In the previous section, we saw that, pressure always acts perpendicular to the surface. In this section, we will see a few more details about pressure
We know that:
• If a body is kept in the interior of a fluid, it will be compressed from all directions
• That means, the body will experience pressure from all directions
• We want to know the relation between those pressures
• That is., we want the answers to these questions:
♦ Are all those pressures equal in magnitude?
♦ Do some of those pressures have greater magnitudes than others?
• The following analysis will help us to find the answers. It is written in 14 steps
1. Consider the right angled wedge shown in fig.10.5(a) below
♦ This wedge is situated at the interior of a fluid
♦ The fluid is at rest
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| Fig.10.5 |
2. The forces acting on the wedge are:
(i) Yellow forces (FY) on the vertical sides
(ii) Blue force (FB) on the rear side
(iii) Green force (FG) on the bottom side
(iv) Red force (FR) on the sloping side
3. In fig.a, some sides of the wedge are hidden from view
• So we give a bit of transparency to the sides. This is shown in fig.b
• Now the sides can be named:
♦ The two vertical sides are: ABC and DEF
✰ Note that, DEF is not visible in fig.a
♦ The rear side is: ABED
✰ Note that, this rear side is not visible in fig.a
♦ The bottom side is BCFE
✰ Note that, this bottom side is not visible in fig.a
♦ The sloping side is ACFD
4. There is a total of four forces. They were mentioned in (2)
• For our present discussion we consider only three forces:
♦ FR which acts on ACFD
♦ FB which acts on ABED
♦ FG which acts on BCFE
5. Let us give specific names to the areas:
♦ The red force acts on ACFD
✰ So we will denote the 'area of ACFD' as: AR
♦ The blue force acts on ABED
✰ So we will denote the 'area of ABED' as: AB
♦ The green force acts on BCFE
✰ So we will denote the 'area of BCFE' as: AG
6. Resolving the forces into components:
• We see that FB is horizontal and FG is vertical
♦ So these two forces need not be resolved into their components
• We will resolve FR into it's 'horizontal and vertical components'
♦ For that, we need the 2D view of the wedge. This is shown in fig.10.6 below:
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| Fig.10.6 |
• The angle of the wedge is 𝛳
♦ So the horizontal component of FR will be FR cos 𝛳
♦ Also, the vertical component of FR will be FR sin 𝛳
7. Next, we apply the conditions of equilibrium:
(i) Since the wedge is in equilibrium, the horizontal components must cancel each other
So we get: FB = FR cos 𝛳
(ii) Since the wedge is in equilibrium, the vertical components must cancel each other
So we get: FG = FR sin 𝛳
8. Now we apply the principles of trigonometry to the areas of the wedge. We get:
(i) AB = AR cos 𝛳
(ii) AG = AR sin 𝛳
9. Let us divide both sides of 7(i) by AB
♦ We get: $\mathbf\small{\rm{\frac{F_B}{A_B}=\frac{F_R \cos \theta}{A_B}}}$
• But from 8(i), we have: AB = AR cos 𝛳
• So we can replace the AB on the right side denominator. We get:
$\mathbf\small{\rm{\frac{F_B}{A_B}=\frac{F_R \cos \theta}{A_R \cos \theta}}}$
$\mathbf\small{\rm{\Rightarrow \frac{F_B}{A_B}=\frac{F_R}{A_R}}}$
10. Let us divide both sides of 7(ii) by AG
♦ We get: $\mathbf\small{\rm{\frac{F_G}{A_G}=\frac{F_R \sin \theta}{A_G}}}$
• But from 8(ii), we have: AG = AR sin 𝛳
• So we can replace the AG on the right side denominator. We get:
$\mathbf\small{\rm{\frac{F_G}{A_G}=\frac{F_R \sin \theta}{A_R \sin \theta}}}$
$\mathbf\small{\rm{\Rightarrow \frac{F_G}{A_G}=\frac{F_R}{A_R}}}$
11.From (9) and (10), we get:
$\mathbf\small{\rm{\frac{F_B}{A_B}=\frac{F_R}{A_R}=\frac{F_G}{A_G}}}$
⇒ PB = PR = PG
■ That means, pressure experienced by the three sides are the same
12. We chose a right angled wedge (fig.10.5) for our analysis
• By choosing another suitable shape, we can include the yellow forces also in our calculations
• But then, the red force will have to be resolved into three components. It is a 3D problem
• However, from such a 3D problem also, we will get the same result:
■ The pressure experienced by all sides of the wedge is the same
13. In the above step (12), we concluded that:
■ The pressure experienced by all sides of the wedge is the same
• We arrived at this result by analyzing the wedge in fig.10.5 above
• But we need to recognize some important points about this wedge
• They can be written in 4 steps
(i) The wedge in fig.5.10 is very very small
♦ We were doing all the analysis on an enlarged shape
(ii) The edges are very close to each other:
♦ The edge AD is very close to edge BE
♦ Also, edge DE is very close to edge AB
• In effect, the side ABED is very small
♦ That is., the area AB is very small
(iii) It is compulsory to do the analysis on such a small area because:
• If the area AB is large, FB will not be uniform
♦ FB will be having smaller magnitudes near the top edge AD
♦ FB will be having larger magnitudes near the bottom edge BE
(We will see the reason for this variation, later in this section)
• If FB varies in such a manner, the ratio ($\mathbf\small{\rm{P_B = \frac{F_B}{A_B}}}$) is meaning less
• If the area is very small, FB can be assumed to be uniform
♦ Then the ratio ($\mathbf\small{\rm{P_B = \frac{F_B}{A_B}}}$) will be meaningful
✰ It will give the actual pressure on the area AB
(iv) In a similar way, all sides of the wedge must be very small
• This can be achieved easily if the 'wedge as a whole' is very small
14. Now we can answer the questions asked at the beginning of this section. The questions were:
♦ Are all those pressures equal in magnitude?
♦ Do some of those pressures have greater magnitudes than others?
• The answer is that:
♦ If the object is very very small, all pressures will be of the same magnitude
♦ If the object is large, some of the pressures will be larger in magnitude than the others
• In the above discussion, we concluded that:
♦ If the object is large, some of the pressures will be larger in magnitude than the others
• That means, a large object will be experiencing different pressures
♦ Some portions of that body will experience greater pressures
♦ Some other portions will experience lesser pressures
• We need to find the reason for such a variation
• Also, we need to find a method to calculate pressure at the various portions of a body
• The following analysis will help us to achieve those goals. It is written in 7 steps:
1. In fig.10.7(a) below, a fluid is taken in a container
♦ A point ‘1’ is marked in the interior of the fluid
♦ Another point ‘2’ is marked vertically below ‘1’
♦ The distance between the two points is h
• We want to find the pressures at points 1 and 2
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| Fig.10.7 |
• This cylinder has 3 properties:
(i) The point 1 is at the exact center of the top surface
(ii) The point 2 is at the exact center of the bottom surface
♦ So height of the cylinder will be h
(iii) The area of cross section is A
3. Consider the yellow cylinder in fig.b
• This cylinder has 4 properties:
(i) The top surface coincides with the free surface of the liquid
(ii) The bottom surface rests on top of the red cylinder
(iii) Height of the cylinder is z
(iv) The area of cross section is the same A as that of the red cylinder
4. Consider the green cylinder in fig.b
• This cylinder has 3 properties
(i) The top surface is shown to be irregular
♦ This is to indicate that, the height extends up to the top of the atmosphere
♦ That is., height of the green cylinder is same as the ‘height of atmosphere’
(ii) The bottom surface rests on top of the yellow cylinder
(iii) The area of cross section is the same A as that of the red and yellow cylinders
5. Next, we calculate the force and pressure at the top of the red cylinder. It can be done in 8 steps:
(i) The top surface of the red cylinder is a circular area A
(ii) This area supports the weight of two cylinders: green and yellow
(iii) Let the weight of the green cylinder be Wa
♦ The subscript ‘a’ indicates that, it is the weight due to atmospheric air
(iv) Weight of the yellow cylinder can be calculated as follows:
Weight = mass × g
⇒ Weight = volume × density × g
⇒ Weight = base area × height × density × g
⇒ Weight = A × z × 𝛒 × g = A𝛒gz
♦ Where 𝛒 is the density of the fluid
(v) So the total force on the top of the red cylinder = Wa + A𝛒gz
(vi) So pressure on the top of the red cylinder = $\mathbf\small{\rm{\frac{Force}{Area}=\frac{W_a+A \rho g z}{A}=\frac{W_a}{A}+\rho gz}}$
(vii) But $\mathbf\small{\rm{\frac{W_a}{A}}}$ is the atmospheric pressure Pa
• Also, 'pressure on the top of the red cylinder' is same as 'pressure at point 1'
• So we get:
Pressure at point 1 = Pa + 𝛒gz
(viii) We can use this equation to find the pressure at any point which is at a depth of 'z' from the surface
■ If Pz is the pressure at a depth 'z' from the free surface, we can write:
Eq.10.1: Pz = Pa + 𝛒gz
6. Next, we calculate the force and pressure at the bottom of the red cylinder. It can be done in 8 steps:
(i) The bottom surface of the red cylinder is a circular area A
(ii) This area supports the weight of three cylinders: green, yellow and red
(iii) We know that weight of the green cylinder is Wa
(iv) We calculated the weight of the yellow cylinder as: A𝛒gz
(v) In a similar way, the weight of the red cylinder will be: A𝛒gh
(vi) So the total force at the bottom of the red cylinder = Wa + A𝛒gz + A𝛒gh
(vii) So pressure at the bottom of the red cylinder = $\mathbf\small{\rm{\frac{Force}{Area}=\frac{W_a+A \rho g z+A \rho g h}{A}=\frac{W_a}{A}+\rho gz+\rho gh}}$
• But $\mathbf\small{\rm{\frac{W_a}{A}}}$ is the atmospheric pressure Pa
• Also, 'pressure at the bottom of the red cylinder' is same as 'pressure at point 2'
• So we get:
Pressure at point 2 = Pa + 𝛒gz + 𝛒gh = Pa + 𝛒g(z + h)
(viii) Note that, the equation in (vii) above is a mere extension of Eq.10.1
• This can be explained in 4 steps:
♦ Eq.10.1 gives the pressure at a depth z
♦ Point 2 is at a depth of (z+h)
♦ So, to find the pressure at point 2, we simply replace z by (z+h)
♦ This will give the equation in (vii)
7. Now we know the method to find the pressure at any given depth
• Next, let us calculate the ‘pressure difference’ between point 1 and point 2
• This can be done in 3 steps:
(i) We have: pressure at point 1 = Pa + 𝛒gz
(ii) We have: pressure at point 2 = Pa + 𝛒g(z + h)
(iii) So the difference in pressure = [Pa + 𝛒g(z + h)]-[Pa + 𝛒gz] = 𝛒gh
■ Thus we can write:
Eq.10.2: Pressure difference between two points at a vertical distance of h apart = 𝛒gh
The above Eq.10.2 can be derived by another method also. It can be written in 5 steps:
1. Consider the red cylinder that we saw in fig.10.7(b) above
• It is shown again in fig.10.8 below:
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| Fig.10.8 |
2. This cylinder will be experiencing pressure from all directions
• First we will consider the vertical direction
♦ Let the pressure at the top surface be P1
♦ Let the pressure at the bottom surface be P2
• Then we get:
♦ Force acting on the top surface = F1 = P1 × A
✰ This force acts vertically downwards
♦ Force acting on the bottom surface = F2 = P2 × A
✰ This force acts vertically upwards
3. We know that weight of the cylinder will be acting vertically downwards
• If m is the mass of the cylinder, the weight will be mg
4. So we have three forces in the vertical direction: F1, F2 and mg
Considering equilibrium in the vertical direction, we get:
F1 + mg = F2
⇒ F1 - F2 = mg
⇒ P1 × A - P2 × A = mg
⇒ (P1-P2)A = mg
5. If 𝝆 is the density of the fluid, we can calculate m in just two steps:
(i) Volume of the cylinder = Ah
(ii) Mass = volume × density = Ah𝝆
• So the equation in (4) becomes: (P1-P2)A = Ah𝝆g
• Thus we get: P1 – P2 = 𝝆gh
• This is same as Eq.10.2 that we derived earlier
Now we will see two solved examples
Solved example 10.2
What is the pressure on a swimmer 10 m below the surface of a lake? Take atmospheric pressure to be 1.01 × 105 Pa and g = 10 Nm-2
Solution:
1. The pressure at a depth z is given by Eq.10.1: Pz = Pa + 𝛒gz
• In our present case, z = 10 m
2. Substituting the known values, we get:
P1 = 1.01 × 105 + 1000 × 10 × 10 = 2.01 × 105 Pa
Solved example 10.3
An open vessel contains water and oil in layers. The bottom water layer has a height of 20 cm. The top oil layer has a height of 10 cm. Find the pressure at (i) the interface of the two liquids (b) at the bottom of the vessel. The relative density of oil is 0.9. Take atmospheric pressure to be 1.01 × 105 Pa and g = 9.8 Nm-2
Solution:
• The arrangement is shown in fig.10.9(a) below. We have to find the pressures at points 1 and 2
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| Fig.10.9 |
• The relative density of oil is given to be 0.9
♦ We have: $\mathbf\small{\rm{Relative \;Density\;=\;\frac{Density\;of\;substance}{Density\;of\;water}}}$
♦ Substituting the values, we get: $\mathbf\small{\rm{0.9\;=\;\frac{Density\;of\;oil}{1000\;(kg\;m^{-3})}}}$
♦ Thus we get: Density of oil (𝛒) = 900 kg m-3
Part (a)
1. The pressure at a depth z is given by Eq.10.1: Pz = Pa + 𝛒gz
• In our present case, z = 10 cm = 0.10 m
2. Substituting the known values, we get:
P1 = 1.01 × 105 + 900 × 9.8 × 0.10 = 101882 Pa
Part (b)
1. We have to find the pressure at the bottom of the red cylinder in fig.10.9(b)
2.So we need to consider the weights of the red, yellow and green cylinders
• Thus there will be three terms
3. The pressure at point 2 will be given by: Pz = Pa + 𝛒ogz1 + 𝛒wgz2
• In our present case:
♦ 𝛒o = 900 kg m-3 and z1 = 0.10 m
♦ 𝛒w = 1000 kg m-3 and z2 = 0.20 m
4. Substituting the known values, we get:
P2 = 1.01 × 105 + 900 × 9.8 × 0.10 + 1000 × 9.8 × 0.20 = 103842 Pa
• Eq.10.1 and 10.2 that we saw above, were derived based on pressures in the vertical direction
• We want to know how they are related to the pressures in the horizontal directions
• We will see it in the next section
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