged in a fluid

Photo credit: Giancoli Physics 6th
of the cylinder. The force due to this pressure on top of the cylinder is 𝐹1 = 𝑃1 𝐴 =
𝜌𝐹 𝑔ℎ1 𝐴, and it is directed downward. Similarly, the fluid exerts an upward force on
the bottom of the cylinder equal to 𝐹2 = 𝑃2 𝐴 = 𝜌𝐹 𝑔ℎ2 𝐴. The net force on the cylinder
exerted by the fluid pressure, which is the buoyant force, ⃗⃗⃗⃗
𝐹𝐵 , acts upward and has the
magnitude

𝐹𝐵 = 𝐹2 − 𝐹1 = 𝜌𝐹 𝑔𝐴(ℎ2 − ℎ1 )

= 𝜌𝐹 𝑔𝐴∆ℎ
= 𝜌𝐹 𝑉𝑔
= 𝒎𝑭 𝒈
where 𝑉 = 𝐴∆ℎ is the volume of the cylinder, the product 𝜌𝐹 𝑉 is its mass, and 𝜌𝐹 𝑉𝑔 =
𝑚𝐹 𝑔 is the weight of fluid which takes up a volume equal to the volume of the cylinder.
Thus, the buoyant force on the cylinder is equal to the weight of the fluid displaced
by the cylinder. This result is valid no matter what the shape of the object.

Buoyancy
Buoyancy is the force that causes objects to float. It is the force exerted on an
object that is partly or wholly immersed in a fluid. It is also known as the buoyant
force. Buoyancy is the phenomenon due to Buoyant Force.
We can define Buoyancy as: “The upward force applied by the fluid on the
object or the body when an object is put in or submerged in the fluid.”
The symbol for the magnitude of buoyancy is B or FB. As with other forces, the
SI unit of buoyancy is the newton (N).

Buoyancy is caused by differences in pressure acting on opposite sides of an

object immersed in a static fluid. A typical situation:
1. The pressure on the bottom of an object is greater than the top (since pressure
increases with depth).
2. The force on the bottom pushes up and the force on the top pushes down (since
force is normal to the surface).
3. The direction of the net force due to the fluid is upward.

KEYPOINT
The buoyant force on an object is equal to the weight of the fluid
displaced by the object.

KEYPOINT
When an object is immersed in water or in any fluid, the pressure exerted on the
lower surface is greater than the pressure at the upper surface. This difference in
pressure leads to an upward force acting on the object due to fluid pressure. This
upward force is called buoyant force.
 Buoyancy arises from the fact that fluid pressure increases with depth and from
the fact that the increased pressure is exerted in all directions (Pascal’s Principle) so
that there is an unbalanced upward force on the bottom of a submerged object.

Figure 11. Water ball

Photo credit: HyperPhysics

Since the “water ball” at left is exactly supported by the difference in pressure
and the solid object at right experiences exactly the same pressure environment, it
follows that the buoyant force on the solid object is equal to the weight of the water
displaced (Archimedes’ Principle).

Objects of equal volume experience equal buoyant forces.

Equal Volumes Feel Equal Buoyant Forces

Suppose you had equal-sized balls of cork, aluminum, and lead, with respective
specific gravities of 0.2, 2.7, and 11.3. If the volume of each is 10 cubic centimeters
then their masses are 2, 27, and 113 gm.

Figure 12. Three equal

sized balls (cork,
aluminum and lead) with
same volumes.

Photo credit: HyperPhysics

Each would displace 10 grams of water, yielding apparent masses of -8 (the

cork would accelerate upward), 17 and 103 grams respectively.

The behavior of the three balls would certainly be different upon release from
rest in the water. The cork would bob up, the aluminum would sink, and the lead
would sink more rapidly. But the buoyant force on each is the same because of
identical pressure environments and equal water displacement. The difference in
behavior comes from the comparison of that buoyant force with the weight of the
object.
Upthrust
As discussed, the buoyant force is the upward force exerted on an object that is
wholly or partly immersed in a fluid. This upward force is also called as Upthrust. It is
due to the buoyant force that a body submerged partially or wholly in a fluid appears to
lose its weight i.e. appears to be lighter.

Figure 13. Body subjected to an

upward force when immersed in fluid.

Photo credit: Byjus.com

An object whose density is greater than that of the fluid in which it is submerged
tends to sink. If the object is either less dense than the liquid or is shaped appropriately
(as in a boat), the force can keep the object afloat. In terms of relative density, if the relative
density of a substance is less than one it floats in water, and substances with a relative
density greater than 1 sink in water.

KEYPOINT
When an object is immersed in water or in any fluid, the pressure exerted on the
lower surface is greater than the pressure at the upper surface. This difference in
pressure leads to an upward force acting on the object due to fluid pressure. This
upward force is called buoyant force.

The buoyant force on a submerged object is equal to the weight of the fluid
displaced. This principle is useful for determining the volume and therefore the
density of an irregularly shaped object by measuring its mass in air and its effective
mass when submerged in water (density = 1 gram per cubic centimeter). This effective
mass underwater will be its actual mass minus the mass of the fluid displaced. The
difference between the real and effective mass, therefore, gives the mass of water
displaced and allows the calculation of the volume of the irregularly shaped object
(like the king’s crown in the Archimedes story).
Sample Problem

What is the buoyant force exerted by olive oil when a 17.8 kg piece of copper is
totally immersed into it? The density of the olive oil is 910 kg/m 3 and the density for
copper is 8 890 kg/m3.

𝑘𝑔 𝑘𝑔
Given: 𝑚𝐶𝑢 = 17.8 𝑘𝑔 𝜌𝑜𝑖𝑙 = 910 𝜌𝐶𝑢 = 8 890
𝑚3 𝑚3
Unknown: FB = ?
Formula:
Remember
𝐹𝐵 = 𝐹2 − 𝐹1 = 𝜌𝐹 𝑔𝐴(ℎ2 − ℎ1 )
= 𝜌𝐹 𝑔𝐴∆ℎ
= 𝜌𝐹 𝑉𝑔
= 𝑚𝐹 𝑔

Solution: When copper piece is totally immersed into olive oil, the volume of
olive oil it displaces is equal to its own volume.
𝑚 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟 15.7 𝑘𝑔
𝑉 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟 = =
𝜌 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟 𝑘𝑔
8 890 3
𝑚
𝑉 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟 = 0.00177 𝑚3
𝑤𝑜𝑖𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 = 𝜌𝑜𝑖𝑙 𝑔 𝑣
𝑘𝑔 𝑚
𝑤𝑜𝑖𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 = (910 3 ) (9.8 2 ) (0.0022 𝑚3 )
𝑚 𝑠
𝑤𝑜𝑖𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 = 15.8 𝑁
𝐹𝐵 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑜𝑖𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 = 𝟏𝟓. 𝟖 𝑵

Applications of Buoyancy
It is due to buoyancy that human swimmers, fish, ships, and icebergs stay
afloat. Some applications of buoyancy are given in the points below.

1. Submarine
A submarine has a large ballast tank, which is used to control its
position and depth from the surface of the sea. A submarine submerges by
letting water into the ballast tank so that its weight becomes greater than the
buoyant force.

Figure 14. Submarine

Photo credit: Byjus.com
2. Hot Air Balloon
The atmosphere is filled with air that exerts a buoyant force on any
object. A hot air balloon rises and floats due to the buoyant force. It descends
when the balloon’s weight is higher than the buoyant force. It becomes
stationary when the weight equals the buoyant force.

Figure 15. Hot Air Balloon

Photo credit: Byjus.com

3. Ship
A ship floats on the surface of the
sea because the volume of water displaced
by the ship is enough to have a weight
equal to the weight of the ship. A ship is
constructed in a way so that the shape is
hollow, to make the overall density of the
ship lesser than the sea water. Therefore,
the buoyant force acting on the ship is
large enough to support its weight.
Figure 16. Cruise Ship

4. Fish
A certain group of fishes
uses Archimedes’ principle to go up and
down the water. To go up to the surface,
the fishes will fill its swim bladder (air
sacs) with gases. The gases diffuse from
their own body to the bladder and thus
making the body lighter. This enables
the fishes to go up.
Figure 17. Clown Fish
 Explore

Direction: Read and analyze the following problems. Answer them properly.

Activity: Immerse yourself!

Problem 1. Pressure
Say you’re in your neighbor’s pool, waiting near the bottom until your neighbors
give up trying to chase you off and go back into the house. You’re near the deep end
of the pool, and using the handy pressure gauge you always carry, you measure the
pressure on the back of your hand as 1.2 x 105 pascals. What force does the water
exert on the back of your hand? The back of your hand has an area of about 8.4x10 -
3 square meters.

Problem 2. Static pressure

An airplane in level flight whose mass is 25,500 kg has a wing area of 68 m2.
What is the pressure difference between the upper and lower surfaces of its wing?
Express your answer in atmospheres.

Problem 3. Pascal’s Principle

A hydraulic lift is supporting a car of mass 1 000 kg. If the cross-sectional area of the
smaller chamber is 1 m2 and that of the bigger chamber is 4 m 2, what force can be
applied on top of the movable piston on the smaller chamber to support the car at the
bigger chamber?

Problem 4. Archimedes’ Principle

What is the net force on a ball of mass 12.3 kg and volume of 0.25 m3 when
𝑘𝑔
it is submerged under water? The density of water is 1000 3 and the acceleration due
𝑚
𝑚
to gravity is 10 2. Note: 𝐹𝑛𝑒𝑡 = 𝐹𝑏 − 𝐹𝑔. (Approach - The buoyant force on the ball is
𝑠
simply the weight of water displaced by the ball.)
 Deepen

Direction: Read and analyze the following problems. Answer them properly.

Activity: Take a depth breath!

Problem 1. Pressure
Pressure at a Faucet.
The surface of the
water in a storage tank is 30
m above a water faucet in the
kitchen of a house, Fig.
Calculate the difference in
water pressure between the
faucet and the surface of the
water in the tank. The
density of the water is
𝑘𝑔
1𝑥103 .
𝑚3

Figure 18. Pressure at a Faucet

Photo Credit: Giancoli Physics 6th

P
Approach – Water is practically incompressible, so 𝜌 is constant even for a
∆ℎ = 30𝑚 when used in equation ∆𝑃 = 𝜌𝑔 ∆ℎ. Only ∆ℎ matters; we can ignore
the “route” of the pipe and its bends.
P
Note: The height ℎ is sometimes called the pressure head. In the figure, the
head of water is 30 m at the faucet. The very different diameters of the tank
and faucet don’t affect the result – only pressure does.

Problem 2. Fluids at Rest

What Force Must a Dam Withstand?
Consider the pressure and force acting on the dam retaining a reservoir of
water. Suppose the dam is 800 m wide and the water is 120.0 m deep at the dam. (a)
What is the average pressure on the dam due to the water? (b) Calculate the force
exerted against the dam?
 Figure 19. Pressure at a Faucet

Photo Credit: Giancoli Physics 6th

The average pressure 𝜌 due to the weight of the water is the pressure
at the average depth ℎ of 60.0 m, since pressure increases linearly with
depth. The force exerted on the dam by the water is the average pressure
times the area of contact, 𝐹 = 𝑃𝐴. P

Problem 3. Pascal’s Principle

A hydraulic press has pistons of area 0.006 m2 and 0.35 m2,
respectively. A force of 25 N is applied to the smaller one. Find the (a)
pressure exerted on the smaller piston and on (b) the big piston. (c) What is
the lifting force on the bigger piston?

Note that the input and output pressure is undiminished throughout

the system.
 Gauge

Directions: Read carefully each item. Write only the letter of the best answer
before the number.

1. What quantity is defined as the magnitude of the force acting perpendicular to

a surface divided by the area of the surface?
A. Pressure C. Density
B. Volume D. Acceleration

2. Which of the following is the standard unit for pressure?

A. atm B. psi C. pascal D. torr

3. Among the following, which object gives a lot of pressure?

A. glass B. knife C. flat stone D. chopping board

4. The downward pressure of a liquid does not depend on the shape of the
container but only on the height of the liquid, what phenomenon is being
implied?
A. Static pressure C. Buoyancy
B. Pascal’s principle D. Hydrostatic paradox

5. What can be inferred about the needle of a syringe and a poke of a finger which
is in contact with the skin?
A. Sharp tip of the needle will have a greater pressure
B. Blunt tip of a finger will have a lesser pressure
C. Sharp tip of the needle will have a lesser pressure
D. Blunt tip of a finger will have a greater pressure

6. What conclusion can be drawn why going up an altitude could cause a popped
in your ears?
A. so much cold temperature
B. lesser motion of air around you
C. weight of air exerted on you decreases
D. air is much denser than water

7. A ball of cork, aluminum, and lead has a volume of 8 cubic centimeters each
and submerge into water. What quantity does these three had the same with
each other?
A. mass B. density C. specific gravity D. buoyancy
 8. During a deep dive in a swimming pool, your ears ached. Why is this so? Choose
the best option that describes the statement.
A. the direction of the net force due to the fluid is downward
B. the pressure on the bottom of an object is greater than the top
C. the force on the bottom pushes sideward and the force on the top pushes
up the surface
D. increased pressure is exerted in all directions

9. The hydraulic lift is a force-multiplying device with a multiplication factor equal

to the ratio of the areas of the two pistons. What is/are true about the
statement?
I. product of force and distance remains unchanged
II. small input force produces a large output force
III. larger area comes with a larger force
A. I, II B. III only C. II, III D. I, III

10. When a body is completely or partially immersed in a fluid, the fluid exerts an
upward force on the body equal to the weight of the fluid displaced by the body.
Which of the following examples manifest the statement above?
I. Submarine V. Hot air balloon
II. Dentist’s chair VI. Hydraulic jacks
III. Fish VII. Ship
IV. Car lifts VIII. Car brakes
A. I, III, V, VII
B. II, IV, VI, VIII
C. I, III, V, VII
D. II, IV, VI, VIII

For numbers 11-15. Answer the problem below.

For a hydraulic device, a piston has a cross-sectional area of 40 square
centimeters moving an incompressible liquid with a force of 75 N. The other end of the
hydraulic pipe is attached to a 2nd piston with a force of 150 N. Determine the cross-
sectional area on the second piston? (Note: 1𝑚2 = 10 000 𝑐𝑚2 = 104 𝑐𝑚2 )
11. Given =
12. Unknown =
13. Formula =
14. Solution =
15. Answer =

Given: Ai = 40 square centimeters = 0.004 𝑚2 Fo = 150 N

Fi = 75 N

Unknown: Ao = ?
𝐹 𝐴
Formula: 𝐴𝑜 = 𝑜 𝑖
𝐹𝑖
𝐹𝑜 𝐴𝑖 (150 𝑁)(0.004 𝑚2 )
Solution: 𝐴𝑜 = =
𝐹𝑖 75 𝑁

Answer: 𝐴𝑜 = 𝟎. 𝟎𝟎𝟖 𝒎𝟐
 𝐹 25 𝑁 𝑵
Solution and Answer: (a) 𝑃𝑖 = 𝐴𝑖 = 0.006 𝑚2 = 𝟒 𝟏𝟔𝟔. 𝟔𝟔 𝒎𝟐
𝑖
𝐹 1 458.33 𝑁 𝑵
(b) 𝑃𝑜 = 𝐴𝑜 = 0.35 𝑚2
= 𝟒 𝟏𝟔𝟔. 𝟔𝟔 𝒎𝟐
𝑜
𝐹𝑖 𝐴𝑜 (25 𝑁 )(0.35𝑚2 )
(c ) 𝐹𝑜 = 𝐴𝑖
= 0.006 𝑚2
= 𝟏 𝟒𝟓𝟖. 𝟑𝟑 𝑵
Note that the input and output pressure is undiminished
throughout the system.
Gauge
1. A
2. C
3. B
4. D
5. A
6. C
7. D
8. B
9. A
10. C
11-15.
Given: Ai = 40 square centimeters = 0.004 𝑚2 Fo = 150 N
Fi = 75 N
Unknown: Ao = ?
𝐹𝑜 𝐴𝑖
Formula: 𝐴𝑜 =
𝐹𝑖
𝐹𝑜 𝐴𝑖 (150 𝑁)(0.004 𝑚2 )
Solution: 𝐴𝑜 = =
𝐹𝑖 75 𝑁
Answer: 𝐴𝑜 = 𝟎. 𝟎𝟎𝟖 𝒎𝟐
Key Answer
 𝐹𝑜 𝐴𝑖
Formula: 𝐹𝑖 =
𝐴𝑜
𝐹𝑜 𝐴𝑖 (9 800 𝑁)(1 𝑚2 )
Solution: 𝐹𝑖 = =
𝐴𝑜 4 𝑚2
Answer: 2 450 N
Problem 4. Archimedes’ Principle
𝑚 𝑘𝑔
Given: 𝑔 = 10 2 𝜌𝑤𝑎𝑡𝑒𝑟 = 1000 3 𝑚 = 12.3 𝑘𝑔 𝑣 = 0.25 𝑚3
𝑠 𝑚
Unknown: FNet = Net force = ?
Formula: 𝐹𝑛𝑒𝑡 = 𝐹𝑏 − 𝐹𝑔
Other formulas: 𝐹𝑏 = 𝑉𝑏𝑎𝑙𝑙 ∙ 𝜌𝑤𝑎𝑡𝑒𝑟 ∙ 𝑔 𝐹𝑔 = 𝑚𝑔
Solution:
Approach - The buoyant force on the ball is simply the weight of water displaced by the ball.
𝑘𝑔 𝑚
𝐹𝑏 = 𝑉𝑏𝑎𝑙𝑙 ∙ 𝜌𝑤𝑎𝑡𝑒𝑟 ∙ 𝑔 = (0.2 𝑚3) (1000
) (10 2 ) = 2500 𝑁
𝑚3 𝑠
The force of gravity on the ball is:
𝑚
𝐹𝑔 = 𝑚𝑔 = 12.3 𝑘𝑔 (10 ) = 123 𝑁
𝑠2
These forces oppose each other, so we can say:
𝐹𝑛𝑒𝑡 = 𝐹𝑏 − 𝐹𝑔 = 2 000 𝑁 − 150 𝑁 = 𝟐𝟑𝟕𝟕 𝑵
Deepen
Activity: Take a depth breath!
Problem 1. Pressure
Pressure at a Faucet.
Approach – Water is practically incompressible, so 𝜌 is constant even for a ∆ℎ = 30𝑚 when used in
equation ∆𝑃 = 𝜌𝑔 ∆ℎ. Only ∆ℎ matters; we can ignore the “route” of the pipe and its bends.
𝑚 𝑘𝑔
Given: ℎ = 30 𝑚 𝑔 = 9.8 𝜌𝑤𝑎𝑡𝑒𝑟 = 1𝑥103
𝑠2 𝑚3
Unknown: difference in water pressure = ?
Formula: ∆𝑃 = 𝜌𝑔 ∆ℎ
Solution:
The same atmospheric pressure acts both at the surface of the water in the storage tank and on the
water leaving the faucet. So, the water
Jumpstart
Activity: Press on me!
1. A
2. A
3. B
4. B
5. C
6. C
7. D
8. D
9. A
10. C
Explore
Activity: Immerse yourself!
Problem 1. Pressure
Given: Pressure on the back of your hand = 1.2 x 10 5 N/m2
Area at the back of your hand = 8.4x10-3 m2
Unknown: Force of the water that exert on the back of your hand = ?
Formula: P=F/A → F=PA
Solution and Answer:
𝑁
𝐹 = 𝑃𝐴 = (1.2 x 105 2) (8.4𝑥10−3 𝑚2 ) = 𝟏 𝟎𝟎𝟖 𝑵 𝑜𝑟 𝟏. 𝟎𝒙𝟏𝟎𝟑 𝑵
𝑚
Problem 2. Static pressure
𝑁
Given: m= 25 500 kg, A=68m2, 𝜌𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑒 = 1.01𝑥105 2
𝑚
Unknown: Pressure difference =?
Formula: P=F/A, F=mg
𝑚
Solution and Answer: 𝐹 = 𝑚𝑔 = (25 500𝑘𝑔) (9.8 ) = 249 900 𝑁
𝑠2
𝐹 249 900 𝑁
𝑃= = = 𝟑𝟔𝟕𝟓 𝑷𝒂
𝐴 68 𝑚2
Conversion
1 𝑎𝑡𝑚
3675 𝑃𝑎 𝑥 = 𝟎. 𝟎𝟑𝟔 𝒂𝒕𝒎
1.013𝑥105 𝑃𝑎
Problem 3. Pascal’s Principle
Given: Ai = 1 m2
Ao = 4 m2
Fo = weight of the 1 000 kg car = 1 000 kg (9.8 m/s2) = 9 800 N
Unknown: Fi = ?
 References

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