differential element
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Applications of Integration to Physics andEngineering
MATH 211,Calculus II
J. Robert Buchanan
Department of Mathematics
Spring 2011
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Mass and Weight
mass: quantity of matter (units: kg or g (metric) or slugs
(English))
weight: a force related to mass throughNewtons Second
Law
weight= (mass)(gravitational acceleration)
gravity: gravitational acceleration (notation,g)
Metric unitsg=9.8 m/s2 org=980 cm/s2.
English units,g=32 ft/s2.
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Work
Workis what is accomplished by moving aforcethrough some
distance.
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Work
Workis what is accomplished by moving aforcethrough some
distance.
If the forceFis constant and is moved in a straight line a
distanced, the work is W =Fd.
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Work
Workis what is accomplished by moving aforcethrough some
distance.
If the forceFis constant and is moved in a straight line a
distanced, the work is W =Fd.
Units of work:
force distance work
pound (lb) foot (ft) foot-pound (ft-lb)
inch (in) inch-pound (in-lb)
Newton (N) meter (m) Newton-meter (N-m)
Note: 1Newton-meter= 1Joule= 1 J.
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Variable Forces
Suppose the forcefdepends on positionx,i.e., the forcef(x)is moved in a straight line fromx=atox=b.
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Variable Forces
Suppose the forcefdepends on positionx,i.e., the forcef(x)is moved in a straight line fromx=atox=b.
Letn N and definex= ban andxi=a+ix fori=1, 2, . . . ,n.
The work done moving the force fromxk1 toxk isapproximatelyWk=f(xk)x.
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V i bl F
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Variable Forces
Suppose the forcefdepends on positionx,i.e., the forcef(x)is moved in a straight line fromx=atox=b.
Letn N and definex= ban andxi=a+ix fori=1, 2, . . . ,n.
The work done moving the force fromxk1 toxk isapproximatelyWk=f(xk)x.
A Riemann sum for the work is then
W n
k=1
f(xk)x
= limn
nk=1
f(xk)x
=
b
a
f(x) dx.
J. Robert Buchanan Applications of Integration to Physics and Engineering
H k L S i
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Hookes Law Spring
Hookes Law: the force required to stretch or compress a
spring beyond its natural length is f(x) =kxwherekis calledthespring constant.
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H k L S i
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Hookes Law Spring
Hookes Law: the force required to stretch or compress a
spring beyond its natural length is f(x) =kxwherekis calledthespring constant.
ExampleA force of 10 lb is required to stretch a spring from its natural
length of 7 in to 7.5 in. Find the work done in stretching thespring. How much work is done stretching the spring from 7.5in to 9 in?
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S l ti
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Solution
Since 10= k(7.5 7)then the spring constantk=20.The work done to stretch the spring the initial 1/2 inch is
W = 1/2
0
20x dx= 10x21/20
=5
2 in-lb.
The work done to stretch the spring the next 1.5 inches is
W = 2
1/2
20x dx= 10x22
1/2
=75
2
in-lb.
J. Robert Buchanan Applications of Integration to Physics and Engineering
Pumping Out a Tank
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Pumping Out a Tank
Example
A hemispherical tank of radius 5 m is full of water with a densityof 103 kg/m3. Find the work done in pumping the water out of
the tank.
-4 -2 2 4x
-4
-2
2
4
y
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Solution (1 of 2)
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Solution (1 of 2)
Imagine a thin layer of water parallel to the surface of thetank at a depth ofy.
The volume of the thin layer of water can be expressed as
(25
y2) dy.
The weight of the thin layer of water is therefore
1000(9.8)(25 y2) dy=9800(25 y2) dy.
The distance this thin layer must be lifted to remove it fromthe tank is 0 y= y.
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Solution (2 of 2)
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Solution (2 of 2)
The work done is
W =
05
9800(25 y2)(y) dy
= 9800 0
5(y3
25y) dy
= 9800
1
4y4 25
2y205
=
98001
4
(
5)4
25
2
(
5)2
= 1531250 J.
J. Robert Buchanan Applications of Integration to Physics and Engineering
Winding a Cable
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Winding a Cable
Example
A 50-ft cable weighing a total of 25 lbs is attached to a 600 lb
object. Find the work done in using the cable to lift the object
30 ft.
J. Robert Buchanan Applications of Integration to Physics and Engineering
Solution
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Solution
The cable weighs 1/2 lb/ft.
When the object has been liftedxfeet, the total weight of
the object and the remaining length of cable is
f(x) =600 +25
x/2= 625
x/2.
The work done is
W =
300
625 x2
dx=
625x x
2
4
30
0
=18525 ft-lb.
J. Robert Buchanan Applications of Integration to Physics and Engineering
Moments and Center of Mass
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Moments and Center of Mass
In mathematics, the physical sciences, and engineering it is
convenient to replace a rigid object of massmby an idealized
point-mass (also ofm).
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Moments and Center of Mass
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Moments and Center of Mass
In mathematics, the physical sciences, and engineering it is
convenient to replace a rigid object of massmby an idealized
point-mass (also ofm).
Question: where do we place the mass?
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Example (1 of 2)
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Example (1 of 2)
Suppose two massesm1 andm2are attached to opposite ends
of a rod.
x1 x2x
x
m1 m2
Where can we support the rod so that the system is balanced?
Letxbe the location of the balance point.
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Example (2 of 2)
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Example (2 of 2)
m1(x x1) = m2(x2 x)x(m1+m2) = m1x1+m2x2
x = m1x1+ m2x2
m1+ m2
Remark:the coordinatexis called thecenter of mass(or
center of gravity) of the system.
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General Case
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General Case
DefinitionIfSdenotes a set of point masses {m1, m2, . . . ,mn} located atthe points {x1, x2, . . . , xn} respectively along thex-axis, then
thetotal massis m=n
i=1
mi,
themoment about the originisM0=n
i=1
mixi,
thecenter of massisx=M
0m .
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Example
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Example
ExampleSupposeSconsists of masses {5, 7,11,13} kg located at{3,1, 1, 2} respectively along thex-axis. Find the center ofmass ofS.
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Example
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p
ExampleSupposeSconsists of masses {5, 7,11,13} kg located at{3,1, 1, 2} respectively along thex-axis. Find the center ofmass ofS.
m = 5 +7+11 +13= 36
M0 = (5)(3) + (7)(1) + (11)(1) + (13)(2) =15
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Distributed Case
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Suppose an object is continuously distributed along thex-axis
in the interval[a, b]and thedensity(mass/length) of the objectis given by(x).
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Distributed Case
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Suppose an object is continuously distributed along thex-axis
in the interval[a, b]and thedensity(mass/length) of the objectis given by(x).
Question: how can we find the center of mass of such an
object?
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Distributed Case
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Suppose an object is continuously distributed along thex-axis
in the interval[a, b]and thedensity(mass/length) of the objectis given by(x).
Question: how can we find the center of mass of such an
object?
Answer:a Riemann Sum! Letn N and definex= ban andxi =a+ix. The mass of the portion of the object in theinterval[xk1, xk]is mk (xk)x. Thus the total mass of theobject is
m
nk=1 (xk)x
= limn
nk=1
(xk)x=
ba
(x) dx.
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Moment
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We can use a Riemann sum to find the moment about the
origin of the distributed object.
M0 n
k=1
xk(xk)x
= lim
n
n
k=1
xk(xk)x
=
ba
x(x) dx
Thus the center of mass is
x=M0
m =
ba
x(x) dx
b
a (x) dx
.
J. Robert Buchanan Applications of Integration to Physics and Engineering
Example (1 of 2)
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Example
Find the mass and center of mass of an object whose density isgiven by the(x) = x5 +1 for 0 x 7.
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Example (1 of 2)
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Example
Find the mass and center of mass of an object whose density isgiven by the(x) = x5 +1 for 0 x 7.
m =
7
0
x5
+1
dx=
x2
10+x
70
=119
10
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Example (1 of 2)
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Example
Find the mass and center of mass of an object whose density isgiven by the(x) = x5 +1 for 0 x 7.
m =
7
0
x5
+1
dx=
x2
10+x
70
=119
10
M0 = 7
0 x2
5 +x dx=
x3
15+
x2
2 7
0
=1421
30
J. Robert Buchanan Applications of Integration to Physics and Engineering
Example (1 of 2)
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Example
Find the mass and center of mass of an object whose density isgiven by the(x) = x5 +1 for 0 x 7.
m =
7
0
x5
+1
dx=
x2
10+x
70
=119
10
M0 = 7
0 x2
5 +x dx=
x3
15+
x2
2 7
0
=1421
30
x =1421
3011910
=203
51
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Example (2 of 2)
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Example
Find the mass and center of mass of an object whose density isgiven by the(x) = x2 x+6 for 3 x 2.
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Example (2 of 2)
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Example
Find the mass and center of mass of an object whose density isgiven by the(x) = x2 x+6 for 3 x 2.
m =
2
3
x2 x+6
dx=
x3
3 x2
2 +6x
23
=125
6
M0 = 2
3 x
3 x2 +6x dx= x4
4 x
3
3 +3x2
2
3
= 12512
x = 12512
1256
= 12
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Hydrostatic Force
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Terminology:pressure: force exerted per unit area (notation,p)
gravity: gravitational acceleration (notation,g)
Metric unitsg=9.8 m/s2 org=980 cm/s2.
English units,g=32 ft/s2.density: mass per unit volume (notation,)
for water = 1000 kg/m3 or = 1 g/cm3.English units,g=62.4 lb/ft3.
depth: distance to the surface of a fluid (notation,h)
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Pascals Principle
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Pascals Principle: the pressure exerted at a depthhin a fluid
is the same in every direction.
If the area of a plate is Athen the force on the plate is ghA,provided the plate is entirely at depthh.
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Pascals Principle
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Pascals Principle: the pressure exerted at a depthhin a fluid
is the same in every direction.
If the area of a plate is Athen the force on the plate is ghA,provided the plate is entirely at depthh.
Question: what if the plate is not oriented horizontally?
J. Robert Buchanan Applications of Integration to Physics and Engineering
Riemann Sum Approach
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Suppose the plate is oriented vertically (parallel to the
xy-plane) so that the width of the plate is a function of y,
call itw(y
).
Suppose the plate lies in the interval (on the y-axis)[a, b].Letn N andy= ban andyi=a+iy, then the forceon the portion of the plate betweenyk1 andyk is
Fk
gh(yk)w(yk)y.
Totalhydrostatic forceis
F n
k=1
gh(yk)w(yk)y
= limn
nk=1
gh(yk)w(yk)y
= g b
a
h(y)w(y) dy
J. Robert Buchanan Applications of Integration to Physics and Engineering
Example
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Example
A dam has a submerged gate in the shape of an equilateral
triangle, two feet on a side with the horizontal base nearest the
surface of the water and ten feet below it. Find the force on the
gate.
-2 -1 1 2x
-2
2
4
6
8
10
y
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Solution
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The edges of the plate are described by the lines with
equationsy= 2+2xandy= 2 2x.The width of the plate isw(y) =2+y
The hydrostatic force is
F = 62.4 02(10 y)(2+y) dy
= 62.4
02
(20+8y y2) dy
= 62.464
3
1331.2 lb.
J. Robert Buchanan Applications of Integration to Physics and Engineering
Example
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Example
A square plate of sides 5 feet is submerged vertically in water
such that one of the diagonals is parallel to the surface of thewater. If the distance from the surface of the water to the center
of the plate is 4 feet, find the force exerted by the water on one
side of the plate.
-3 -2 -1 1 2 3
x
-3
-2
-1
1
2
3
4
y
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Solution
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The edges of the plate are described by the lines with
equationsy x=5/
2 andx+y=5/
2 (fory 0) andbyx+y= 5/
2 andy x= 5/
2 (fory
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Read Section 5.6
Exercises: 113 (work), 2130 (center of mass), 3540
(hydrostatic force)
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