Physics 1408-002
Principles of Physics
Lecture 20
– Chapter 12 & 13 –
March 31, 2009
Announcement I
Lecture note is on the web
Handout (6 slides/page)
http://highenergy.phys.ttu.edu/~slee/1408/
*** Class attendance is strongly encouraged and will be
taken randomly. Also it will be used for extra credits.
Sung-Won Lee
Sungwon.Lee@ttu.edu
HW Assignment #8 is placed on
MateringPHYSICS, and is due by
11:59pm on Wednesday, 4/1
Announcements II
Announcement III
3rd Exam
4/9 (Next week)
SI session by
Reginald Tuvilla
9: 30 am – 10:50 am
Chapters 10, 11, 12, 13
Rotation motion, Angular momentum, Statics,
Fluids
Thursday 4:00 - 5:30pm - Holden Hall 106
Next week, test review
will be on Monday 04/06
in the BA room 55 from
4:30 - 7:30.
Chapter 12
Static Equilibrium;
Elasticity and Fracture
•!The Conditions for Equilibrium
•!Solving Statics Problems
•!Stability and Balance
•!Elasticity; Stress and Strain
•!Fracture
•!Trusses and Bridges / Arches and Domes
Solving Statics Problems
Hanging Lamp
Hanging Lamp
•! A lamp of mass M hangs from the end of plank of mass m and
length L. One end of the plank is held to a wall by a hinge, and
the other end is supported by a massless string that makes an
angle with the plank. (The hinge supplies a force to hold the
end of the plank in place.)
–!What is the tension in the string?
–!What are the forces supplied by the
hinge on the plank?
•! First use the fact that
!!
m
!"
hinge
L
M
in both x and y directions:
x:
T cos ! + Fx = 0
y:
T sin ! + Fy - Mg - mg = 0
"
y
x
!
Now use ! = 0 in the z direction.
"!If we choose the rotation axis to
be through the hinge then the
hinge forces Fx and Fy will not
enter into the torque equation:
L
LMg + mg - LT sin ! = 0
2
Hanging Lamp
m
L/2
M
Fy
!"
L/2
Fx
mg
Mg
12-4 Elasticity; Stress and Strain
Stress is defined as the force per unit area. (F/A)
•! So we have three equations and three unknowns:"
! T cos ! + Fx = 0
T sin ! + Fy - Mg - mg = 0
LMg +
!
!F = 0
Strain is defined as the ratio of the change in length to
the original length. (#l/l0)
y
L
mg - LT sin ! = 0
2
x
Therefore, the elastic modulus (E) is equal to the
stress divided by the strain:
which we can solve to find:
m%
"
m$
!
!$ M + ' g
#" M + &% g
#
2& ˆ
2
Fx =
i
T =
tan (
sin '
1
Fy = mg ĵ
2
m
L/2
M
Fy
!"
L/2
Fx
mg
Mg
Young’s Modulus: Elasticity in Length
•! Tensile stress is the ratio of the
external force (F) to cross-sectional
area (A)
Young’s Modulus
A mass M hangs from a steel wire of cross sectional
area A and length L and Young’s modulus Y.
By what #L amount does the wire stretch?
–! For both tension and compression
•! The elastic modulus is called
Young’s modulus
Y=
•! SI units of stress: Pascals, [Pa]
F
tensile stress
F Lo
= A =
tensile strain !L A !L
Lo
Answer: stress = F/A= Mg/A, so
–! 1 Pa = 1 N/m2
•! The tensile strain is the ratio of
F
the change in length to
tensile stress
F Lo
Y=
= A =
the original length
!L
tensile strain
A !L
Lo
#L
M
#L = (1/Y)(FL/A) = (1/Y)(MgL/A)
Note: Y(steel) = 200*109 N/m2
12-4 Elasticity; Stress and Strain
12-4 Elasticity; Stress and Strain
Example 12-7: Tension in piano wire.
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 0.25 cm
when tightened?
12-4 Elasticity; Stress and Strain
12-4 Elasticity; Stress and Strain
Compressional stress is exactly the opposite of tensional
stress. These columns are under compression.
The three types of stress for rigid objects:
This Greek temple, in Agrigento, Sicily, built 2500 years ago, shows the
post-and-beam construction. The columns are under compression.
12-4 Elasticity; Stress and Strain
The shear strain, where G
is the shear modulus:
The fatter book (a) shifts
more than the thinner
book (b) with the same
applied shear force.
Shear Modulus: Elasticity of Shape
•! Forces may be parallel to
one of the objects faces
•! The stress is called a shear
stress; F/A
•! The shear strain is the
ratio of horizontal
displacement &
the height of the object;
#x/h
F
shear stress
Fh
GS =
= A =
!x
shear
strain
A
!x
•! The shear modulus: G
h
•! A material having a large
shear modulus is difficult
to bend
12-4 Elasticity; Stress and Strain
If an object is subjected to inward forces on all
sides, its volume changes depending on its bulk
modulus. This is the only deformation that
applies to fluids. (see next)
or
Bulk Modulus: Volume Elasticity
•! Bulk modulus characterizes the response of an object
to uniform squeezing
–! Suppose the forces are perpendicular to, and acts on, all the
surfaces as when an object is immersed in a fluid
•! The object undergoes a change in volume without a
change in shape
•! Volume stress (#P) = the ratio of the force to the surface
area; This is also the Pressure
•! Volume strain = Ratio of the change in
volume to the original volume
B=
12-5 Fracture
"F
volume stress
A = ! "P
=!
"V
"V
volume strain
V
V
12-5 Fracture
If the stress on an object is too great, the object will
fracture. The ultimate strengths of materials under
tensile stress, compression stress, and shear stress
have been measured.
When designing a
structure, it is a
good idea to keep
anticipated stresses
less than 1/3 to 1/10
of the ultimate
strength.
12-5 Fracture
Example 12-8: Breaking the piano wire.
A steel piano wire is 1.60 m long with a diameter of
0.20 cm. Approximately what tension force would
break it?
12-5 Fracture
A horizontal beam will be under both tensile and
compressive stress due to its own weight. Therefore, it
must be made of a material that is strong under both
compression and tension.
12-5 Fracture
A uniform pine beam, 3.6 m long and 9.5 cm x 14 cm in cross
section, rests on two supports near its ends, as shown. The
beam’s mass is 25 kg and two vertical roof supports rest on it,
each one-third of the way from the ends. What maximum load
force FL can each of the roof supports exert without shearing
the pine beam at its supports? Use a safety factor of 5.0.
12-6 Trusses and Bridges
One way to span a wide space is to use a
truss—a framework of rods or struts
joined at their ends into triangles.
A truss bridge
A roof truss
12-6 Trusses and Bridges
On a real bridge, the load will not be centered.
The maximum load rating for a bridge must
take this into account.
A uniform pine beam, 3.6 m long and 9.5 cm x 14 cm in cross section, rests on two supports
near its ends, as shown. The beam’s mass is 25 kg and two vertical roof supports rest on it,
each one-third of the way from the ends. What maximum load force FL can each of the roof
supports exert without shearing the pine beam at its supports? Use a safety factor of 5.0.
12-6 Trusses and Bridges
Each truss member is under either tension or
compression; if the mass is small, these forces act
along the strut.
(a) Each massless rod of a truss is assumed to be under tension or
compression. (b) The two equal and opposite forces must be along
the same line or a net torque would exist. (c) Real struts have mass,
so the forces F1 and F2 at the joints do not act precisely along the
strut. (d) Vector diagram of part (c).
12-6 Trusses and Bridges
For larger bridges, trusses are too heavy.
Suspension bridges are one solution; the
roadway is suspended from towers by
closely spaced vertical wires.
(a) Truss with truck of
mass m at center of strut
AC. (b) Forces on strut
AC.
Suspension bridges (Brooklyn and
Manhattan bridges, NY).
12-7: Arches and Domes
12-7: Arches and Domes
The stones or bricks in a round arch are mainly
under compression, which tends to strengthen
the structure.
The Romans developed
the semicircular arch
about 2000 years ago.
This allowed wider
spans than could be
built with stone or brick
slabs.
12-7: Arches and Domes
Unfortunately, the
horizontal forces
required for a
semicircular arch can
become quite large. The
pointed arch was an
improvement, but still
needed external
supports, or “flying
buttresses.”
12-7: Arches and Domes
A dome is similar to
an arch, but spans a
two-dimensional
space.
12-7: Arches and Domes
The pointed arches require considerably less
horizontal force than a round arch.
Chapter 13
Fluids
•!Density and Specific Gravity
•!Pressure in Fluids
•!Atmospheric Pressure and Gauge Pressure & Measurement
•!Pascal’s Principle
•!Buoyancy and Archimedes’ Principle
•!Fluids in Motion; Flow Rate and the Equation of Continuity
•!Bernoulli’s Equation & its Applications!
Fluids
13-1 Phases of Matter
The three common phases of matter are solid,
liquid, and gas.
•!A solid has a definite shape and size.
•!A liquid has a fixed volume but can be any shape
•!A gas can be any shape and also can be easily
compressed.
Liquids and gases both flow, and are called fluids.
•! A fluid is a collection of molecules that are randomly
arranged and held together by weak forces and by forces
exerted by the walls of a container.
•! Both liquids and gases are fluids.
Statics & Dynamics with Fluids
•! Fluid Statics
–! Describes fluids at rest.
•! Fluid Dynamics
–! Describes fluids in motion.
•! The same physical principles that have applied to
statics and dynamics will also apply to fluids.
13-2 Density & Specific Gravity
The density ! of a substance is its
mass per unit volume:
The SI unit for density: kg/m3.
Density is also sometimes given
13-2 Density and Specific Gravity
Example 13-1: Mass, given volume and
density.
What is the mass of a solid iron wrecking
ball of radius 18 cm?
in g/cm3; to convert g/cm3 to kg/m3,
multiply by 1000.
Water at 4°C has a density of
1 g/cm3 = 1000 kg/m3.
The specific gravity of a substance is
the ratio of its density to that of water.
13.3 Pressure
•! The pressure P of the fluid at the
level to which the device has been
submerged is the ratio of the
force to the area
•! Pressure is a scalar quantity
–! Because it is proportional to the magnitude of the force
•! If the pressure varies over an area,
evaluate dF on a surface of area(dA)
as dF = P dA
•! Unit of pressure: Pascal (Pa)
2
1 Pa = 1 N/m
Pressure vs. Force
•! Pressure is a scalar and force is a vector.
•! The direction of the force producing a pressure is
perpendicular to the area of interest.
Measuring Pressure
•! The spring is calibrated by a known
force.
•! The force due to the fluid presses
on the top of the piston and
compresses the spring.
•! The pressure on the piston is then
measured.
13-3 Pressure in Fluids
Example 13-2: Calculating pressure.
The two feet of a 60-kg person cover an
area of 500 cm2.
Determine the pressure exerted by the two
feet on the ground.