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Module No. 8
STUDY GUIDE FOR MODULE NO. 8
PRESSURE, DENSITY, AND ELASTICITY
MODULE OVERVIEW
This module provides a discussion on pressure, density, stress, strain, and modulus of
elasticity of different particles. The elastic properties usually have a larger effect that
the density so it is important to both material properties. Moreover, density is directly
proportional to pressure.
MODULE LEARNING OBJECTIVES
At the end of this chapter, the student should be able to:
1. Define pressure, density, stress, strain, and modulus; and
2. Solve problems involving pressure, density, and elasticity.
LEARNING CONTENTS (Pressure, Density, Specific Gravity, Elasticity)
Pressure, Density, Specific Gravity
Definition 8.1.1 — Pressure.
Pressure P exerted by a force with magnitude F over an area A (the area and force
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are understood to be perpendicular) is defined as the amount of force per unit area.
In symbols:
𝑃=
𝐹
(8.1)
𝐴
The SI unit for pressure is known as Pascal (Pa) and a Pascal is equal to 1 Newton
per square meter. Note that since F and A are scalars, then P is also scalar.
Definition 8.1.2 — Density.
Density, or to be more precise for this discussion, mass density, of a material is the
amount of mass per unit volume of the material. Mathematically,
𝜌=
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
=
𝑚
𝑉
(8.2)
Definition 8.1.3 — Specific Gravity.
Specific gravity of a substance refers to the ratio of the density of the substance to
the density of a base (standard) substance. In symbols:
𝑆𝑝𝑔 =
𝜌
𝜌𝑠
(8.3)
where 𝜌 is the density of the substance and 𝜌𝑠 is the density of the base substance.
For solids and liquids, the common (but not always) standard substance is water while
for gases it is usually air. Given at the right are the densities of some common
substances.
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8.2
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Elasticity
Consider a rod suspended vertically. Suppose a force is exerted on the rod in such a
way that it elongates for some sufficiently small distance Δl. Experiments show that
the applied force F is proportional to the elongation or vice versa. This gives the
famous expression known as Hooke’s law:
𝐹 = 𝑘 ∆𝑙
(8.4)
Note that this expression is only applicable up to some point, which we call
proportional limit. Beyond this point, the expression no longer holds. As the object
elongates further, it will reach a point called elastic limit. The part from the starting
point to the elastic point is called the elastic region while region after the elastic point
is called the plastic region. At the end of the plastic region is the breaking point or
failure point.
If the applied force is removed while elongation is still in the elastic region, the object
reverts to its original form. If the applied force is removed while elongation is in the
plastic region, the object remains deformed. If the minimal applied force (maximum
non-breaking force/area) to reach the breaking point is called the ultimate strength
of the rod. If ultimate strength is exceeded, the rod breaks or fractures.
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Intuitively, the amount of elongation is not solely dependent on the applied force, it
also relies on the material and its dimensions. This means k is actually a combination
of these factors.
Moreover, observations tell us that given same material and same applied force:
1. Longer objects have larger elongations.
2. Objects with larger cross-sections have smaller elongations.
These, in combination with equation (8.4) gives the following relation:
∆𝑙 =
𝑙𝐹
𝑙
(8.5)
𝛾𝐴 𝑜
where A is the cross-sectional area, 𝑙𝑜 is the original length, and 𝛾 is the what we call
elastic modulus otherwise known as the Young’s modulus.
Clearly, 𝛾 depends only upon the material. Now, if we rewrite 𝛾 as follows:
𝐹⁄
𝛾 = ∆𝑙 𝐴
⁄∆𝑙
𝑜
(8.6)
One may find it interesting to look at the numerator and denominator. We define the
applied force per unit area as stress (notice that this is a vector, in reality, this is an
internal force).
We denote stress by 𝜎. Thus,
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𝜎=
Module No. 8
𝐹
(8.7)
𝐴
Moreover, if we take the ratio of the elongation to the original length, we have what we
call elastic strain. Strain in general is the ratio of the change in dimension to the
original dimension of an object. Here, we denote elastic strain as 𝜀 , thus
𝜀=
∆𝑙
(8.8)
𝑙𝑜
and
𝛾=
𝜎
(8.9)
𝜀
Stress causing length deformation or strain or linear strain can be classified into
three. Object is said to be under Tension or tensile stress when there is elongation
while it is under compression or compressive stress when the material is
compressed. Fortunately, the previous formulas apply to both cases.
On the other hand, the object is said to be under shear stress if it is subjected to
equal opposite forces (known as shearing forces) acting on opposite faces. In this
case, the distortion is for the shape but not for volume.
We denote the shearing stress as 𝜎𝑠 , the shearing strain as 𝜀𝑠 , and the shear
modulus as 𝜉 .Hence:
𝐹
𝜎𝑠 = ,
𝐴
𝜀𝑠 =
∆𝑙
𝑙𝑜
,
𝜎
𝜉= 𝑠
𝜀𝑠
(8.10)
where F; A; ∆𝑙; 𝑙𝑜 are the magnitude of shearing forces, area parallel to the shearing
forces, shear change, and distance between the parallel surfaces respectively, and for
sufficiently small strains with θ the shear angle,
𝜀𝑠 = 𝑡𝑎𝑛𝜃 ≈ 𝜃
(8.11)
Consider an object having inward forces applied to it. Evidently, in such scenario, the
volume of the object will decrease.
Technically, we have the following observations provided all other things constant:
1. The change in volume is directly proportional to the original volume.
2. The change in volume is directly proportional to the change in pressure (with
negative relationship).
Hence, we arrive at the following relation:
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∆𝑉 = − ∆𝑃𝑉𝑜
(8.12)
𝛽
where 𝑉𝑜 is the original volume, ∆𝑃 is the change in pressure, ∆𝑉 is the change in
volume, and 𝛽 is the volume elasticity modulus or the bulk modulus. We also have
∆𝑉
𝜎
𝛽 = 𝜀𝑉
𝜎𝑉 = ∆𝑃, 𝜀𝑉 = − 𝑉 ,
0
(8.12)
𝑉
where 𝜎𝑉 ; 𝜀𝑉 are the volume stress and strain, respectively. Fracture or breaking
happens when the applied force per unit area exceeds the ultimate strength of the
material.
In consideration of this notion, especially in buildings, materials need to maintain a
safety factor of from three (3) to say 10 (or more) - that is, actual stress on material
must not exceed one-tenth to one-third of the material’s ultimate strength.

Example 8.1
A 50-kg man stands on a cube with negligible mass. If the box has edges measuring
5.0 cm, what pressure does the box exert on the floor?
SOLUTION:
The cubic box has a square face which interacts with the floor. In this case, the
pressure is then
𝑃=

𝐹
𝐴
=
9.8 ×50 𝑁
6 5.0 ×10−2
2
= 32, 666. 67 𝑁/𝑚2
Example 8.2
The bulk modulus of water is 2:1 x 109 N/m. Suppose 200 L of water is exposed to a
pressure of 1.5 MPa, how large is the volume contraction of water?
SOLUTION:
Using formula (8.12), we have
1
∆𝑉 = ∆𝑃𝑉𝑜 = −0.1429 𝑚𝐿
𝛽
That is, the volume of water contracts by -0.1429 mL.

Example 8.3
Find the density and specific gravity of gasoline if 51 g occupies 75 cm 3.
SOLUTION:
𝜌=

𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
=
𝑚
𝑉
=
51 𝑔
75 𝑐𝑚3
𝑔
= 0.68 ⁄ 3
𝑐𝑚
Example 8.4
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What volume does 300 g of mercury occupy? The density of mercury is 13 600 kg/m 3.
SOLUTION:
𝜌=
𝑉=
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
𝑚
𝜌
=
=
𝑚
𝑉
0.3 𝑘𝑔
= 2.206 × 10−5 𝑚3
𝑘𝑔⁄
13,600
𝑚3
Example 8.5
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a load
of 8.0 kg is hung on its end. Find the stress, the strain, and the Young’s modulus for
the material of the wire.
SOLUTION:
Given:
lo = 75.0 cm = 0.75 m
d = 1.3 x 10 -3 m
r = 6.5 x10 -4 m
Δl = 0.0350 cm = 3.5 x 10 -4 m
m = 8.0 kg
For the stress,
From the formula of stress,
𝜎=
𝐹
𝐴
We need to find the force F and A,
𝐹 = 𝑚𝑔 = 8.0 𝑘𝑔 × 9.8 𝑚⁄ 2 = 78.4 𝑁
𝑠
𝐴 = 𝜋𝑟 2 = 𝜋(6.5 × 10−4 )2 = 1.33 × 10−6 𝑚2
Then,
𝜎=
𝐹
𝐴
=
78.4 𝑁
1.33 × 10−6 𝑚2
= 5.89 × 107 𝑁⁄ 2
𝑚
𝟕
The stress is 𝟓. 𝟖𝟗 × 𝟏𝟎 𝑵⁄ 𝟐 .
𝒎
For the strain,
𝜀=
∆𝑙
𝑙𝑜
=
3.5 x 10 −4 m
0.75 m
= 4.67 × 10−4
The strain is 𝟒. 𝟔𝟕 × 𝟏𝟎−𝟒 .
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For the Young’s modulus,
5.89× 107 𝑁⁄ 2
𝑚
𝛾= =
= 1.26 × 1011 𝑁⁄ 2
𝜀
4.67 ×10−4
𝑚
𝜎
The Young’s Modulus of elasticity is 𝟏. 𝟐𝟔 × 𝟏𝟎𝟏𝟏 𝑵⁄ 𝟐 .
𝒎

Example 8.6
The mass of a calibrated flask is 30 g when empty, 81 g when filled with water, and
68 g when filled with oil. Find the density of the oil.
SOLUTION:
Given:
mf = 30 g = 0.030 kg
mw = 81 g – 30 g = 0.051 kg
mo = 68 g - 30 g = 0.038 kg
Find: density of the oil
We first find the volume of the flask, 𝜌 = 𝑚⁄𝑉, using the water data:
𝑉=
(0.051 𝑘𝑔)
𝑚
=
= 5.1 × 10−5 𝑚3
𝜌 1000 𝑘𝑔⁄
𝑚3
Then, for the oil,
𝜌𝑜𝑖𝑙 =
𝑚𝑜𝑖𝑙
0.038 𝑘𝑔
𝑘𝑔
=
= 745 ⁄ 3
−5 3
𝑚
𝑉
5.1 × 10 𝑚
For further discussion visit the links listed:
.https://www.labxchange.org/library/items/lb:LabXchange:3df83c41-b306-3ba4-915c04e3057659e3:html:1
https://www.labxchange.org/library/items/lb:LabXchange:53777f0e-2aa2-3249-abc69e9f02b4273b:html:1
https://www.labxchange.org/library/items/lb:LabXchange:b0b0851b-9af5-3a1e-9925e305fba14544:html:1
LEARNING ACTIVITY 1
Solve the following problems completely:
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1. An iron rod 4.00 m long and 0.500 square cm in cross section stretches 1.00 mm when
a mass of 225 kg is hung from its lower end. What is the Young’s modulus for iron?
2. Compute the volume change of a solid copper cube, 40 mm on each side when
subjected to a pressure of 20 MPa. The bulk modulus for the metal is 125 x 109 N/m.
3. The compressibility of water is 5.0 x 10-10 m2/N. Find the decrease in volume of 100
mL of water when subjected to a pressure of 15 MPa.
4. A load of 50 kg is applied to the lower end of a steel rod 80 cm long and 0.6 cm in
diameter. How much will the rod stretch? 𝛾 = 190 x 109 N/m for steel.
5. A vertical wire 5.0 m long and of 0.0088 square cm cross-sectional area has
𝛾= 200 x109 N/m. A 2.0 kg object is fastened to its end and stretches the wire
elastically. If the object is now pulled down a little and released, the object undergoes
vertical SHM. Find the period of its vibration.
SUMMARY
Pressure P exerted by a force with magnitude F over an area A (the area and force
are understood to be perpendicular) is defined as the amount of force per unit area.
Density, or to be more precise for this discussion, mass density, of a material is the
amount of mass per unit volume of the material.
Specific gravity of a substance refers to the ratio of the density of the substance to
the density of a base (standard) substance.
Stress is a quantity proportional to the force producing a deformation; strain is a
measure of the degree of deformation. Strain is proportional to stress, and the constant
of proportionality is the elastic modulus:
𝑺𝒕𝒓𝒆𝒔𝒔
𝑬𝒍𝒂𝒔𝒕𝒊𝒄 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 = 𝑺𝒕𝒓𝒂𝒊𝒏
Three common types of deformation are represented by (1) the resistance of a solid
to elongation under a load, characterized by Young’s modulus Y; (2) the resistance
of a solid to the motion of internal planes sliding past each other, characterized by the
shear modulus S; and (3) the resistance of a solid or fluid to a volume change,
characterized by the bulk modulus B.
REFERENCES




Buffa, Anthony J. & Wilson, Jerry D. Physics, Fourth Edition. Pearson
Education, Inc., Prentice Hall Inc., 2000.
Cutnell, John D. Introduction to Physics. John Wiley & Sons, Inc., Singapore.
2010.
Serway, Raymond A. College Physics. Thomson Brooks/Cole, Singapore.2003.
Tipler, Paul A. Physics for Scientists and Engineers. W. H. Freeman and
Company, New York. 2004.
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





Module No. 8
Weber, Robert L., Manning, Kenneth V., White, Marsh W., Weygand, George
A. College Physics. McGraw-Hill book Company, New York.1977.
Weber, Robert L., Manning, Kenneth V., White, Marsh W., Weygand, George
A. College Physics. McGraw-Hill book Company, New York.1977.
Young, Hugh D. and Freedman Roger A. Sear’s and Zemansky’s University Physics
with Modern Physics. San Francisco, CA. 2004
https://www.youtube.com/watch?v=5GyJqQSebt0
https://www.youtube.com/watch?v=HALbtyDUjp0&t=10s
https://www.youtube.com/watch?v=KxxTTf7kUTM
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