6.5 Work and Fluid Forces

Work

Work=Force · Distance

Units

Force Distance WorkNewton meter Joule (J)pound foot foot-pound (ft· lb)

In order to do work, the force must be parallel to the motion.

Work

Work=Force · Distance

Units

Force Distance WorkNewton meter Joule (J)pound foot foot-pound (ft· lb)

In order to do work, the force must be parallel to the motion.

Work

Work=Force · Distance

Units

Force Distance WorkNewton meter Joule (J)pound foot foot-pound (ft· lb)

In order to do work, the force must be parallel to the motion.

Is It Work?

If you walk across the room carrying a book, are you doingwork?

No; the force on the book is vertical but the motion of the bookis horizontal.

It is work, however, if we lift a book from the floor to the table.

Is It Work?

If you walk across the room carrying a book, are you doingwork?

No; the force on the book is vertical but the motion of the bookis horizontal.

It is work, however, if we lift a book from the floor to the table.

Is It Work?

If you walk across the room carrying a book, are you doingwork?

No; the force on the book is vertical but the motion of the bookis horizontal.

It is work, however, if we lift a book from the floor to the table.

Example 1

Example

How much work is done on an object if a force of 12 newtonsmoves an object 3m?

36J

If force is constant, we can just multiply. If not, we needintegrals.

Example 1

Example

How much work is done on an object if a force of 12 newtonsmoves an object 3m? 36J

If force is constant, we can just multiply. If not, we needintegrals.

Example 1

Example

How much work is done on an object if a force of 12 newtonsmoves an object 3m? 36J

If force is constant, we can just multiply. If not, we needintegrals.

Example 2

Example

Hooke’s Law says that the force, F, required to compress aspring from its equilibrium position a distance of x meters, isgiven by F = kx, for some constant k. Find the work done incompressing a spring by .1 m if k = 8 nt

m .

What is the force? F = 8x.

But, the force varies with x, so we break the distance movedinto small increments, ∆x.

The work through one increment ≈ F∆x = 8x∆x J. Total work,then, would be ∑ 8x∆x.

Example 2

Example

Hooke’s Law says that the force, F, required to compress aspring from its equilibrium position a distance of x meters, isgiven by F = kx, for some constant k. Find the work done incompressing a spring by .1 m if k = 8 nt

m .

What is the force?

F = 8x.

But, the force varies with x, so we break the distance movedinto small increments, ∆x.

The work through one increment ≈ F∆x = 8x∆x J. Total work,then, would be ∑ 8x∆x.

Example 2

Example

Hooke’s Law says that the force, F, required to compress aspring from its equilibrium position a distance of x meters, isgiven by F = kx, for some constant k. Find the work done incompressing a spring by .1 m if k = 8 nt

m .

What is the force? F = 8x.

But, the force varies with x, so we break the distance movedinto small increments, ∆x.

The work through one increment ≈ F∆x = 8x∆x J. Total work,then, would be ∑ 8x∆x.

Example 2

Example

Hooke’s Law says that the force, F, required to compress aspring from its equilibrium position a distance of x meters, isgiven by F = kx, for some constant k. Find the work done incompressing a spring by .1 m if k = 8 nt

m .

What is the force? F = 8x.

But, the force varies with x, so we break the distance movedinto small increments, ∆x.

The work through one increment ≈ F∆x = 8x∆x J. Total work,then, would be ∑ 8x∆x.

Example 2

Example

Hooke’s Law says that the force, F, required to compress aspring from its equilibrium position a distance of x meters, isgiven by F = kx, for some constant k. Find the work done incompressing a spring by .1 m if k = 8 nt

m .

What is the force? F = 8x.

But, the force varies with x, so we break the distance movedinto small increments, ∆x.

The work through one increment ≈ F∆x = 8x∆x J.

Total work,then, would be ∑ 8x∆x.

Example 2

Example

Hooke’s Law says that the force, F, required to compress aspring from its equilibrium position a distance of x meters, isgiven by F = kx, for some constant k. Find the work done incompressing a spring by .1 m if k = 8 nt

m .

What is the force? F = 8x.

But, the force varies with x, so we break the distance movedinto small increments, ∆x.

The work through one increment ≈ F∆x = 8x∆x J. Total work,then, would be

∑ 8x∆x.

Example 2

Example

Hooke’s Law says that the force, F, required to compress aspring from its equilibrium position a distance of x meters, isgiven by F = kx, for some constant k. Find the work done incompressing a spring by .1 m if k = 8 nt

m .

What is the force? F = 8x.

But, the force varies with x, so we break the distance movedinto small increments, ∆x.

The work through one increment ≈ F∆x = 8x∆x J. Total work,then, would be ∑ 8x∆x.

Example 2

limn→∞

n

∑ 8x∆x

=∫ .1

08x dx

= 4x2∣∣∣.10

= .04 J

In general,

Work

W =∫ b

aF(x) dx

Example 2

limn→∞

n

∑ 8x∆x

=∫ .1

08x dx

= 4x2∣∣∣.10

= .04 J

In general,

Work

W =∫ b

aF(x) dx

Example 2

limn→∞

n

∑ 8x∆x

=∫ .1

08x dx

= 4x2∣∣∣.10

= .04 J

In general,

Work

W =∫ b

aF(x) dx

Example 2

limn→∞

n

∑ 8x∆x

=∫ .1

08x dx

= 4x2∣∣∣.10

= .04 J

In general,

Work

W =∫ b

aF(x) dx

Example 2

limn→∞

n

∑ 8x∆x

=∫ .1

08x dx

= 4x2∣∣∣.10

= .04 J

In general,

Work

W =∫ b

aF(x) dx

Example 3

Example

How much work is required to lift a 1000 kg satellite to analtitude of 2 · 106 m above the Earth’s surface?

We need to know some constants to solve this problem.M=mass of Earth = 6 · 1024kgm=mass of the satelliteG = 6.67× 10−11 (gravitational constant)r = 6.4× 106 m

We have to be careful with the limits of integration here as well.

Example 3

Example

How much work is required to lift a 1000 kg satellite to analtitude of 2 · 106 m above the Earth’s surface?

We need to know some constants to solve this problem.M=mass of Earth = 6 · 1024kg

m=mass of the satelliteG = 6.67× 10−11 (gravitational constant)r = 6.4× 106 m

We have to be careful with the limits of integration here as well.

Example 3

Example

How much work is required to lift a 1000 kg satellite to analtitude of 2 · 106 m above the Earth’s surface?

We need to know some constants to solve this problem.M=mass of Earth = 6 · 1024kgm=mass of the satellite

G = 6.67× 10−11 (gravitational constant)r = 6.4× 106 m

We have to be careful with the limits of integration here as well.

Example 3

Example

How much work is required to lift a 1000 kg satellite to analtitude of 2 · 106 m above the Earth’s surface?

We need to know some constants to solve this problem.M=mass of Earth = 6 · 1024kgm=mass of the satelliteG = 6.67× 10−11 (gravitational constant)

r = 6.4× 106 m

We have to be careful with the limits of integration here as well.

Example 3

Example

How much work is required to lift a 1000 kg satellite to analtitude of 2 · 106 m above the Earth’s surface?

We need to know some constants to solve this problem.M=mass of Earth = 6 · 1024kgm=mass of the satelliteG = 6.67× 10−11 (gravitational constant)r = 6.4× 106 m

We have to be careful with the limits of integration here as well.

Example 3

Example

How much work is required to lift a 1000 kg satellite to analtitude of 2 · 106 m above the Earth’s surface?

We need to know some constants to solve this problem.M=mass of Earth = 6 · 1024kgm=mass of the satelliteG = 6.67× 10−11 (gravitational constant)r = 6.4× 106 m

We have to be careful with the limits of integration here as well.

Example 3

W =∫ 8.4×106

6.4×106

GMmr2 dr

= −GMmr

∣∣∣8.4×106

6.4×106

= 1.489× 1010 J

Example 3

W =∫ 8.4×106

6.4×106

GMmr2 dr

= −GMmr

∣∣∣8.4×106

6.4×106

= 1.489× 1010 J

Example 3

W =∫ 8.4×106

6.4×106

GMmr2 dr

= −GMmr

∣∣∣8.4×106

6.4×106

= 1.489× 1010 J

Example 4

Example

A worker on a scaffolding 76 feet above ground needs to lift a500 lb bucket of cement from the ground to a point 30 feetabove the ground by pulling a rope that weighs 5 lbs

ft . Howmuch work is required?

Example 4

Example

A worker on a scaffolding 76 feet above ground needs to lift a500 lb bucket of cement from the ground to a point 30 feetabove the ground by pulling a rope that weighs 5 lbs

ft . Howmuch work is required?

Example 4

Let x be the distance from the ground to the bucket.

At height x, if the bucket is lifted by ∆x, the work done is500 ∆x + ∆x · (5)︸︷︷︸

rope weight

· (76− x)︸ ︷︷ ︸rope length

W =∫ 30

0500 dx + 5

∫ 30

0(76− x)dx

= 500x∣∣∣30

0+ 5

[76x− x2

2

]∣∣∣30

0

= 24150 ft · lbs

Example 4

Let x be the distance from the ground to the bucket.

At height x, if the bucket is lifted by ∆x, the work done is

500 ∆x + ∆x · (5)︸︷︷︸rope weight

· (76− x)︸ ︷︷ ︸rope length

W =∫ 30

0500 dx + 5

∫ 30

0(76− x)dx

= 500x∣∣∣30

0+ 5

[76x− x2

2

]∣∣∣30

0

= 24150 ft · lbs

Example 4

Let x be the distance from the ground to the bucket.

At height x, if the bucket is lifted by ∆x, the work done is500 ∆x + ∆x · (5)︸︷︷︸

rope weight

· (76− x)︸ ︷︷ ︸rope length

W =∫ 30

0500 dx + 5

∫ 30

0(76− x)dx

= 500x∣∣∣30

0+ 5

[76x− x2

2

]∣∣∣30

0

= 24150 ft · lbs

Example 4

Let x be the distance from the ground to the bucket.

At height x, if the bucket is lifted by ∆x, the work done is500 ∆x + ∆x · (5)︸︷︷︸

rope weight

· (76− x)︸ ︷︷ ︸rope length

W =∫ 30

0500 dx + 5

∫ 30

0(76− x)dx

= 500x∣∣∣30

0+ 5

[76x− x2

2

]∣∣∣30

0

= 24150 ft · lbs

Example 4

Let x be the distance from the ground to the bucket.

At height x, if the bucket is lifted by ∆x, the work done is500 ∆x + ∆x · (5)︸︷︷︸

rope weight

· (76− x)︸ ︷︷ ︸rope length

W =∫ 30

0500 dx + 5

∫ 30

0(76− x)dx

= 500x∣∣∣30

0+ 5

[76x− x2

2

]∣∣∣30

0

= 24150 ft · lbs

Example 4

Let x be the distance from the ground to the bucket.

At height x, if the bucket is lifted by ∆x, the work done is500 ∆x + ∆x · (5)︸︷︷︸

rope weight

· (76− x)︸ ︷︷ ︸rope length

W =∫ 30

0500 dx + 5

∫ 30

0(76− x)dx

= 500x∣∣∣30

0+ 5

[76x− x2

2

]∣∣∣30

0

= 24150 ft · lbs

Example 5

Example

Find the work required to pump all of the oil out of a full tankif the tank is cylindrical with a height of 10 feet and a radius of5 feet. Oil weighs 50 lbs

ft3 .

Example 5

Example

Find the work required to pump all of the oil out of a full tankif the tank is cylindrical with a height of 10 feet and a radius of5 feet. Oil weighs 50 lbs

ft3 .

Example 5

Let x be the distance to the bottom of the tank from the slice.

Vs = πr2∆x = 25π∆x

Ws = 25π∆x · (10− x)︸ ︷︷ ︸how far to go

·(50)

Example 5

Let x be the distance to the bottom of the tank from the slice.

Vs = πr2∆x = 25π∆x

Ws = 25π∆x · (10− x)︸ ︷︷ ︸how far to go

·(50)

Example 5

Let x be the distance to the bottom of the tank from the slice.

Vs = πr2∆x = 25π∆x

Ws = 25π∆x · (10− x)︸ ︷︷ ︸how far to go

·(50)

Example 5

W =∫ 10

025π(50)(10− x)dx

= 1250π∫ 10

0(10− x)dx

= 1250π

[10x− x2

2

]∣∣∣10

0

= 1250π(50)≈ 19634.54 ft · lbs

Example 5

W =∫ 10

025π(50)(10− x)dx

= 1250π∫ 10

0(10− x)dx

= 1250π

[10x− x2

2

]∣∣∣10

0

= 1250π(50)≈ 19634.54 ft · lbs

Example 5

W =∫ 10

025π(50)(10− x)dx

= 1250π∫ 10

0(10− x)dx

= 1250π

[10x− x2

2

]∣∣∣10

0

= 1250π(50)≈ 19634.54 ft · lbs

Example 5

W =∫ 10

025π(50)(10− x)dx

= 1250π∫ 10

0(10− x)dx

= 1250π

[10x− x2

2

]∣∣∣10

0

= 1250π(50)≈ 19634.54 ft · lbs

Example 6

What is the tank was only half full?

W = 1250π∫ 5

0(10− x)dx

= 1250π

[10x− x2

2

]∣∣∣50

≈ 147262.16 ft · lbs

Example 6

What is the tank was only half full?

W = 1250π∫ 5

0(10− x)dx

= 1250π

[10x− x2

2

]∣∣∣50

≈ 147262.16 ft · lbs

Example 6

What is the tank was only half full?

W = 1250π∫ 5

0(10− x)dx

= 1250π

[10x− x2

2

]∣∣∣50

≈ 147262.16 ft · lbs

Example 6

What is the tank was only half full?

W = 1250π∫ 5

0(10− x)dx

= 1250π

[10x− x2

2

]∣∣∣50

≈ 147262.16 ft · lbs

Example 7

Example

A rectangular water tank is underground with the top 3 feetbelow the surface of the ground. If the tank is 10× 20× 10, findthe work needed to pump out the water if the tank is full.Water weighs 62.4 lbs

ft3 .

Example 7

Example

A rectangular water tank is underground with the top 3 feetbelow the surface of the ground. If the tank is 10× 20× 10, findthe work needed to pump out the water if the tank is full.Water weighs 62.4 lbs

ft3 .

Example 7

What is the volume of each slice?

Vs = 20 · 10 · ∆x

How far does each slab need to travel?

Each slab needs to travel x feet.

Example 7

What is the volume of each slice?

Vs = 20 · 10 · ∆x

How far does each slab need to travel?

Each slab needs to travel x feet.

Example 7

What is the volume of each slice?

Vs = 20 · 10 · ∆x

How far does each slab need to travel?

Each slab needs to travel x feet.

Example 7

What is the volume of each slice?

Vs = 20 · 10 · ∆x

How far does each slab need to travel?

Each slab needs to travel x feet.

Example 7

∫ 13

3200 · 62.4 · x dx

= 12480∫ 13

3x dx

= 12480[

x2

2

]∣∣∣13

3

= 998400 ft · lbs

What if the tank was on the surface of the ground and we werepumping to a height of 3 feet above the top of the tank? Howmuch of a difference is there in the amount of work done?THERE IS NO DIFFERENCE.

Example 7

∫ 13

3200 · 62.4 · x dx

= 12480∫ 13

3x dx

= 12480[

x2

2

]∣∣∣13

3

= 998400 ft · lbs

What if the tank was on the surface of the ground and we werepumping to a height of 3 feet above the top of the tank? Howmuch of a difference is there in the amount of work done?THERE IS NO DIFFERENCE.

Example 7

∫ 13

3200 · 62.4 · x dx

= 12480∫ 13

3x dx

= 12480[

x2

2

]∣∣∣13

3

= 998400 ft · lbs

What if the tank was on the surface of the ground and we werepumping to a height of 3 feet above the top of the tank? Howmuch of a difference is there in the amount of work done?THERE IS NO DIFFERENCE.

Example 7

∫ 13

3200 · 62.4 · x dx

= 12480∫ 13

3x dx

= 12480[

x2

2

]∣∣∣13

3

= 998400 ft · lbs

What if the tank was on the surface of the ground and we werepumping to a height of 3 feet above the top of the tank? Howmuch of a difference is there in the amount of work done?THERE IS NO DIFFERENCE.

Example 7

∫ 13

3200 · 62.4 · x dx

= 12480∫ 13

3x dx

= 12480[

x2

2

]∣∣∣13

3

= 998400 ft · lbs

What if the tank was on the surface of the ground and we werepumping to a height of 3 feet above the top of the tank? Howmuch of a difference is there in the amount of work done?

THERE IS NO DIFFERENCE.

Example 7

∫ 13

3200 · 62.4 · x dx

= 12480∫ 13

3x dx

= 12480[

x2

2

]∣∣∣13

3

= 998400 ft · lbs

What if the tank was on the surface of the ground and we werepumping to a height of 3 feet above the top of the tank? Howmuch of a difference is there in the amount of work done?THERE IS NO DIFFERENCE.