mechanics chapter 1
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Introduction to Statics
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What is mechanics?
Mechanics is a science which describes and predicts the conditions of rest
and motion of bodies under the action of forces. It is the foundation of most
engineering sciences.
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Basic Concepts in Mechanics
There are four basic concepts in mechanics:
Space: position of a point P
Time: time of an event
Mass: used to characterize and compare bodies of certain
mechanical experiments.
Force: Represents the action of one body on another.
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Space, (x,y,z)
The position of a point P can
be defined by three lengths: x,
y, and z, measured from a
certain reference point or
origin (O), in three given
directions X, Y, and Z
These lengths (x,y,z) are
known as the coordinate of P:
P(x,y,z)
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Time, t
The time of the event should be
given.
For example :
At time t0
, point P coordinates
are (x0,y0,z0)
At time t1, point P coordinates
are (x1,y1,z1)
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Mass, m
Two bodies of the same mass
will be attracted by the earth in
the same manner.
These two bodies will alsooffer the same resistance to a
change in translational motion.
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Force, F
Force can be exerted by actualcontact, or it can be exerted ata distance such asgravitational and magneticforces.
Forces is characterized by itspoint of application, itsmagnitude, and its direction.
A force is represented by a
vector.
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Example of Force Vector
Force F1 has:
a magnitude of 10 lbs, its point of application is A it is in the vertical direction.
Force F2 has:
a magnitude of 25 lbs, its point of application is B, its direction is at 45 degree
from horizontal (or 45
degree from vertical).
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Newtonian Mechanics
In newtonian mechanics space, time, and mass are absolute
concepts i.e. independent of each other.
However, the concept of force is not independent of the other three.
dt
dvmamF ==
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Definition of Particle and Rigid Bodies
A particle is a very small amount
of matter which may be assumed
to occupy a single point in space.
A rigid body is a combination of a
large number of particlesoccupying fixed positions with
respect to each other.
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Fundamental Principles of Newtonian Mechanics
The parallelogram Law of additionof forces- figure 1. (Chapter 2)
The principle of transmissibility:
Equivalent Forces - figure 2
(Chapter 3)
Newtons Three fundamental laws :
First law Inertia (Chapter 2)
Second law: F= ma (dynamics)
Third law: action=-reaction
(Chapter 6)
Newtons law of Gravitation
Figure 1
Figure 2
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Newtons Law of Gravitation
Two particles of mass M and m
are mutually attracted with equal
and opposite forces F and F of
magnitude given by the formula:
r is the distance between two
particles.
G is a universal constant calledthe constant of gravitation.
2rMmG=F
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What happens if M = mass of the earth?
The force F exerted by the
earth on a particle located on
its surface is defined as the
weight W of the particle.
The weight W of a particle of
mass m may be expressed as
W= mg
g= 9.81 m/s2 or 32.2 ft/s2
* We assume the Earth to be spherical and
neglect the radius of the object relative to the
radius of the Earth in this discussion
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Importance of Dimensions and Units
Any physical quantity can be characterized by dimensions.
The magnitudes assigned to the dimensions are called units.
Basic dimensions such as mass m, length L, time t, and areselected as primary or fundamental dimensions.
While others such as velocity v, volume V, and force F are
expressed in terms of the primary dimensions and are called
secondary dimensions, or derived dimensions.
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System of Units
The metric SI ( Le Systme International dUnits) is a simple and
logical system based on a decimal relationship between the various
units and its is being used for scientific and engineering work in
most of the industrialized nations.
E.g.: 1m = 10 dm = 100 cm = 1000 mm
The english system which is also known as the United States
Customary System (USCS) has no apparent systematic numerical
base and various units are related to each other rather arbitrarily.
E.g.: 1 ft= 12 in, 1mile = 5280 ft etc..
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Primary Dimensions and their base units
The SI units are said to be absolute units, i.e. the three base units
chosen are independent of the location where measurements are made.
The USCS units do not form an absolute system of units because of theirdependence upon the gravitational attraction of the earth. They form agravitational system of units.
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Example of Secondary Dimensions and their
Derived Units
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SI unit prefixes (table 1.1)
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USCS units frequently encountered
1 mile (mi) = 5280 ft
1 inch (in) = ft
1 kilopound (kip) = 1000 lb
1 ton = 2000 lb
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Mass
Mass is the property of a body of fluid that is a measure of its inertiaor resistance to a change in motion.
It is a physical quantity expressing the amount of matter in a body.Therefore, the mass of an object is not dependent on gravity and isdifferent from its weight.
The mass symbol is m
a
Fm
onaccelerati
forcemass:LawSecondsNewton ==
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Force-Mass Relationship
In the S.I. units:
1 kg is the mass which
receives an acceleartion of 1
m/s2 when a force of 1N is
applied to it.
F= ma where F= 1N, a=1m/s2
In the U.S.C.S. units:
1 slug is the mass which
receives an acceleartion of 1
ft/s2 when a force of 1lb is
applied to it.
F= ma where F= 1lb, a=1 ft/s2
2
m/skg1N1 =2
ft/sslug1lb1 =
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Weight
Weight is the amount that a body of fluid weighs, that is, the forcewith which the fluid is attracted toward Earth by gravitation. Itssymbol is W
When acceleration is equal to the acceleration due to gravity, g,then Newtons law becomes:
The mass of a body remains the same whether on earth or on themoon. However its weight will change.
g
WmgmWeight ==
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Weight-Mass Relationship
In the S.I. units:
g = 9.81 m/s2
The mass of 1 kg weighs:w = m x g= 1kg x 9.81m/s2 =
9.81kgm/s2 = 9.81 N
1 kg weighs 9.81 N
In the U.S.C.S. units:
g= 32.2 ft/s2
The mass of 1 slug weighs:
w = m x g= 1slug x 32.2ft/s2 =
32.2 slugsft/s2 = 32.2 lb
1 slug weighs 32.2 lb
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Mass expressed as lbm(pounds-Mass)*
A fluid having a weight of 1.0 lb (pound-force) has a mass of 1.0lbm (pound-mass)i.e.: 1lb= 1lbm x 32.2 ft/s2 = 32.2 lbmft/s2 =1 slugft/s2
1 slug = 32.2 lbm
1 lbm will not receive an acceleration of 1 ft/s2 when a force of 1lb isapplied to it.
This is not a coherent system. This system will be avoided.
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1st Example using SI conversion: What is the volume in cubic meters
(m3) of a rectangular prism having a length l = 9 dm, and a width w= 5
cm, and a height h= 2 mm?
Solution:
- What we know:l = 9 dm= 0.9 m
w = 5 cm= 0.05 m
h = 2 mm = 0.002 m
- Formula:
Volume(m3
) = l(m)xw(m)xh(m)- Solving:
Volume = 0.9 m x 0.05 m x 0.002 m = 0.00009 m3 or 9x10-5 m3
2nd Example : convert 15 cm3 to m3
Solution:
What we know: 1cm = 0.01 m = 10-2 m
Solving: 15 cm3 = 15 (10-2 m) 3 = 15 x10-6 m3
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Example in the USCS Conversion
The magnitude of a velocity is 30 mi/h. Convert it to ft/s
Solution:
What we know:
1s3600
1hs3600h1
1mi1
ft5280ft5280mi1
==
==
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Conversion From One System of Units to Another
Units of length:
Units of force:
Per definition, 1 lb has a mass of 0.4536 kg at sea level and altitude of 45
where g= 9.807 m/s2
w = mg 1 lb = 0.4536kg x 9.807 m/s2 = 4.448 kg.m/s2 = 4.448 N
Units of mass:
kg14.59/ms14.59Nm0.3048
ft
lb1
N4.448
ft
slb11slug
/fts1lb1slugft/sslug1lb1
22
22
==
=
==
1 ft = 0.3408 m
1 lb = 4.448 N
1 slug = 14.59 kg
1 lbm = 0.4536 kg
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Example of conversion from one system of units to another
Convert the moment of a force from USCS units which was found to be 47
lbin, to SI units of Nm
Solution:
We know to convert: 1lb = 4.448 N and 1 ft = 0.3048 m
We need to convert inch (in) to meter (m)
1 in = ft= (0.3048 m)=0.0254 m
We can solve :
( ) N.m315in1
m0.0254lb1
N4.448lbin47lbin47 .==
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Numerical Accuracy
The accuracy of the solution of a problem depends upon two items:
The accuracy of the given data
The accuracy of the computations performed.
The solution cannot be more accurate than the less accurate ofthese two items.
In engineering problems the data are seldom known with an
accuracy greater than 0.2 percent.
Practical rule: use 4 significant figures to record numbers beginning
with 1 and 3 significant figures in all other cases.
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How should the following numbers be reported?
729000 lb = 729 kip729134.65 lb
16950 N = 16.95 kN16947 N
0.0429 m = 42.9 10-3 m =
42.9 mm
0.0428571429 m
Reported resultsComputed results
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Example: the loading of a bridge is known to be 75,000 lb with a
possible error of 100 lb. To what numerical accuracy should the
reaction at one of the bridge supports be recorded if the calculator
gave an answer of 14,322.225 lb.
Solution
lb2014,320asanswertherecordshouldWe
lb20~18.6x0.001314,322.225
:islb14,322.225onerrorpossibletheTherefore
0.13%or00130
lb75,000
lb100:isaccuracyofdegreeThe
=
= .
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