PC1433 MECHANICS and WAVES Module Lecturers Prof Wang Chien Ming
Engineering Science Programme and Department of Civil Engineering, National University of Singapore E-mail:
[email protected] Office: E3A-04-17
Asst Prof Andrew Bettiol
Engineering Science Programme and Department of Physics, National University of Singapore E-mail:
[email protected] Office: E3A-04-21 PC1433 Mechanics and Waves
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PART A: MECHANICS 1. INTRODUCTION 1.1 What is Mechanics? Mechanics is that branch of science which describes and predicts the conditions of rest or motion of bodies under the action of forces Classical Mechanics Mechanics of rigid bodies
Statics deals with bodies at rest
Mechanics of deformable bodies
Dynamics deals with bodies in motion
Mechanics of fluids
Incompressible fluids
Compressible fluids
Hydraulics PC1433 Mechanics and Waves
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1.2 Fundamental Concepts of Mechanics Basic Quantities
x t m F
space time mass force
In Newtonian (classical) mechanics, these are absolute quantities, independent of each other. But force is dependent on the three
Not true in relativistic mechanics because t
= f ( x); m = f (v)
¾ Particle – relatively small amount of matter which may be assumed to occupy a single point in space ¾ Rigid body – combination of large amount of particles occupying fixed position with respect to each other
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1.3 Superstars in Mechanics
Archimedes (287-212 BC)
Discovered the law of hydrostatics (or also know as Archimedes' Principle) Discovered the principles of lever Invented the hydraulic screw Invented the catapult Discovered a mirror system that burned invader's boats and ships by focusing the sun's rays (perhaps a myth) Discovered Pi Wrote 5 books such as Floating Bodies, Measurement of the Circle, and Sphere and
Cylinder
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Leonardo da Vinci (1452-1519) A renaissance painter, architect,
engineer, mathematician and philosopher, a genius the world has never seen again so far.
In mechanics, for example, he was the first person to state the two basic laws of friction (200 years before Newton even defined what force is). Da Vinci stated that (1) the areas in contact have no effect on friction and (2) if the load of an object is doubled, its friction will also be doubled (i.e. F is proportional to N).
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Galileo Galilei (1564 -1642)
• Figured out how to use a pendulum to regulate clocks • Disproved Aristotle's notion that heavier bodies fall faster and proved the law of uniformly accelerated motion • First man to use a telescope to study the skies • Proved that the earth revolved around the sun and that the earth was not the centre of the universe • Discovered the moons of Jupiter and constructed accurate tables for the revolution of Jupiter’s moon • Discovered sun spots
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Sir Isaac Newton (1642-1727) He discovered the nature of gravity and the laws of motion. He wrote the book Mathematica Principia, which provided a detailed explanation of the laws of gravity and motion, particularly as they applied to astronomy. He invented calculus. He invented the first reflecting telescope. Newton discovered the origin of colour. Quotable Quotes “If I have seen further...it is by standing upon the shoulders of giants.“ Once Newton was asked how he came up with all his brilliant ideas, "By always thinking about them," he answered. Albert Einstein once said of Newton, "Nature to him was an open book whose letters he could read without effort." PC1433 Mechanics and Waves
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John Bernoulli (1667-1748)
– principle of virtual displacements (virtual work) Principle of Virtual Work If a structure, subjected to a system of forces in equilibrium is given small compatible virtual displacements, then the total virtual work vanishes.
Equilibrium Equation via Principle of Virtual Work P
Virtual Work
P d
R2
aL
(1-a)L
R1
= Pad – R1d = 0 R1 = Pa
R2 R1 PC1433 Mechanics and Waves
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Leonhard Euler (1707-1793)
– rigid body dynamics (moment of inertia) Euler’s Buckling Load for Pinned-Ended Rods / Columns
P L
Q
L Δ P = Kπ2 / L2
K = QL3 / (3Δ)
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Max Planck (1858-1947) Father of quantum mechanics Discovered the energy of an oscillator is quantized
E=hv E – Energy h – Planck’s Constant V – Frequency
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Albert Einstein (1879-1955)
Formulated the Theory of Relativity Albert Einstein’s Quotes "When you sit with a nice girl for two hours, it seems like two minutes. When you sit on a hot stove for two minutes, it seems like two hours that's relativity." "I know not with what weapons World War III will be fought, but World War IV will be fought with sticks and stones.” “Imagination is more important than knowledge.”
“There is not the slightest indication that nuclear energy will ever be obtainable. It would mean that the atom would have to be shattered at will.” (WCM: What a boo-boo) PC1433 Mechanics and Waves
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Stephen Hawking (1942-?) Stephan Hawking and Roger Penrose showed that Einstein's General Theory of Relativity implied space and time would have a beginning in the Big Bang and an end in black holes. These results indicated it was necessary to unify General Relativity with Quantum Theory. Stephen Hawking wrote two famous books: A Brief History of Time, and Black Holes and
Baby Universes and Other Essays. PC1433 Mechanics and Waves
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1.4 Fundamental Principles of Elementary Mechanics (a) Parallelogram Law for Addition of Force F1 + F2 F2 F1 (b) The Principle of Transmissibility
FF F F FF F FFFF (c) First Law of Newton (Law of Inertia) Every body continues in its state of rest, or in uniform motion in a right (straight) line unless it is compelled to change that state by forces impressed upon it.
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(d) Second Law of Newton
F = ma Work-Energy Principle
Impulse-Momentum Principle dv dt
dv F = mv ds
F =m
∫ Fds = ∫ mvdv
∫ Fdt = ∫ mdv
1 1 mv 22 − mv12 2 2 = change in kinetic energy
Work done =
Impulse = mv 2 − mv1 = change in momemtum
(e) Third Law of Newton
"To every action there is always opposed an equal reaction” Philosophiae Naturalis Principia Mathematica (1687)
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(f)
Newton’s Law of Gravitation
F =G
Mm , where G is the universal gravitatio nal consta nt 2 r r
m
F
F
M
This principle leads to the concept of weight
W = mg Me where g = G 2 = gravitational acceleration, re M e and re are the mass and the radius of the Earth respectively.
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These quizzes are based on ideas from Eric Mazur, Peer Instruction, Prentice Hall, Saddle River, N.J. 1997 1. A collision occurs between a bus and a small sports car. During impact: a) There is no force between the bus and the car b) The bus exerts a force on the car, but the car does not exert any force on the bus. c) The bus exerts a much larger force on the car than the car does on the bus. d) The force that the bus exerts on the car is equal to the force the car exerts on the bus. e) The car exerts a larger force on the bus than the bus on the car. 2. If you moved to a planet of the same radius, but a greater mass your weight would a) decrease b) increase c) remain the same 3. If you moved to a planet of the same mass, but a larger radius, your mass would a) decrease b) increase c) remain the same PC1433 Mechanics and Waves
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2. STATICS OF PARTICLES 2.1 Definition of a Force Force is a vector. It possesses both magnitude and direction.
F
In SI Units, the force unit is Newton (N). This is the force which, applied to a body of mass 1 kilogram, causes an acceleration of 1 meter per second2 in the direction of application of the force.
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2.2 Addition of Forces The addition of forces obeys the parallelogram law of vector addition. Magnitude and direction of c may be obtained by the law of cosines +b α b a = c2 = a2 + b2 – 2abcos γ c or the law of sines γ β a b c = = a sin α sin β sin γ The addition of 3 or more forces is carried out by arranging the given vectors in tip-to-tail fashion and connecting the tail of the first vector with the tip of the last one - polygon rule for addition of vectors
b a
d
c
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d=a+b+c 18
2.3 Force Components The component of a force along a given axis may be obtained using the dot (or scalar) product of vectors, a ● b = a b cos θ For example, to find the component of force P along the direction of vector Q
B
P cos θ
O
θ
P
Q
The component of P along Q direction is OB = OA cos θ = P cos θ
A Q
λQ
.
As , P Q = PQ cos θ P Q P cos θ = Q
.
Q
P
But Q = λ Q is a unit vector along Q. Therefore the component of P along Q = P λ Q
.
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Any force in space can be represented as the sum of its component force vectors along three convenient axes, normally mutually perpendicular (rectangular Cartesian coordinates system)
y j
k θz
z
θy
i
F
θx
Fy
Fx
In terms of the unit vectors i, j, k the force F expressed as
x Fz may be
F = Fx i + Fy j + Fz k PC1433 Mechanics and Waves
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in which the Cartesian components Fx = F . i = F cos θx , Fy = F . j = F cos θy , Fz = F . k = F cos θz are the projections of F onto the coordinate axes. The unit vector in the direction of F is given by
λF = F / F = cos θx i + cos θy j + cos θz k It follows that the unit vector has the direction cosines of that vector as its Cartesian components. Note that cos2 θx + cos2 θy + cos2 θz = 1
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Addition of forces by summing up their Cartesian components Earlier, the addition of vectors was done graphically using the polygon rule. The same addition may be done analytically by summing algebraically the corresponding scalar components of the vectors. For example,
R = P+Q+S Rx i + R y j + Rz k = Px i + Py j + Pz k + Qx i + Q y j + Qz k + Sx i + S y j + Sz k
= (Px + Qx + S x )i + (Py + Q y + S y ) j + (Pz + Qz + S z )k
⇒ R x = Px + Q x + S x ,
R y = Py + Q y + S y ,
R z = Pz + Q z + S z
The magnitude of the resultant vector R is given by
R = Rx2 + R y2 + Rz2 PC1433 Mechanics and Waves
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Concept Quiz 1. Can you resolve a 2-D vector along two directions, which are not at 90° to each other? A) Yes, but not uniquely. B) No. C) Yes, uniquely. 2. Can you resolve a 2-D vector along three directions (say at 0°, 60°, and 120°)? A) Yes, but not uniquely. B) No. C) Yes, uniquely.
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2.4 Forces on a Particle Forces passing through a single point (or a particle) are called concurrent forces. The resultant of concurrent forces, may be obtained by the polygon rule (graphically) or by the addition of the scalar components of the forces.
y F2
Fn
0 z
R
F1 x
n R = ∑ Fi i=1 i.e Rx = ∑ Fx 2D 3D Ry = ∑ Fy Rz = ∑ Fz
}
}
R = Rx 2+ Ry 2+ R2z
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Example 2.1
450
1150
Given: TAB = 1425 N
y
TAC = 2130 N
A Length dimensions in mm
750
O
D
z 1125
C
Cable AB. AB = 1125i − 750 j − 450k
Forces exerted at A. 450
1150
y
A 750
O
D
z 1125
C
Find the resultant of forces x exerted at A by the two cables B
AB = 1425 mm AB T AB = TAB λ AB = TAB AB ⎛ 1125i − 750 j − 450k ⎞ x ⎟⎟ = 1425⎜⎜ 1425 B ⎝ ⎠ T AB = 1125i − 750 j − 450k
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Cable AC. AC = 1125i − 750 j + 1150k and AC = 1775 mm T AC = TAC λ AC = TAC Resultant
⎛ 1125i − 750 j + 1150k ⎞ AC ⎟⎟ = 1350i − 900 j + 1380k = 2130⎜⎜ AC 1775 ⎠ ⎝
R = Rx i + R y j + Rz k Rx = ∑ Fx = 1125 + 1350 = 2475 N
R y = ∑ Fy = −750 − 900 = −1650 N Rz = ∑ Fz = −450 + 1380 = 930 N
Magnitude of Resultant R = Rx2 + R y2 + Rz2 = 24752 + 1650 2 + 930 2 = 3116.6 N Direction of Resultant Rx 2475 = = 0.7941 ⇒ θ x = 37.4o R 3116.6 R y − 1650 = = −0.5294 ⇒ θ y = 122.0 o cosθ y = R 3116.6 R 930 = 0.2984 ⇒ θ z = 72.6o cosθ z = z = R 3116.6
cosθ x =
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2.5 Equilibrium of a Particle If the resultant force on a particle is zero, the particle is said to be in equilibrium. ⎧ ⎪ Rx = ⎪ ⎪ ⇒ i.e. R = 0 ⎨Ry = ⎪ ⎪ ⎪ Rz = ⎩
∑F
x
=0
∑F
y
=0
∑F
z
=0
Note: With 3 equations, 3 unknowns maybe solved for 3-D problems. However, only 2 equations are only available for 2-D problems.
The concept of equilibrium will be used not to solve for the resultant, which is zero, but to solve for the required forces that act on the particle and maintain its equilibrium.
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2.6 Free Body Diagram A sketch showing the physical conditions of the problem is known as a space diagram as shown in the figure below. FB θ = 40o
FA
A (a) Space diagram
To calculate the tensile forces in the chains, consider a free body diagram of the joint A as shown in Figure (b).
500 N For equilibrium,
FB
FA
40o
30o
A
W = 500 N
FB 500 FA (c) Force triangle
FA FB 500 = = sin 40o sin30o sin110o FA = 342N and FB = 266N
(b) Free body diagram PC1433 Mechanics and Waves
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A free-body diagram is a sketch of an object or a connected group of objects, modeled as a single particle/rigid body that is completely isolated from its environment or surrounding bodies and represents the interactions of its environment by appropriate external forces (and moments). Drawing a free-body diagram is an art, and can be learned only by practice. If a correct free-body diagram is constructed, then the balance of the forces can be carried out in a very systematic manner. No equilibrium problem should be solved without first drawing the free-body diagram, so as to account for all the forces and couple moments that act on the free-body.
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Construct a free-body diagram for member AB
B R
R
α
R A
3R R
α
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R α
α
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Construct a free-body diagram for member ABC Ball and Socket Support
y
E
Cables 2m
D
F
A 4m
A
2m
G B
4m
C
4m
x
B
z
y
C
G z
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x
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Two smooth pipes, each weighing W, are supported by the forks of a tractor as shown in the figure. Draw free body diagrams for each pipe and for both pipes.
Pipe B Pipe A
A
FBD of pipe A
The weight of a body is an external force and its effect is shown as a single resultant force acting vertically down through the body’s centre of gravity.
B
y
C
G z
FBD of pipe B
PC1433 Mechanics and Waves
x
FBD of both pipes
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Example 2.2 y 7m C
7m
A
B D
7m
Q = 36.4 kN 4m z
Determine the tension in cables AB, AC and AD.
12 m
3m 4m
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x
33
Free Body. Point A is chosen as a free body. Force Q.
AE = 12 i − 3 j + 4k and AC = 13 m
Q = Qλ AE
⎛ 12 i − 3 j + 4k AE =Q = 36.4⎜⎜ 13 AE ⎝
Cable AD.
AD = −4 i − 3 j and AD = 5 m
T AD = TAD λ AD = TAD Cable AB.
⎛ − 4i − 3 j AD = TAD ⎜⎜ 5 AD ⎝
⎞ ⎟⎟ ⎠
AB = −4 i + 4 j + 7 k and AB = 9 m
T AB = TAB λ AB = TAB Cable AC.
⎞ ⎟⎟ = 33.6i − 8.40 j + 11.20k ⎠
⎛ − 4 i + 4 j + 7k AB = TAB ⎜⎜ 9 AB ⎝
⎞ ⎟⎟ ⎠
AC = −4 i + 4 j − 7 k and AC = 9 m
T AC = TAC λ AC = TAC
⎛ − 4 i + 4 j − 7k AC = TAC ⎜⎜ 9 AC ⎝ PC1433 Mechanics and Waves
⎞ ⎟⎟ ⎠ 34
Equilibrium Condition
∑F = 0:
T AB + T AC + T AD + Q = 0
Substitute for T AB , T AC , T AD , and Q and set the coefficients of i, j , k equal to zero. 4 4 4 i : − TAB − TAC − TAD + 33.6 = 0 9 9 5 4 4 3 j : TAB + TAC − TAD − 8.40 = 0 9 9 5 7 7 k : TAB − TAC + 11.20 = 0 9 9 Solving these equations, we obtain TAB = 14.4kN , TAC = 28.8kN , TAD = 18.0kN PC1433 Mechanics and Waves
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3. EQUILIBRIUM OF RIGID BODIES 3.1 Definition of a Moment When dealing with a rigid body, both forces and moments have to be considered. A moment is a vector (i.e. it possesses both magnitude and direction). The moment M of a force F about a point is the product of the magnitude of the force and the distance between the point and the line of action of the force, measured perpendicular to the line of action.
About A,
M = Fd
F
d
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A 36
Moment of a Force F about a Point may be obtained using Cross (or Vector) Product, a x b = a b sin θ ệ
y
Moment of a force F about a point A,
F z axis out towards viewer
r
θ
A
M A = r×F
B x
d
B’
where r is a vector from A to ANY point B on the force F
Proof: By definition, moment of a force about point A = force x perpendicular distance from A = F x d = F x r sinθ Also, by definition of a vector (or cross) product
r × F = rF sin θ k also a vector, along the perpendicular to the plane containing r and F, positive as r rotates into F by the right hand rule. PC1433 Mechanics and Waves
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Example 3.1 200 N
y 60 mm
30o 60o
25 mm
B
C
z x
50 mm A
Determine the moment of the 200-N force about point A
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Solution to Example 3.1 F = 200 N
y 0.06 m
30o 60o
0.025 m
B
C
z
r
0.05 m
A
Force F F = − F sin 60 o j + F cos 60 o k = −100 3 j + 100k N
x
Position vector r r = 0.06i + 0.075 j m Moment about A i j k 0 M A = r × F = 0.06 0.075 − 100 3 100 0 = 7.5i − 6 j − 6 3 k Nm
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Moment of a Force F about an Axis Consider the problem of finding the moment of the force F about an axis AL. This moment is simply the component of MA axis AL, i.e. the projection of MA on AL:
along the
y MA
M AL = λ AL • M A 0 Note: A is any point along the axis and B is any point along the force vector.
A
L F λ AL r
B x
z
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Determine the moment of the force P (a) about A point A; (b) about the edge AB; (c) about the diagonal AG of the cube with side length L. Using the result of (c) determine the perpendicular distance from AG to FC z E D C (a) Moment about point A
r
y
P G
x
E
B P G F
r = L j − Lk
B
A
C
D
Example 3.2
F
(b) Moment about edge AB
⎛−i +k ⎞ P = P⎜ ⎟ ⎝ 2 ⎠ PL (i + j + k ) M A = r×P = 2
M AB
PL = j•M A = 2
Since AB is parallel to the y axis, MAB is also the y component of moment MA PC1433 Mechanics and Waves
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z D
(c) Moment about diagonal AG
AG (− i + j − k ) = λ= AG
PL M AG = λ • M A = − 6 PL (i − j + k ) or M AG = 3 2
B
A
3
C
λ
d y
P G x
E
F
(d) Perpendicular distance d from AG to FC Observe that P is perpendicular to AG. Check by taking scalar product P.λ = 0. Therefore MAG = - Pd where d is the perpendicular distance. In part (c), we found that M = − PL . Thus d = L AG 6 6
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3.2 Moment of a Couple Two forces F and –F having the same magnitude, parallel lines of action and opposite in sense are said to form a couple
y rB
-F z
B
rA A
d
M the sense is defined by the right hand screw rule
F
x M O = r A × F + r B × (− F )
= (r A − r B ) × F = r × F The moment developed by the couple at any point is simply given by M = r x F and is always perpendicular to the plane containing the forces. The magnitude of the moment of a couple is M = rF sin θ = Fd where d is the perpendicular distance between the lines of action PC1433 Mechanics and Waves
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Equivalent couples -- Couples having the same magnitude and direction. F
F
F
F
d
d
A couple can be translated to a parallel position in its plane or to any parallel plane. F d
1 2
F
1 2
F
2d
F
A couple can be rotated in its plane. PC1433 Mechanics and Waves
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3.3 Equivalent Force-Couple A single force F acting eccentrically about a point A may be considered as a force F acting through the point A plus a couple or a moment M about the point.
F
F d
d
=
F
A
A
=
M = Fd
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A
F
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Example 3.3 400 N
Replace the loading on the frame by a single resultant force. Specify where its line of action intersects the member AB, measured from A.
300 N 0.4 m
0.3 m
B
A
0.2 m
50 Nm
200 N
j 0.7 m
i
C
Solution We first replace the given loading by a force couple at point A R = ∑F = - 200 i - 700 j M = 50 k - 300x0.3 k - 200x0.2 k = - 80 k 728 N
A
74o
80 Nm
C PC1433 Mechanics and Waves
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Next, we replace the force-couple at A by a single equivalent resultant force R. 728 N 74o
728 N
AA
74o
A
A
80 Nm
x
C
C MA = - 80 k = 700x k x = 80/700 = 0.114 m R = 728 N
74o at x = 0.114 m from A PC1433 Mechanics and Waves
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Reduction of a system of forces to one force and a couple ---- equivalent force-couple system at a given point O M2 = r2 x F2 ~ ~
F2 ~
F1 ~
r ~1 r ~n
R ~
F2 ~
M
F1
r2 ~ O
~
O
~
R
O
O
F
n M1 = r1 x F1 ~ ~ ~ Mn = r n x F n ~ ~
Fn ~
Equivalent force-couple system
R=
n
∑F i =1
i
M
R 0
=
n
∑M
oi
i =1
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=
n
∑ (r
i
× Fi)
i =1
48
Equivalent system of forces
( ( ( Two systems of forces F 1 , F 2 , F 3 ,.... and F 1 , F 2 , F 3 ,.... acting on a rigid body are equivalent if and only if the sums of the forces and the sums of the moments about a given point O of the forces of the two systems are, respectively, equal.
( ∑F = ∑F
and
( ∑M O = ∑M O
( ∑ Fx = ∑ Fx and ( ∑ Fy = ∑ Fy and ( ∑ Fz = ∑ Fz and
( ∑Mx = ∑M x ( ∑My = ∑M y ( ∑Mz = ∑M z
or
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3.4 Equilibrium Equations External forces and moments acting on a rigid body may be reduced to a force-couple system at some arbitrary point O. When the force and the couple are both equal to zero, the rigid body is said to be in equilibrium. Therefore, for the equilibrium of a rigid body, n
n
n
i =1
i =1
i =1
R = ∑ F i = 0 ; M O = ∑ (r i × F i ) + ∑ M j = 0 or
∑F = 0 ; ∑F = 0 ; ∑F = 0 ∑M = 0 ; ∑M = 0 ; ∑M = 0 x
y
x
z
y
z
• 6 equations to solve 6 unknowns PC1433 Mechanics and Waves
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For a two-dimensional rigid body contained in the x-y plane
Fz = 0 ; ⇒ ∑ Fx = 0 ;
M x = M y = 0;
M z = MO
∑F = 0 ;
∑M = 0
y
O
• 3 equations to solve 3 unknowns Alternatively,
∑F =0 ; ∑M = 0 ; x
or
O
∑M ∑M
O O'
=0; =0;
∑M ∑M
O'
=0
O"
=0
in which points O, O’ and O” are not in a straight line. PC1433 Mechanics and Waves
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3.5 Reactions at Supports and Connections For 2-dimensional bodies, the various kinds of supports are
52 and Waves PC1433 Mechanics
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Reactions at Supports and Connections (continued)
In sum, if a support prevents translation of a body in a particular direction, then the support exerts a force on the body in that direction. If a rotation is prevented, then the support exerts a moment on the body. PC1433 Mechanics and Waves
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For 3-dimensional bodies, the various kinds of supports are
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Supports for 3-dimensional bodies (continued)
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3.6 Equilibrium of a Two-Force Body A special case of equilibrium which is of interest is that of a rigid body subjected to two forces (or forces acting at only two points). If this so-called two-force body is in equilibrium, the two forces must have the same magnitude, same line of action, and opposite sense. F3
F2
B
FB = - FA
F1 FA
A A
FA
F6 F5
FA
F4 B
FB = - FA FB = - FA
FA
A
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B
FB = - FA 56
Proof of forces being equal, opposite in direction and acting along the line joining the two points of force application in a twoforce body FB FB B B
FA
FA
A
FB
A
Equilibrium of moments about A
B A FA
Equilibrium of moments about B. Equilibrium of forces along AB
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⇒
FA = FB
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3.7 Equilibrium of a Three-Force Body A three-force body is a rigid body subjected to forces acting at only three points. If the body is in equilibrium, the lines of the three forces must be either concurrent or parallel.
F2
F1
F3
F1 (a) Concurrent Forces
F3
F2 (b) Parallel Forces
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Proof of forces being concurrent in a three-force body Fc
Fc
C
FB
C
FB B
B FA
A
FA
A
D
Assume the lines of action of forces FA and FB intersect at D. Equilibrium of moments about D ⇒ FC must pass through the point of concurrence D
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Example 3.4
Determine the components of the reactions at A and B. +
Σ MA = 0:
B(1.5) - (9.81)(2) -(23.5)(6) = 0 ⇒ B=107.1 kN Free body diagram
+
Ax + B = 0 Ax = -107.1 kN or 107.1 kN
Ay AX
23.5 kN
1.5 m B
+
9.81 kN 2m
17.3°
107.1 kN
Σ Fy = 0:
Ay - 9.81 - 23.5 = 0 Ay = 33.3 kN
4m
A =112.2 kN
Σ Fx = 0:
33.3 kN
Check
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Σ MB = 0 60
Example 3.5 Determine the reaction at the fixed end E, given that the tension in the cable is 150 kN. Free body diagram D
+
4.5 (150) = 0 7.5 ⇒ E x = −90 kN 90 kN 6 ∑ = − − F 0 : E 4 ( 20 ) (150) = 0 + y y 7.5 ⇒ E y = 200 kN ∑ Fx = 0 : E x +
A
+ ∑ M E = 0 : 20(7.2) + 20(5.4) + 20(3.6) + 20(1.8) − M E = 180 kNm PC1433 Mechanics and Waves
FD =7.5 m
B
6m E
EX
F
4.5 m
ME Ey
150 kN
6 (150)(4.5) + M E = 0 7.5 61
Example 3.6 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C.
A
150 mm C
150 mm
160mm B
170N
The rod AB is chosen as a free body.
Free Body
Three-Force Body Reaction at C to AB. Forces concurrent at D. 160 α = 28.07° tan α = 300
160mm
A
E
A
150 mm
α
α
α 150 mm
D
C
C
α B
170N
Since AC=CB,
CAD = a, and
direction of A is
2α
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Example 3.6 cont’d Force Triangle
90o - 2α = 33.9o 2α A
170 N 90o - α
β
C α = 28.07o
(90º - α) + 2α + β = 180º β = (90º - α) This is the same as angle between C and 170-N force; triangle is isosceles and we have; A = 170 N C = 2(170) sin α = 160 N A = 170 N
33.9 º
C = 160 N
28.1 º
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Quiz Problem on Equilibrium of Rigid Body
l=
0 .1
A θ
6m
C
B
k = 12.5 kN/m O W = 400 N
r = 0.06 m
When θ = 0 , the spring is unstretched. Determine θ for the equilibrium state.
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Solution to the Quiz l sin θ
Undeformed position
A
s
θ
B O
W W = 400 N
F = ks = krθ r
Rx Ry
r = 0.06 m
∑Μo = 0 ⇒ Wlsin θ - krθ .r = 0 400 x 0.16 sin θ - 12.5 x 103 x 0.062 .θ = 0 ⇒ θ = 0o, θ = 80.3o PC1433 Mechanics and Waves
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Example 3.7 Cable G
E
B
2m
2m
A and D are ball and socket joints
D C
4m
450 N 2m
A
Determine the position of G so that the tension in cable EG is a minimum and the tension force value
4m
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Solution to Example 3.7
Dyj
y
G
T
B
Considering free body diagram ABCD
E
2m
Dzk
Dxi 2m
C
D 4m
W = -450j
z
Axi Azk
A
∑ M = 0 : λ • (AE × T )+ λ • (AC × W ) = 0 AD
Now
x 2m
4m Ayj
AC × W = (4i + 4 j )× (− 450 j ) = −1800k
(1) ( 2)
1 AD 4i + 4 j − 2k 2 2 λ= = = i+ j− k 6 3 3 3 AD In view of Eq. (2), we have
(
)
1 ⎞ ⎛2 2 λ • AC × W = ⎜ i + j − k ⎟ • (− 1800k ) = 600 3 ⎠ ⎝3 3 PC1433 Mechanics and Waves
(3) 67
(
)
(
)
Note that λ • AE × T = T • λ × AE by commutative property for mixed triple product.
(4)
(
For minimum T , the direction of T must be parallel to the direction of λ × AE
(λ × AE ) = ⎛⎜⎝ 23 i + 23 j − 13 k ⎞⎟⎠ × (2i + 4 j ) = 43 i − 23 j + 23 k
)
2 ⎞ ⎛2 1 T min = T ⎜ i − j + k ⎟ (5) 3 ⎠ ⎝3 3 By substituting Eq. (5) into Eqs. (1), (3) and (4), we obtain T min = −400i + 200 j − 400k ⇒ T min = 600 N
Therefore
In view that EG is parallel with T min , the location of G is given by x−2 y−4 0−2 = = ⇒ x = 0, y = 5m − 400 200 − 400
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Quiz on Writing Equilibrium Equations Consider the triangular playground supported by three springs A, B and C. A child with weight W stands at D. Determine the spring forces FA, FB and FC. y
C D B
A
x
Plan View
Equilibrium equations ΣFz = 0
⇒
FA + FB + FC = W
(1)
ΣMA = 0
⇒
FB(AB) + FC(AC) - W (AD) = 0
(2)
ΣMC = 0
⇒
FA(CA) + FB(CB) - W (CD) = 0
(3)
What do you think of the above equations? Can you see that Eqs. (2) and (3) are incorrect? Why? Because the moments are not added in a vectorial manner. PC1433 Mechanics and Waves
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3.8 Dry (or Coulomb) Friction Until now, the surfaces in contact are assumed to be either frictionless or rough. This view is a simplified one. Actually, no perfectly frictionless surface exists (always have friction forces in reality). On the other hand, these friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied. 3.8.1 Experimental illustration of frictional laws
F equilibrium motion Fm Fk
W P A B
N
F P PC1433 Mechanics and Waves
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3.8.2 Laws of Dry Friction
At impending motion
Fm = μ s N
Coefficient of static friction
Fk = μ k N
Coefficient of kinetic friction
In motion
Both μs and μk do not depend on contact area of the surfaces but depend on the nature of the surfaces in contact Typically μk is about 25% smaller than μs
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Coefficients of Static Friction Metal on metal Metal on wood Metal on stone Metal on leather Wood on wood Wood on leather Stone on stone Earth on earth Rubber on concrete
0.15 – 0.60 0.20 – 0.60 0.30 – 0.70 0.30 – 0.60 0.25 – 0.50 0.25 – 0.50 0.40 – 0.70 0.20 – 1.00 0.60 – 0.90
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3.8.3 Four Situations which may occur when a rigid body is in contact with a horizontal surface
P
1. Forces applied to body do not tend to move it along surface of contact
there is no friction force
W
F =0 N = P +W
N P
W
Py Px
F
N
F = Px F < μs N N = Py + W
2. Applied forces tend to move body along surface of contact but not large enough to set it in motion friction force develops and can be found by solving equations of equilibrium no evidence that F has reached its limiting value; thus F = μsN cannot be used to find F
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P
W
Py Px
N
P
3. Applied forces are such that body is just about to slide
motion is impending
Friction force F reaches its maximum value Fm Fm Fm = Px
Equations of equilibrium and the equation Fm = μ s N F = μsN can be used N = Py + W
Friction force has a sense opposite to sense of impending motion
W
Py Px
Fk < Px Fk = μ k N N N = Py + W
Fk
4. Body is sliding under action of applied forces equations of equilibrium not applicable F is now equal to kinetic-friction and the equation F = μkN may be used sense of F is opposite to sense of motion PC1433 Mechanics and Waves
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3.8.4 Angles of Friction
P
P
W
tan φs =
W
Py
tan φs = μ s
Px N
R=N
Py
φ < φs
(b) No motion
W
P
R
Angle of static friction
F = Px
(a) No friction
W
P
Py Px N R Fm = Px
φ = φs
(c) Motion impending →
Fm μ s N = N N
tan φk =
Fk μ k N = N N
tan φk = μ k
Px N Fk < Px
R
Angle of kinetic friction
φ = φk
(d) Motion →
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W sin θ
W
W
W cos θ
θ < φs θ =0
θ
R
N = W cos θ R F = W sin θ
(a) No friction
(b) No motion
W sin θ W W cos θ
θ = φs
W sin θ W W cos θ
θ = φs
θ > φs
φk
N = W cos θ R F = W sin θ (c) Motion impending
θ
R
N = R cos φ = W cos θ F = R sin φ < W sin θ
(d) Motion
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3.8.5 Problems Involving Dry Friction Generally, 3 groups of problems involving dry friction: First Group: All applied forces and coefficient of friction known; determine whether body remains at rest or slide Friction force required to maintain equilibrium is unknown – its magnitude is not equal to Fm = μsN Friction force F and normal force N should be determined by drawing a FBD & solving equations of equilibrium Magnitude of F found should be compared with the maximum value Fm
If F <= Fm , body remains at rest
If F > Fm , equilibrium cannot be maintained and motion takes place; actual magnitude of friction is then Fk = μkN
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Second Group: All applied forces are given and motion is known to be impending; determine coefficient of static friction Friction force required to maintain equilibrium is known and equal to Fm = μsN Friction force F and normal force N should be determined by drawing a FBD & solving equations of equilibrium Equating F found to value of Fm, solve for μs Third Group: Coefficient of static friction given and motion is known to be impending in a given direction; determine magnitude or direction of one of the applied forces Friction force should be shown in FBD with a sense opposite to that of impending motion and equal to Fm = μsN Write and solve equations of equilibrium for unknown applied force
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Example 3.8 A support block is acted upon by two forces as shown. Knowing that the coefficients of friction between the block and the incline are μs = 0.35 and μk = 0.25, determine the force P required (a) to start the block moving up the incline, (b) to keep it moving up, (c) to prevent it from sliding down.
25
P
0
800 N
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Solution
Free-Body Diagram
800 N
a. Force P to start block moving up P
R
φs
25
0
tan φs = μ s = 0.35 ⇒ φ s = 19.29 0
P = (800N ) tan(φs + 250 ) = 800 tan 44.290
P 800 N
ur P = 780 N ← R
# (Ans)
φ s + 25 0 PC1433 Mechanics and Waves
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b. Force P to keep block moving up at constant velocity (zero acceleration, block is in dynamic equilibrium):
800 N P
P 800 N
R φk
R
25
0
φk + 250
c. Force P to prevent sliding 800 N
R
0
# (Ans)
P
P 25
P = (800N ) tan(φk + 250 ) ur P = 649 N ←
800 N
φs
tan φk = μ k = 0.25 ⇒ φk = 14.04 0
R
25 0 − φ s
P = (800N ) tan(250 − φs ) ur P = 80.0 N ←
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# (Ans)
81
Example 3.9
x
P
60 mm
30 mm
The movable bracket shown may be placed at any height on the 30 mm diameter pipe. If the coefficient of static friction between the pipe and bracket is 0.25, determine the minimum distance x at which the load P can be supported. Neglect the weight of the bracket.
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Solution
x FA ΝA
x − 15
P
A
60 mm Free-Body Diagram
B 30 mm
When P is placed at min. distance x from axis of pipe, bracket is just about to slip, and forces of friction at A & B have reached their max. values:
FA = μs Ν A = 0.25 Ν A
ΝB
FB = μs Ν B = 0.25 Ν B
FB
Equilibrium Equations.
+ → ∑ Fx = 0 : + ↑ ∑ Fy = 0 :
NB − N A = 0 ⇒ NB = N A
(1)
FA + FB − P = 0 ⇒ 0.25 N A + 0.25 N B = P
Solve (1) & (2) simultaneously,
(2)
N A = 2P
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x
x − 15
FA
NA A 60 mm
P
FB B
NB
30 mm
∑M
B
= 0 : N A (60mm) − FA (30mm) − P ( x − 15mm) = 0 60 N A − 30(0.25 N A ) − Px + 15P = 0 60(2 P ) = 7.5(2 P ) − Px + 15 P = 0 ∴ x = 120mm # (Ans)
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