View
1
Download
0
Category
Preview:
Citation preview
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.
Recommended