Chp.5 elasticity & fluid

PHY082 (PHYSICS II)
CHAPTER 5: MATTER
1

Lecture Outline
2
5.1 Elasticity
Concept and Definition
5.2 Hooke’s Law
5.2.1 Definition
5.2.2 Force – Extension Graph
5.2.3 Work done & elastic potential energy
5.3 Buoyancy in fluid
5.3.1 Archimedes’ Principle
5.3.2 Problem solving involving buoyant force
5.3.3 Bernoulli’s Principle and applications

5.1 Elasticity
• Every body will change its shape or size when
external force acting on it – deformation.
• If the body returns to original shape or size when
the external force is removed, then it is said to
be elastic.
• Example is a spring.
• Objects which are elastic obeys Hooke’s law.
3

5.2 Hooke’s Law
5.2.1 Definition:
• An object exerted by force, such as pulling force
will cause the length of the change.
• If the amount of elongation is small compare to
the length of the object, Hooke’s Law state that:
The change in length is proportional to the
applied force.
4
F =kDl

• F = force (or weight) pulling on the
object,
" D L = increase in length,
• k = proportionality constant
5
F =kDl ….(Eq. 5.1)
• First noted by Robert Hooke (1635 –
1703)
• Valid for almost any solid material.
• But applicable for only to a certain
point – force is not too strong.
• Up to a point – Hooke’s Law is not
valid.
Figure 5.1

Example 5.1
• A 0.7 kg wooden block is used to stretch a spring as
shown in figure below. When a force of 100 N is exerted
on the spring, it is stretch 0.2 m to the right direction.
Calculate the spring constant. (PHY082 - APR2011).
6
F = kDl
F =kx
500 N/m
= =100 N =
x
0.2 m
k F
 Hooke’s Law
 Hooke’s Law for
Spring

7
5.2.2 Force – Extension Graph
• This proportionality, holds until the force reaches
the proportional limit.
• Beyond that, the object will still return to its original shape up to
the elastic limit.
• Beyond the elastic limit, the material is permanently deformed.
• The maximum elongation
is reached at the breaking
point.
• Ultimate Strength of the
material – Maximum force
that can be applied without
breaking.
Figure 5.2

Stress and Strain
• Stress is defined as the force per unit area.
= = F
area
(Eq. 5.2)… Stress, s force [Unit: N/m2]
• Strain is defined as the ratio of the change in length
to the original length. (how much an object is
deform).
8
A
e = = Dl (Eq. 5.3)…
Strain, change in length
l
0 original length

• Therefore, the elastic modulus is equal to
the stress divided by the strain:
• E = Young’s modulus
• Unit: Newton/ meter2 (N/m2)
9

Example 5.2
• Tension in a piano wire (Giancoli).
A 1.60 m long steel piano wire has a diameter of 0.20
cm. How great is the tension in the wire if it stretches
0.30 cm when tightened?
11
F = E D
l
(2.0 1011N/m2 )(0.0030m 6 2
= ´ ´
1200N
)(3.1 10 m )
1.60m
0
=
-
A
l

5.2.3 Work done & Elastic potential energy
Work done, W
• Review:
[Source: Giancoli]
12
W =F.s
F =kx
ò
=
W =
F x dx
0
x
ò
x
p
kx dx
0
( )
2
W =1 kx
2
….(Eq. 5.4)

Work done =
Area under the
curve
W =1 x kx = kx
( )( ) 1
2
[Source: Giancoli] 13
2
2
[Unit: Joule (J)]

Potential Energy, U
U =W =1 kx ….(Eq. 5.5)
Notes:
• From principle of conservation energy:
14
2
2
Potential Energy = Kinetic Energy
1 kx = 1
mv
2 2
2
2

Example 5.3
• Work done on a spring. (Giancoli)
A person pulls on the spring in Figure below, stretching it
3.0 cm, which requires a maximum force of 75 N. How
much work does the person do?
15
75N
0.030m
max
k F
= =
x
max
= ´ 3
2.5 10 N/m
1
W = kx
2
1
= ´
(2.5 10 N/m)(0.030m)
2
1.1 J
3 2
2
=

Example 5.4
• A 0.7 kg wooden block is used to stretch a spring as
shown in figure below. When a force of 100 N is exerted
on the spring, it is stretch 0.2 m to the right direction.
Calculate the elastic potential energy stored in the
spring. Given the k = 500 N/m. (PHY082 - APR2011).
16
1
U = kx
2
1
(500N/m)(0.2)
2
10 J
2
2
=
=

5.3 Buoyancy in Fluid
• Buoyant force, FB: The
force acting perpendicularly
upward on a body when a
part or the whole body is
immersed in a fluid.
17
Figure 5.6
….(Eq. 5.6)

18
5.3.1 Archimedes’ Principle:
The buoyant force on an object immersed in
a fluid is equal to the weight of the fluid
displaced by that object.
Figure 5.7

Example 5.5
• A block of dimensions 5 m x 2 m x 20 m floats in
water with half its volume immersed in water. If
density of water 1000 kgms-3, calculate the buoyant
force acting on the block? Assume g = 10ms-2.
19
=r gV
(1000)(10)(1
= ´ ´
1.0 10 N
(5 2 20))
2
Buoyant force
6
W
= ´

Example 5.6
• A metal block is hung against a Newton scale.
When in air, the scale reads 5.5 N but when fully
submerged in water the reading becomes 4.8 N.
What is the factor that contributes to the difference?
Calculate the buoyant force that acts on the block.
Buoyant force = Actual weight in air – weight in water
20
5.5 - 4.8
=
0.7 N
=
Solution:

Example 5.7
• A hot air balloon floats in air. If the total mass of
the balloon is 270 kg, what is the buoyant force
on the balloon? Assume g = 10 ms-2.
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Buoyant force = Weight of object
g
=m
(270 kg)(10 ms )
2700 N
-2
=
=
Solution:

5.3.3 Bernoulli’s Principle &
applications
• Bernoulli’s Principle states that where the
velocity of a fluid is high, the pressure is low,
and where the velocity is low, the pressure is
high.
P + 1 rv + rgy = P + rv + rgy ….(Eq. 5.7)
22
, ¯ fluid v P ¯ , fluid v P
2
2
1
1 2 2
2
1 2
1 2
P + 1 rv2 + rgy =
2 constant

Applications:
1. Torricelli’s (fig. 13-25)
2. Atomizer (a)
3. Ping-pong ball in jet of air (b)
4. Airplane wing (c)
23

Chp.5 elasticity & fluid

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