Calculus Of One Real Variable – By Pheng Kim Ving

12.9

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1. Pressure And Force Exerted By A Fluid 

According to Pascal's principle, the molecules in a fluid
interact in such a way that the pressure of the fluid at any depth h
acts equally in all directions. For example, the pressure on a horizontal plane
region and that on a vertical one at the
same depth are the same.
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2. Finding The Force Exerted By A Fluid 
For a horizontal plane region submerged in a fluid, the
pressure of the fluid on it is constant throughout the region. So the
total force F exerted
by the fluid is found by multiplying the pressure p of the fluid by the area A of the region: F =
For a nonhorizontal plane region, the pressure isn't constant throughout the
region, and thus the total force can't be
found so easily.
Fig. 2.1
Total force is approximated by sum of
forces on strips.

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3. Total Force As Integral Of Differential Force 
Noting that p
dA is the differential force dF exerted by the fluid on the
differential strip of area dA
at depth h we see
that
the total force is the integral of the differential force:

Remark that Eq. [3.1] also applies if the pressure is constant throughout the plane region:
where A is the area of the plane region.
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4. Practical Treatment 
In practice, Eq. [3.1] allows us to
find the force exerted by a fluid on a plane region by simply considering
differential
elements and forming the appropriate integral, without going thru all the steps
that lead to that equation. This aspect is
similar to that of work done by a force as discussed in Section
12.8 Part 2. For example, let's again find the force F in
the problem posed in Part 2, this time using the
practical treatment.
Example 4.1
Solution
Fig. 4.1
Total force is integral of differential
force.

EOS
Note that the differential rectangle is horizontal. For a
vertical differential rectangle, the pressure isn't the same from 1
point on it to another at a different depth, so we can't use the formula (force
on rectangle) = (pressure) x (area of
rectangle), thus the vertical rectangle can't be used. Remark that for it, the
variable would be L
and its width would be
dL.
As shown in Fig. 4.2, a section of a dike has length of 100
m, vertical height of 20 m, and slanted surface inclining at an
angle of 30^{o}
from the vertical and holding back the water. The water comes up to the top of
the dike. Find the total force
exerted by the water on the slanted surface.
Fig. 4.2
A Section Of The Dike. 
Solution
Fig. 4.3
A Section Of The Dike. 
Refer to Fig. 4.3. Let the vertical haxis be directed downward and have origin at
the top of the dike. Let h
be an
arbitrary point in [0, 20]. A horizontal strip at h has width dh.
The corresponding rectangle on the slanted surface has
length 100 and width say dw
equal to the hypotenuse of the little solid black right triangle T, whose vertical side equals
dh and, by similar triangles,
top angle equals 30^{o}.
In T, as cos 30^{o} = dh/dw we have dw = dh/cos 30^{o}. The area dA of
the rectangle is:
EOS
Fig. 4.4
A Section Of The Dike. 
This approach yields the same answer. Remark that the depth
of the water isn't measured along the slanted side. Of
course it's as always measured along a vertical axis.
Problems & Solutions 
1. A rectangular tank has a height of 2 m and a
bottom measuring 4 m long by 3 m wide and is filled with water. Find the
force exerted by the water on the bottom
of the tank.
The pressure of the water on the bottom of the tank is (1
ton/m^{3})(2
m) = 2 ton/m^{2}.
The force exerted by the water on the
bottom of the tank is F
= (pressure) x (area of bottom) = (2 ton/m^{2})(4 m)(3 m) = 24 tons.
Note
Since the bottom of the tank is horizontal, the pressure is
the same everywhere on it, so there's no need to do integration.
Let's do integration here to see if it yields the same answer.
The above figure represents the bottom of the tank. Let the
horizontal laxis be
along a side of length 4, directed
rightward, and with origin at the left corner. Let l be any point in [0, 4]. A strip at l is a rectangle with length 3
and width
dl. The water pressure on it
is the same everywhere on it and is (1 ton/m^{3})(2 m) = 2 ton/m^{2}, and the water
force on it is
dF = (2 ton/m^{2})(3 m)(dl m) = 6 dl tons. The force exerted by the water on the
bottom of the tank is:
Indeed integration yields the same answer. Don't use it if you're not asked to do so; you'll waste your time if you do.
2. A sheet of metal has the shape of a semidisk of
radius 3 m. It's immersed in water with its curved edge down so that
its straight edge is level with the
surface of the water. The density of water is 1000 kg/m^{3}. Find the
force exerted by
the water on one side of the sheet.
(As a matter of fact, the force on the other side is equal in magnitude but
opposite
in direction. That's why the sheet
doesn't move.)
Solution
3. A circular plate of radius 2 dm is suspended vertically
in water such that its highest point is at a depth of 4 dm. The
density of water is 1 kg/dm^{3}. Find the
force of the water on one side of the plate.
Solution
It follows that:
Therefore:
4. A dam is 25 m high and is in the shape of a region
bounded by the curves y
= x^{2} and y = 25. The density of water
is
1 ton/m^{3}. Find the total force exerted
by the water on the dam when the water comes up to the top of the dam.
Solution
5. A swimming pool 30 m long and 10 m wide has a
sloping plane bottom so that the depth of the pool at 1 end is 1 m and
at the other end is 3 m. Find the
total force exerted on the bottom of the pool if the pool is full of water.
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