applications of integration to physics and...

Applications of Integration to Physics andEngineering

MATH 211, Calculus II

J Robert Buchanan

Department of Mathematics

Spring 2018

Mass and Weight

mass: quantity of matter (units: kg or g (metric) or slugs(English))

gravity: gravitational acceleration (notation, g)I Metric units g = 9.8 m/s2 or g = 980 cm/s2.I English units, g = 32 ft/s2.

weight: a force related to mass through Newton’s SecondLaw

weight = (mass)(gravitational acceleration)1 pound = (1 slug)(32 ft/s2)

1 Newton = (1 kg)(9.8 m/s2)

Work

Work is what is accomplished by moving a force through somedistance.

If the force F is constant and is moved in a straight line adistance d , the work is W = F d .

Units of work:

force distance workpound (lb) foot (ft) foot-pound (ft-lb)

inch (in) inch-pound (in-lb)Newton (N) meter (m) Newton-meter (N-m)

Note: 1 Newton-meter = 1 Joule = 1 J.

Variable ForcesSuppose the force f depends on position x , i.e., the force f (x)is moved in a straight line from x = a to x = b.

Let n ∈ N and define ∆x = (b − a)/n and xi = a + i∆x fori = 1,2, . . . ,n.

The work done moving the force from xk−1 to xk isapproximately ∆Wk = f (xk )∆x .

A Riemann sum for the work is then

W ≈n∑

k=1

f (xk )∆x

= limn→∞

n∑k=1

f (xk )∆x

W =

∫ b

af (x) dx .

Hooke’s Law Spring

Hooke’s Law: the force required to stretch or compress aspring beyond its natural length is f (x) = k x where k is calledthe spring constant.

ExampleA force of 10 lb is required to stretch a spring from its naturallength 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?

Solution

I Since 10 = k(7.5− 7) then the spring constant k = 20.I The work done to stretch the spring the initial 1/2 inch is

W =

∫ 1/2

020x dx =

[10x2

]x=1/2

x=0=

52

in-lb.

I The work done to stretch the spring the next 1.5 inches is

W =

∫ 2

1/220x dx =

[10x2

]x=2

x=1/2=

752

in-lb.

Work Done Against a Spring

Suppose that 2 J of work is done to stretch a spring from itsnatural length of 30 cm to a length of 42 cm.

1. How much work is done to stretch the spring from 35 cm to45 cm?

2. How far beyond its natural length will a force of 35 Nstretch the spring?

Solution (1 of 2)

Recall that W =∫ x2

x1k x dx , where k is the spring constant.

2 =

∫ (42−30)/100

0k x dx =

[k2

x2]x=3/25

x=0=

9k1250

k =2500

9N/m

1. How much work is done to stretch the spring from 35 cm to45 cm?

W =

∫ (45−30)/100

(35−30)/100

25009

x dx =

[1250

9x2]x=3/20

x=1/20

=1250

9

(9

400− 1

400

)=

259

J

Solution (2 of 2)

2. How far beyond its natural length will a force of 35 Nstretch the spring?

F = k x

35 =2500

9x

x =63

500m =

635

cm = 12.6 cm

Pumping Out a Tank

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 ofthe tank.

-4 -2 2 4x

-4

-2

2

4

y

Solution (1 of 3)

Imagine a thin layer of water parallel to the surface of the tankat a depth of y .

x2+y2=25

-4 -2 2 4x

-4

-2

2

4

y

The radius of “slice” of the water is r =√

25− y2.

Solution (2 of 3)

I The volume of the thin layer of water can be expressed as

π(25− y2) dy .

I The weight of the thin layer of water is therefore

1000(9.8)π(25− y2) dy = 9800π(25− y2) dy .

I The distance this thin layer must be lifted to remove it fromthe tank is 0− y = −y .

Solution (3 of 3)

The work done is

W =

∫ 0

−59800π(25− y2)(−y) dy

= 9800π∫ 0

−5(y3 − 25y) dy

=

[9800π

(14

y4 − 252

y2)]y=0

y=−5

= −9800π(

14

(−5)4 − 252

(−5)2)

W = 1531250π J.

Emptying a Tank

The tank shown below is full of water. Find the work done topump the water out of the spout.

Solution (1 of 2)

-3 -2 -1 0 1 2 3

0

1

2

3

4

5

6

x

y

y=2xy=-2x

width=y

distance=5-y

The “slice” of water at altitude y has a volume of

l w h = 8y dy .

Solution (2 of 2)

The work done to pump out the tank is

W =

∫ 3

01000(9.8)8y(5− y) dy

= 78400∫ 3

0(5y − y2) dy

= 78400[

5y2

2− y3

3

]y=3

y=0

= 78400 · 272

= 1,058,400 J.

Winding a Cable

A 50-ft cable weighing a total of 25 lbs is attached to a 600 lbobject. Find the work done in using the cable to lift the object30 ft.

Solution

I The cable weighs 1/2 lb/ft.I When the object has been lifted x feet, the total weight of

the object and the remaining length of cable isf (x) = 600 + 25− x/2 = 625− x/2.

I If this weight is lifted a distance ∆x , the increment of workdone to lift this weight is ∆W = (625− x/2)∆x .

I The work done is

W =

∫ 30

0

(625− x

2

)dx =

[625x − x2

4

]x=30

x=0= 18525 ft-lb.

Stopped Elevator

The elevator in Wickersham Hall has stopped between the firstand second floors. The cable attached to the elevator carweighs 6 pounds/foot and is 40 feet long. The elevator car byitself weighs 1000 pounds and there are two students weighing100 pounds each inside. To rescue the students, the elevatorcar must be manually lifted 5 feet by winding the cable onto apulley at the top of the building. How much work must be done?

Solution

The force which must be moved can be expressed as

F = 1000 + 2(100) + 6(40− x) = 1440− 6x

when x feet of the cable have been wound onto the pulley.

If this force is moved a distance dx , the differential work done is

F dx = (1440− 6x) dx .

If the elevator must be lifted 5 feet, then

W =

∫ 5

0(1440− 6x) dx =

[1440x − 3x2

]x=5

x=0= 7125 ft-lb.

Moments and Center of Mass

In mathematics, the physical sciences, and engineering it isconvenient to replace a rigid object of mass m by an idealizedpoint-mass (also of m).

Question: where do we place the mass?

Example (1 of 2)

Suppose two masses m1 and m2 are attached to opposite endsof a rod.

m1 m2

x1 x2xx

Where can we support the rod so that the system is balanced?

Let x be the location of the balance point.

Example (2 of 2)

m1(x − x1) = m2(x2 − x)

x(m1 + m2) = m1x1 + m2x2

x =m1x1 + m2x2

m1 + m2

Remark: the coordinate x is called the center of mass (orcenter of gravity) of the system.

General Case

DefinitionIf S denotes a set of point masses {m1,m2, . . . ,mn} located atthe points {x1, x2, . . . , xn} respectively along the x-axis, then

I the total mass is m =n∑

i=1

mi ,

I the moment about the origin is M0 =n∑

i=1

mixi ,

I the center of mass is x =M0

m.

Example

Suppose S consists of masses {5,7,11,13} kg located at{−3,−1,1,2} respectively along the x-axis. Find the center ofmass of S.

m = 5 + 7 + 11 + 13 = 36

M0 = (5)(−3) + (7)(−1) + (11)(1) + (13)(2) = 15

x =1536

=512

Example


m = 5 + 7 + 11 + 13 = 36

M0 = (5)(−3) + (7)(−1) + (11)(1) + (13)(2) = 15

x =1536

=5

12

Example


m = 5 + 7 + 11 + 13 = 36M0 = (5)(−3) + (7)(−1) + (11)(1) + (13)(2) = 15

x =1536

=5

12

Distributed CaseSuppose an object is continuously distributed along the x-axisin the interval [a,b] and the density (mass/length) of the objectis given by ρ(x).

Question: how can we find the center of mass of such anobject?

Answer: a Riemann Sum! Let n ∈ N and define∆x = (b − a)/n and xi = a + i∆x . The mass of the portion ofthe object in the interval [xk−1, xk ] is ∆mk ≈ ρ(xk )∆x . Thus thetotal mass of the object is

m ≈n∑

k=1

ρ(xk )∆x

m = limn→∞

n∑k=1

ρ(xk )∆x =

∫ b

aρ(x) dx .

MomentWe can use a Riemann sum to find the moment about theorigin of the distributed object.

M0 ≈n∑

k=1

xkρ(xk )∆x

= limn→∞

n∑k=1

xkρ(xk )∆x

M0 =

∫ b

ax ρ(x) dx

Thus the center of mass is

x =M0

m=

∫ ba x ρ(x) dx∫ ba ρ(x) dx

.

Example (1 of 2)

Find the mass and center of mass of an object whose density isgiven by the ρ(x) = x/5 + 1 for 0 ≤ x ≤ 7.

m =

∫ 7

0

(x5

+ 1)

dx =

[x2

10+ x

]x=7

x=0=

11910

M0 =

∫ 7

0

(x2

5+ x

)dx =

[x3

15+

x2

2

]x=7

x=0=

142130

x =1421

3011910

=20351≈ 3.98039

Example (1 of 2)

Find the mass and center of mass of an object whose density isgiven by the ρ(x) = x/5 + 1 for 0 ≤ x ≤ 7.

m =

∫ 7

0

(x5

+ 1)

dx =

[x2

10+ x

]x=7

x=0=

11910

M0 =

∫ 7

0

(x2

5+ x

)dx =

[x3

15+

x2

2

]x=7

x=0=

142130

x =142130

11910

=20351≈ 3.98039

Example (2 of 2)

Find the mass and center of mass of an object whose density isgiven by the ρ(x) = −x2 − x + 6 for −3 ≤ x ≤ 2.

m =

∫ 2

−3

(−x2 − x + 6

)dx =

[−x3

3− x2

2+ 6x

]x=2

x=−3=

1256

M0 =

∫ 2

−3

(−x3 − x2 + 6x

)dx =

[−x4

4− x3

3+ 3x2

]x=2

x=−3= −125

12

x =−125

121256

= −12

Hydrostatic Force

Terminology:pressure: force exerted per unit area (notation, p)

gravity: gravitational acceleration (notation, g)I Metric units g = 9.8 m/s2 or g = 980 cm/s2.I English units, g = 32 ft/s2.

density: mass per unit volume (notation, ρ)I for water ρ = 1000 kg/m3 or ρ = 1 g/cm3.I English units, ρg = 62.4 lb/ft3.

depth: distance to the surface of a fluid (notation, h)

Pascal’s Principle

Pascal’s Principle: the pressure exerted at a depth h in a fluidis the same in every direction.

If the area of a plate is A then the force on the plate is ρg h A,provided the plate is entirely at depth h.

Question: what if the plate is not oriented horizontally?

Riemann Sum ApproachI Suppose the plate is oriented vertically (parallel to the

xy -plane) so that the width of the plate is a function of y ,call it w(y).

I Suppose the plate lies in the interval (on the y -axis) [a,b].I Let n ∈ N and ∆y = (b − a)/n and yi = a + i∆y , then the

force on the portion of the plate between yk−1 and yk is

∆Fk ≈ ρg h(yk )w(yk )∆y .

I Total hydrostatic force is

F ≈n∑

k=1

ρg h(yk )w(yk )∆y

= limn→∞

n∑k=1

ρg h(yk )w(yk )∆y

F = ρg∫ b

ah(y)w(y) dy

ExampleA dam has a submerged gate in the shape of an equilateraltriangle, two feet on a side with the horizontal base nearest thesurface of the water and ten feet below it. Find the force on thegate.

y= 3 (x-1)y=- 3 (x+1)

-2 -1 1 2x

-2

2

4

6

8

10

y

Solution

I The edges of the plate are described by the lines withequations y =

√3(x − 1) and y = −

√3(x + 1).

I The width of the plate is w(y) = 2 + 2y/√

3I The hydrostatic force is

F = 62.4∫ 0

−√

3(10− y)(2 + 2y/

√3) dy

= 124.8∫ 0

−√

3

(10 +

10√3

y − y − 1√3

y2)

dy

= 124.8[10y +

5√3

y2 − 12

y2 − 13√

3y3]y=0

y=−√

3

≈ 1143.2 lb.

ExampleA square plate of sides 5 feet is submerged vertically in watersuch that one of the diagonals is parallel to the surface of thewater. If the distance from the surface of the water to the centerof the plate is 4 feet, find the force exerted by the water on oneside of the plate.

x+y=5

2

-x+y=-5

2

-x+y=5

2

x+y=-5

2

-3 -2 -1 1 2 3x

-3

-2

-1

1

2

3

4

y

SolutionI The edges of the plate are described by the lines with

equations y − x = 5/√

2 and x + y = 5/√

2 (for y ≥ 0) andby x + y = −5/

√2 and y − x = −5/

√2 (for y < 0).

I The width of the plate is

w(y) =

{10√

2− 2y if y ≥ 0,

10√2

+ 2y if y < 0.

I The hydrostatic force is

F = 62.4∫ 5/

√2

0(4− y)

(10√

2− 2y

)dy

+ 62.4∫ 0

−5/√

2(4− y)

(10√

2+ 2y

)dy

= 62.4(

50− 1256√

2

)+ 62.4

(50 +

1256√

2

)= 6240 lb.

Homework

I Read Section 6.4I Exercises: see WebAssign/D2L

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