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Applications of Integration to Physics andEngineering

MATH 211,Calculus II

J. Robert Buchanan

Department of Mathematics

Spring 2011

J. Robert Buchanan Applications of Integration to Physics and Engineering
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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


distance.

If the forceFis constant and is moved in a straight line a

distanced, the work is W =Fd.

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Work


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


Letn N and definex= ban andxi=a+ix fori=1, 2, . . . ,n.

The work done moving the force fromxk1 toxk isapproximatelyWk=f(xk)x.


V i bl F
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Variable Forces


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.


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.


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?


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.


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


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.


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.


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.


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.


Moments and Center of Mass
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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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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?


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.


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.


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 .


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).


Distributed Case
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Question: how can we find the center of mass of such an

object?


Distributed Case
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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.


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

.


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.


Example (1 of 2)
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Example


m =

7

0

x5

+1

dx=

x2

10+x

70

=119

10


Example (1 of 2)
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Example


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


Example (1 of 2)
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Example


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


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.


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


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)


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.


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?


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


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


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.


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


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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