## Physics 8 Lesson 8: Liquid pressure- Connecting vessels

## 1. Theoretical summary

### 1.1. The existence of liquid pressure

– Due to its weight, the liquid exerts pressure in all directions on the bottom of the vessel, the walls and the objects inside it.

*Eg:*Divers when diving in the deep sea must wear a diving suit that can withstand the high pressure caused by the pressure of the sea water above.

### 1.2. Formula for calculating liquid pressure

– Formula: p = dh

In which:

- h is the height of the liquid column (m)
- d is the specific gravity of the liquid (N/m
^{3}) - p is the bottom pressure of the liquid column (N/m2 or Pa)

– In a liquid at rest, the pressure at points on the same horizontal plane (with the same height h) has the same magnitude.

**Note:**

– If the tank contains two insoluble liquids, the pressure at a point at the bottom of the vessel is calculated by the formula: p = d_{first}.H_{first} + d_{2}.H_{2}

In which:

- H
_{first}and h_{2}are the heights of the first and second liquid columns. - d
_{first }and d_{2 liters}is the specific gravity of the first and second liquid columns.

### 1.3. Interconnecting vases

– Interconnected flask is a vessel consisting of two or more branches of any shape, the mouth part is open to the air, the bottom part is connected to each other.

– In an interconnected vessel containing the same stationary liquid, the free surfaces of the liquid in different branches are all at the same height (regardless of the shape of the branches).

**– Note: **

- One of the basic applications of communication vessels and the transfer of pressure in fluids is in hydraulic machines.
- When a force f is applied to a small piston of area s, this force exerts pressure on the liquid. This pressure is transmitted intact by the liquid in all directions to the large piston of area S and causes a lift force F on this piston:

\(\begin{array}{l} F = pS = \frac{{fS}}{s}\\ \Rightarrow \frac{F}{f} = \frac{S}{s} \end{array} \)

## 2. Illustrated exercise

### 2.1. Form 1: Determine the depth of the object at two points in time

A submarine is moving under the sea. The pressure gauge placed on the outside of the hull shows the pressure of 2020000 N/m^{2}. After a while the barometer reads 860000 N/m^{2}. Calculate the depth of the submarine at the above two times knowing the specific gravity of sea water is 10300 N/m^{2}.

**Solution guide**

Apply the formula: p = dh

We have: h = pd

Depth of the submarine at the moment before emerging: h_{first }= p_{first}d = 2020000/10300 196 m

Submarine depth at time after surface: h_{2 }= p_{2}d = 860000/10300 83.5 m

### 2.2. Form 2: Determine the pressure of water

A bucket 1.2 m high is filled with water. Calculate the pressure of the water on the bottom of the barrel and up a point 0.4 m from the bottom of the barrel.

**Solution guide**

The pressure acting on the bottom of the tank is:

p = dh_{first} = 10000.1.2 = 12000 N/m^{2}

The pressure acting on a point 0.4 m from the bottom of the barrel is:

p = dh_{2} = 10000.(1.2 – 0.4) = 8000 N/m^{2}

## 3. Practice

### 3.1. Essay exercises

**Question 1: **A submarine is moving under the sea. The pressure gauge placed on the outside of the hull shows a pressure of 2.02.10^{6 }N/m^{2}.

a) Did the ship rise or fall? Why is it so confirmed?

b) Calculate the depth of the submarine at these two times. Given that the specific gravity of sea water is 10300 N/m^{2}.

**Verse 2:** An interconnected vessel holds sea water. People add gasoline to a branch. The two open surfaces in the two branches differ by 18 mm. Calculate the height of the fuel column. Given that the specific gravity of sea water is 10300 N/m .^{3} and that of gasoline is 7000 N/cm^{3}.

**Question 3: **Why is it that when we are quiet we always feel chest tightness and the deeper the silence, the more the feeling of chest tightness increases?

**Question 4: **A ship has a hole in it at a depth of 2.8 m. A patch is placed against the hole from the inside. What is the minimum force required to hold the patch if the hole is 150 cm . wide?^{2} and the specific gravity of water is 2 N/m^{3}.

### 3.2. Multiple choice exercises

**Question 1:** A cylindrical glass tube containing liquid is being placed vertically. If the tube is tilted so that the liquid does not flow out of the tube, the liquid pressure exerted at the bottom of the vessel:

A. increase. B. decrease. C. unchanged. D. zero.

**Verse 2:** The two vessels have the same cross-section. The first vessel contains a liquid of specific gravity d ._{first}, height h_{first}; The first vessel contains a liquid of specific gravity d ._{2 }= 1.5d_{first}, height h_{2} = 0.6h_{first}. If the liquid pressure acting on the bottom of tank 1 is called p_{first}, to the bottom of the second jar is p_{2} then:

A. p_{2 }= 3p_{first}. B. p_{2} = 0.9p_{first}. C. p_{2 }= 9p_{first }D. p_{2 }= 0.4p_{first}.

**Question 3:** Which of the following is true about fluid pressure?

A. Liquids exert pressure in all directions.

B. The pressure acting on the wall of the vessel does not depend on the area being pressed.

C. The pressure exerted by the weight of the liquid on a point is inversely proportional to the depth.

D. If the same depth, the pressure is the same in all different liquids.

**Question 4:** The formula for calculating liquid pressure is:

A. p = d/h B. p = dh C. p = dV D. p = h/d

## 4. Conclusion

Through this lesson, students will be acquainted with the knowledge related to the fluid pressure of the connecting vessel along with related exercises at many levels from easy to difficult…, they need to understand:

- Know that liquids not only exert pressure on the walls of the vessel, the bottom of the vessel, and the objects in the liquid.
- Build the formula to calculate the liquid pressure through the pressure calculation formula.
- State the common sense principle and use it to explain some common phenomena.

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