In the previous section, we saw the pressures in the vertical direction. In this section, we will see pressures in the horizontal directions
• Eqs.10.1 and 10.2 that we saw in the previous section, are related to 'pressures in the vertical direction'
• We can relate those equations to the 'pressures in the horizontal directions' also. This can be written in 11 steps:
1. The red cylinder that we saw in fig.10.7 of the previous section, is shown again in fig.10.10(a) below
• We already know how to calculate the pressure at it’s top and bottom points
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| Fig.10.10 |
2. Let us now consider intermediate points along that cylinder
• For that, a small slice is marked in that cylinder
♦ This slice is in the form of a disc
♦ This disc has a very small thickness
♦ It is situated at a depth of z1 from the free surface
3. We have already seen that:
♦ A downward pressure will be acting at the top of the red cylinder
♦ A upward pressure will be acting at the bottom of the red cylinder
4. In a similar way:
♦ A downward pressure will be acting at the top of the disc
♦ A upward pressure will be acting at the bottom of the disc
5. But there is a major difference between (3) and (4). This can be explained in 2 steps:
(i) The height of the disc is very small
(ii) So the two pressures mentioned in (4) will be the same
(iii) The two pressures mentioned in (3) will be different because, there is a large height difference between their points of applications
■ We can write:
Both the pressures mentioned in (4) will be equal to: Pa + 𝛒gz1
6. At the beginning of the previous section, we saw that, if the element is very small, the pressure experienced by it in all directions will be the same
• We proved this using figs.10.5 and 10.6
7. So it is clear that, the disc will be experiencing the pressure of ‘Pa + 𝛒gz1’ in the horizontal directions also
• So in fig.10.10(a) above, two horizontal arrows are drawn on either sides of this disc
♦ Note that, there will be horizontal arrows all around the disc. We have drawn only two of them
♦ Those arrows will try to compress the disc
8. In a similar way we can draw horizontal arrows for another disc at a depth z2
♦ The horizontal arrows at z2 will be larger than those at z1
♦ This is because, as depth increases, pressure also increases
9. Assuming the red cylinder to be made up of a large number of discs, we can draw a correspondingly large number of horizontal arrows
♦ This is shown in fig.10.10(b)
♦ We can see that, as depth increases, the size of arrows increases
10. We can find the magnitude of any arrow that we see in fig.b
• All we need are:
♦ the depth (z) of that arrow from the free surface
♦ the density of the fluid
■ Then the magnitude of that arrow will be equal to: Pa + 𝛒gz
11. The walls of the container will also be experiencing the same horizontal forces
• That means, the fluid will be trying to push the walls outwards
♦ The upper portions of the walls will be experiencing lower pressures
♦ The bottom portions of the walls will be experiencing higher pressures
Now we will see some solved examples
Solved example 10.4
A submarine is at a depth of 1000 m below the surface of the ocean. The interior of the submarine is maintained at sea level atmospheric pressure. What is the force acting on a window of the submarine, if the area of the window is 400 sq.cm? The relative density of sea water is 1.03. Take sea level atmospheric pressure to be 1.01 × 105 Pa and g to be 10 m s-2.
Solution:
1. The pressure at a depth of z is given by: Pz = Pa + 𝛒gz
2. This Pz acts equally in all directions
♦ So the ocean will exert an inward pressure of Pz on the window
♦ So the inward force experienced by the window = (Pz × area) = 0.04Pz
3. But there is a pressure of Pa inside the submarine
♦ The air inside, will exert an outward pressure of Pa on the window
♦ So the outward force experienced by the window = (Pa × area) = 0.04Pa
4. So net force = (0.04Pz - 0.04Pa)
= 0.04[Pz-Pa] = 0.04[(Pa + 𝛒gz) - Pa] = 0.04𝛒gz = 0.04 × 1.03 × 103 × 10 × 1000 = 412000 N
Solved example 10.5
A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2 . The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.
Solution:
1. The arrangement is shown in fig.10.11 below:
2. The pressure at the bottom of the right side compartment = Pright(4 m) = Pa + 𝛒acid × g × 4
• So the force experienced by the door from the right side = (Pright(4 m) × area) = 0.002 × Pright(4 m)
3. The pressure at the bottom of the left side compartment = Pleft(4 m) = Pa + 𝛒water × g × 4
• So the force experienced by the door from the left side = (Pleft(4 m) × area) = 0.002 × Pleft(4 m)
3. So net force = 0.002[Pright(4 m) - Pleft(4 m)]
= 0.002[(Pa + 𝛒acid × g × 4) - (Pa + 𝛒water × g × 4)]
= 0.002 × g × 4[𝛒acid - 𝛒water]
= 0.002 × g × 4 × 103[1.7 - 1] = 54.88 N
4. This net force will be acting from right to left
So this force will push the door towards the left
We will have to provide a force of 54.88 N from right to left so that, the door remains closed
5. Note that, the base area of the tank is not coming into the calculations
• That means:
♦ We can provide the partition in such a way that, the acid compartment is small
♦ Even then, if the depth is 4 m in both compartments, we will have to provide the same force of 54.88 N
In the next section, we will see hydrostatic paradox
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