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Engineering Mathematics Ⅰ
呂學育 博士
Oct. 20, 2004
1
1.7 Applications to Mechanics,
Electrical Circuits, and Orthogonal Trajectories
1.7.1 Mechanics
Motion of a Chain on a Pulley
1.7.2 Electric Circuits
Kirchhoff's Voltage Law
1.7.3 Orthogonal Trajectories
Orthogonal families
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The ends of the chain have
the same speed as its
center of mass.
WHY ???
The acceleration of the chain
at its center of mass is the same
as it is at its ends.
The motion is of one dimension.
X=0
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Figure 1.16 Chain on a pulley
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The motion is of one dimension.
The mass is constant.
Newton’s law of motion is
dv
F m
dt
The self weight is the only
external force.
X=0
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Figure 1.16 Chain on a pulley
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Mass and Weight
• MKS---kilogram, m/s2 , 1kg× m/s2 =1N
• CGS--- gram, cm/s2 , 1gm× cm/s2 =1dyn
1N= 1kg× m/s2 =1000gm× 100cm/s2
= 100000dyn
1 slug=(1 pound force)/(1 ft/s2 acceleration)
Weight=16 ft × ρ pound/ft=16 ρ pound
Mass=weight/ g=16 ρ pound/32 ft/s2
= ρ /2 pound/ ft/s2 = ρ /2 slug
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Systems of Units
•
SI units (used mostly in physics):
– length: meter (m)
– mass: kilogram (kg)
– time: second (s)
•
•
This system is also referred to as the mks sytem for meter-kilogram-second.
Gaussian units (used mostly in chemistry):
– length: centimeter (cm)
– mass: gram (g)
– time: second (s)
•
•
This system is also referred to as the cgs system for centimeter-gramsecond.
British engineering system:
– length: foot (ft)
– mass: slug
– time: second (s)
•
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The equation of motion is
ρ dv
 2 xρ 
2 dt
dv
 4x
dt
By the chain rule
dv dv dx
dv

v
dt dx dt
dx
Then
X=0
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dv
v  4x ,
dx

vdv  4 xdx
 vdv  
4 xdx
 v2  4x2  K
Figure 1.16 Chain on a pulley
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v2  4x2  K
For such a problem
we need initial boundary condition
involving the independent variable
x
In the beginning, the chain is located
with one of its ends at x=1

x 1 v  0
, so
K  4
X=0
v2  4x2  4
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Figure 1.16 Chain on a pulley
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v2  4x2  4
The chain leaves the pulley when x=8.
 v 2  4(64  1)  252
v  252  6 7 ft / s
Calculate the time t f required for
the chain to leave the pulley
X=0
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Figure 1.16 Chain on a pulley
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v2  4x2  4

v  2 x2  1
Compute
tf
tf 
6 7
 dt  
0
0
8

1
tf 
21
dt
dv
dv
8


1
dt
dx 
dx
8

1
1
dx
v
8
1
1

2
dx

ln
x

x

1
 2

x2  1
1
1
 ln( 8  63 )
2
X=0
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x 1~ 8
v0~6 7
Figure 1.16 Chain on a pulley
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Electric Current
• Electric current is the rate of charge flow past a
given point in an electric circuit, measured in
coulombs/second which is named amperes. In
most DC electric circuits, it can be assumed that
the resistance to current flow is a constant so
that the current in the circuit is related to voltage
and resistance by Ohm's law.
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OHM'S LAW
V=IxR
•
Where:
•
• V = Voltage
• I = Current
• R = Resistance
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OHM'S LAW
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OHM'S LAW
• Ohm's Law defines the relationships between (P) power, (E)
voltage, (I) current, and (R) resistance. One ohm is the
resistance value through which one volt will maintain a
current of one ampere.
( I ) Current is what flows on a wire or conductor like water
flowing down a river. Current flows from points of high
voltage to points of low voltage on the surface of a
conductor. Current is measured in (A) amperes or amps.
( E ) Voltage is the difference in electrical potential between
two points in a circuit. It's the push or pressure behind
current flow through a circuit, and is measured in (V) volts.
( R ) Resistance determines how much current will flow
through a component. Resistors are used to control voltage
and current levels. A very high resistance allows a small
amount of current to flow. A very low resistance allows a
large amount of current to flow. Resistance is measured in
ohms.
( P ) Power is the amount of current times the voltage level
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at a given point measured in wattage or watts
Kirchhoff's Laws
• Ohm's Law described the relationship between current, voltage, and
resistance. These circuits have been relatively simple in nature.
Many circuits are extremely complex and cannot be solved with
Ohm's Law. These circuits have many power sources and branches
which would make the use of Ohm's Law impractical or impossible.
• Through experimentation in 1857 the German physicist Gustav
Kirchhoff developed methods to solve complex circuits. Kirchhoff
developed two conclusions, known today as Kirchhoff's Laws.
• Law 1: The sum of the voltage drops around a closed loop is equal
to the sum of the voltage sources of that loop (Kirchhoff's Voltage
Law).
• Law 2: The current arriving at any junction point in a circuit is equal
to the current leaving that junction (Kirchhoff's Current Law).
• Kirchhoff's laws can be related to conservation of energy and
charge if we look at a circuit with one load and source. Since all of
the power provided from the source is consumed by the load,
energy and charge are conserved. Since voltage and current can be
related to energy and charge, then Kirchhoff's laws are only
restating the laws governing energy and charge conservation.
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Kirchhoff's Current Law (KCL)
• KCL states that the algebraic sum of the
currents at any juncture of a circuit is zero
• As a direct consequence of the conservation of
charge, namely charge can neither be created
nor destroyed.
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Kirchhoff's Voltage Law
• Kirchhoff's Voltage Law (or Kirchhoff's Loop Rule) is a
result of the electrostatic field being conservative. It
states that the total voltage around a closed loop must
be zero.
• We can adopt the convention that potential gains (i.e.
going from lower to higher potential) is taken to be
positive. Potential losses (such as across a resistor) will
then be negative.
Around a closed loop, the total voltage should be zero
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Kirchhoff's Voltage Law
• Kirchhoff's Voltage Law - KVL - is one of two
fundamental laws in electrical engineering, the
other being Kirchhoff's Current Law (KCL).
• KVL is a fundamental law, as fundamental as
Conservation of Energy in mechanics, because
KVL is really conservation of electrical energy.
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Kirchhoff's Voltage Law
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1.7.2 Electric Circuits
Charge q(t ) and current i (t )
are related by i (t )  q ' (t )
By Kirchhoff’s voltage law
E  iR  Li '  0
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Figure 1.18 RL circuit

E
E
i  i
R
L

E
i(t )   Ke Rt / L
R
'
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1.7.2 Electric Circuits
Charge q(t ) and current i (t )
are related by i (t )  q ' (t )
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Figure 1.19 RC circuit
1
iR  q  E
C
1
'
 Ri  q  E
C
1
E

'
q
q
RC
R

q(t )  EC 1  e  t / RC 
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Two curves intersecting at a
point P are said to be orthogonal
If their tangents are perpendicular
(orthogonal) at P.
dy
 tan(θ)
dx
tan(θ  π / 2)  sin( θ  π / 2) / cos(θ  π / 2)
sin(θ) cos(π / 2)  cos(θ) sin( π / 2)
cos(θ) cos(π / 2)  sin(θ) sin( π / 2)
cos(θ)  1/ tan(θ)

 sin(θ)

Two lines are orthogonal if and
only if their slopes are negative
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reciprocals.
Figure 1.20 Orthogonal families: circles and lines
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1.1.5 Direction Fields
Recall
y
Slope=
dy
 y'
dx
x
• Short tangent segments suggest the
shape of the curve
輪廓
23
1.7.3 Orthogonal Trajectories
How to determine the family of orthogonal trajectories
of a given family of curves ?
Given a family F of curves in the plane, we want to
construct a second family R of curves.
Such that every curve in F is orthogonal to every
curve in
R
wherever an intersection occurs.
F ( x, y, k )  0

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Thomson Learning™ is a trademark used herein under
license.
R( x, y, C )  0
Figure 1.20 Orthogonal families: circles and lines
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1.7.3 Orthogonal Trajectories
How to determine the family of orthogonal trajectories
of a given family of curves ?
giving a different curve for
F ( x, y, k )  0
each choice of constant k
as integral curves of a differential equation
y '  f ( x, y )
solve the following differential
equation
1
y'  
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Thomson Learning™ is a trademark used herein under
license.
for the curves in
Figure 1.20 Orthogonal families: circles and lines
f ( x, y)
R
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1.7.3 Orthogonal Trajectories
Example 1.29
Consider the family
F
of curves that are graphs of
a family of parabolas
F ( x, y, k )  y  kx 2  0
get the differential equation of
differentiate

F
y  kx 2  0
to get
y
k 2
x
y
y  2  f ( x, y )
x

y '  2kx  0
'
the differential equation of the family
F
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Thomson Learning™ is a trademark used herein under license.
Figure 1.21 Orthogonal families: Parabolas and ellipses.
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1.7.3 Orthogonal Trajectories
Example 1.29
Curves in F are integral curves of the following
differential equation
y
'
y 2
x
 f ( x, y )
The family R of orthogonal trajectories therefore
has differential equation
1
x
'
y 
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Thomson Learning™ is a trademark used herein under license.

f ( x, y)
2y
 2 ydy   xdx
1 2
2
 y  x C
2
1 2
2
x

y
C
The is a family of ellipses
2
Figure 1.21 Orthogonal families: Parabolas and ellipses.
27
1.1.5 Direction Fields
Recall
y
Slope=
dy
 y'
dx
x
• Short tangent segments suggest the
shape of the curve
輪廓
28