Cap.V. Fluid statics.

• Solved problems.
• Exercises.
• Self-assessment test.

Solved problems.

🔓 Solved problems - Presiunea.

5.1. A 400 g parallelepiped body has the following dimensions:

L = 0,003 hm

l = 15 cm

h = 100 mm

Find out the three pressures the body exerts on a surface.

Solution:

We write the problem data and turn it into SI.
m = 400 g = 0,4 kg
G = m ∙ g = 0,4 kg ∙ 10 N/kg = 4 N
L = 0,003 hm = 0,3 m
l = 15 cm = 0,15 m
h = 100 mm = 0,1 m.

We apply the pressure formula and replace the problem data:

5.2. A man hits a nail with a force of 600 N in the wall, which makes an angle α = 30° with the wall. The tip of the nail is 2 cm². Find out the human pressure on the wall.

Solution:

We write the problem data and turn it into SI. F = 600 N
S = 2 cm² = 2/10000 m²

Standard:
1 cm : 100 N

We calculate the modulus of normal force on the wall

F_n = F ∙ cosα = 600 ∙ 1/2 = 300 N

We write the pressure formula and replace the problem data:

5.3. A parallelepiped aluminum body with a density of 2700 kg/m³ and a height of 20 cm is placed on an inclined plane with an angle of 60° (cosine 60° = 0,5). Calculates the pressure exerted by the body on the inclined plane.

Solution:

We write the problem data and turn it into SI. ρ = 2700 kg/m₃
h = 20 cm = 0,2 m
α = 60°
p = ?

We write the pressure formula and calculate the physical parameter values in the formula:

🔓 Solved problems - Hydrostatic pressure.

5.4. Determine what pressure exerts water with a density of 1000 kg/m³ at a depth of 100 m?

Solution:

We write the problem data:
ρ = 1000 kg/m³
h = 100 m
p = ?

We write the hydrostatic pressure formula:

p = ρ ∙ g ∙ h = 1000 kg/m³ ∙ 10 N/kg ∙ 100 m = 1.000.000 Pa

5.5. The Dead Sea of Jordan has the lowest water surface in the world (427 m below sea level). It is the saltiest water in the world, 9,6 times saltier than the planet's ocean. The extremely salty content is very unfavorable to life and hence the name Dead Sea. Other records: water with the highest concentration of bromine in the world and the deepest hypersaline lake on earth (306 meters).

Is required: At what depth in the Dead Sea with a density of 1240 kg/m³, the water exerts a pressure of 3.720.000 Pa?

Solution:

We write the problem data:
ρ = 1240 kg/m³
h = ?
p = 3.720.000 Pa

We write the hydrostatic pressure formula and remove the unknown, h:

p = ρ ∙ g ∙ h

5.6. A cylindrical beaker with a base area S = 20 cm² contains mercury up to a height of 10 cm. Put 400 g of water over the mercury. The density of mercury is 13600 kg/m³, and that of water is 1000 kg/m³.

Calculate: a) The height of the water column, knowing that it is immiscible with mercury (do not mix). b) The pressure exerted by the two liquids on the bottom of the vessel.

Solution:

Write the problem data and turn it into SI:
S = 20 cm² = 0,002 m²
h₁ = 10 cm = 0,1 m
ρ₁ = 13600 kg/m³
ρ₂ = 1000 kg/m³
m₂ = 400 g = 0,4 kg
h₂ = ?
p = ?

With the density formula, we find the volume of water added:

The liquids taking the shape of the vessel, have the same area of the base as of the glass, so the volume of water is:

V₂ = S ∙ h₂

We calculate the hydrostatic pressures of the two liquids:

p₁ = ρ₁ ∙ g ∙ h₁ = 13.600 kg/m³ ∙ 10 N/kg ∙ 0,1 m = 1.3600 Pa

p₂ = ρ₂ ∙ g ∙ h₂ = 1.000 kg/m³ ∙ 10 N/kg ∙ 0,2 m = 2.000 Pa

p = p₁ + p₂ = 13.600 Pa + 2000 Pa = 15.600 Pa

5.7. In a long-necked bottle, put water up to the middle of the bottle. Then turn the bottle over so that it rests on the stopper. What will the water pressure be like when the bottle is on the bottom compared to the water pressure when the bottle is on the stopper?

Solution:

When the bottle is on the stopper, the height of the water is higher due to the thinner neck than when it is on the bottom. So the water pressure is lower when it is on the bottom than on the stopper. The hydrostatic pressure, for the same liquid, does not depend on the area of the bottom of the vessel, but only on the height of the column of liquid in the vessel.

5.8. The liquid in a pipette flows only if it is lightly pressed on the plastic tube. Why does the liquid not flow on its own when the pipette is upright?

Solution:

The air pressure above the liquid in the pipette (p) is lower than the atmospheric pressure (p₀) at the tip of the pipette. Atmospheric pressure is equal to the pressure of the air in the pipette plus the pressure of the liquid in the pipette (ρ ∙ g ∙ h).

p₀ = p + ρ ∙ g ∙ h > p

🔓 Probleme rezolvate - Atmospheric pressure.

5.9. Calculate the compressive force that the atmospheric air exerts on a window with a length of 1 m and a width of 60 cm, knowing that the air pressure is 10⁵ Pa.

Solution:

We write the problem data: L = 1 m
l = 60 cm = 0,6 m
p = 100.000 Pa
F = ?

We write the pressure formula and remove the unknown, F:

We calculate the window area and replace the data:
S = L ∙ l = 1 m ∙ 0,6 m = 0,6 m²
F = p ∙ S = 100.000 Pa ∙ 0,6 m² = 60.000 N

We notice an enormous force from the air on the glass and yet the glass does not break, because the air acts both externally and internally with the same force, having a resulting force equal to zero.

5.10. How long should Torricelli's tube be if we used water instead of mercury? The normal atmospheric pressure is 100.000 Pa and the water density is 1000 kg/m³.

Solution:

We write the problem data:
h_{col. apă} = ? m
p = 100.000 Pa
ρ = 1000 kg/m³

We write the atmospheric pressure formula and remove the unknown, h_{water col.}:

5.11. A glass of 400 cm³ with a mouth diameter of 6 cm, filled with water, is covered with a cardboard and overturned. The density of the water is 1000 kg/m³.

Is required:

a) What force does water exert on the cardboard?

b) What is the compressive force exerted by the atmospheric air on the cardboard, knowing the air pressure of 100.000 Pa?

Solution:

We write down the problem data and turn it into SI:
F_{water col.} = ?
F_{air col.} = ?
p = 100.000 Pa
ρ = 1000 kg/m³
V = 400 cm³ = 400 ∙ (0,01m)³ = 0,0004 m³
D = 6 cm
R_circle = D/2 = 3 cm = 0,03 m

The force of the water is the weight of the water in the glass:

F_{water col.} = G_{water col.} = m ∙ g = ρ_water ∙ V ∙ g = 1000 kg/m³ ∙ 0,0004 m³ ∙ 10 N/kg = 4 N

The force of the air is found from the pressure formula:

🔓 Solved problems - Pascal's law.

5.12. A 1000 kg machine is lifted using a hydraulic lift. Knowing the radii of the pistons of 10 cm and 50 cm, respectively, calculate:

a) the force necessary to lift the machine;

b) the stroke of the large piston, knowing the stroke of the small piston of 30 cm.

Solution:

We write down the data of the problem and turn it into SI:
m = 1000 kg
R₁ = 10 cm = 0,1 m
R₂ = 50 cm = 0,5 m
h₁ = 30 cm = 0,3 m
F₁ = ?
h₂ = ?

We calculate the weight of the machine, which is the force exerted by the liquid on the large piston, ie F₂:
F₂ = G = m ∙ g = 1.000 kg ∙ 10 N/kg = 10.000 N

We write the equation of the operating principle of the hydraulic press:

We calculate the cross-sectional areas of the two cylinders, which are circles:

S₁ = π ∙ R₁² = π ∙ (0,1 m)² = π ∙ 0,01 m²
S₂ = π ∙ R₂² = π ∙ (0,5 m)² = π ∙ 0,25 m²

We extract the unknown, F₁ from the equation of the operating principle of the hydraulic press and replace the data:

b) According to Pascal's law, the volume of liquid in the small cylinder is equal to the volume of liquid in the large cylinder:

5.13. Into a U tube is poured water with a density of 1000 kg/m₃ and then a column of oil with a height h₁ = 10 cm is poured into the left branch. As the oil is immiscible with water, there is an unevenness in the left branch of 2 cm. Determines the density of the oil used. Give the air pressure, p₀ = 10.000 Pa

Solution:

Write down the problem data and turn it into SI:
ρ_water = 1000 kg/m³
h₁ = 10 cm = 0,1 m (height of the oil column)
Δh = 2 cm = 0,02 m (unevenness of liquids in the two branches) p₀ = 10.000 Pa
ρ_oil = ?

The external pressure exerted by the oil, p_A, is transmitted entirely to the water in the right branch, p_B, according to Pascal's Law:
p_A = p_B

We calculate the two pressures, taking into account the atmospheric pressure exerted on the two branches of the U tube:
Δh = h₁ – h₂ = 0,02 m
h₂ = h₁ – Δh = 0,1 - 0,02 = 0,08 m (height of the water column above level B, which is the same as level A)
p_A = p₀ + ρ_oil ∙ g ∙ h₁ = 100.000 + ρ_oil ∙ 10 ∙ 0,1 = 100.000 + ρ_oil
p_B = p₀ + ρ_water ∙ g ∙ h₂ = 100.000 + 1000 ∙ 10 ∙ 0,08 = 100.000 + 800 = 100.800
p_A = p_B
100.000 + ρ_oil = 100.800
ρ_oil = 100.800 – 100.000 = 800 kg/m³

🔓 Solved problems - Law of Archimedes.

5.14. A cube made of cork, with a side of 0,3 dm and a density of 200 kg/m³ is immersed in water, which has a density of 1000 kg/m³.

Is required:

a) The value of the Archimedean force.

b) The value of the resulting force acting on the body in water. What is this force called?

c) What is the height of the portion of the cube that is under water?

Solution:

We write down the problem data and turn it into SI: l = 0,3 dm = 0,03 m
ρ_cork = 200 kg/m³
ρ_water = 1000 kg/m³
F_A = ?
R = ?
h = ?

a) We calculate the volume of the cube: V_cube = l³ = (0,03 m)³ = 0,000027 m³

We apply the formula of the Archimedean force in the Law of Archimedes:
F_A = ρ_water ∙ V_cube ∙ g = 1000 kg/m³ ∙ 0,000027 m³ ∙ 10 N/kg = 0,27 N

b) We calculate the body weight: G = m ∙ g = ρ_cork ∙ V_cube ∙ g = 200 kg/m³ ∙ 0,000027 m³ ∙ 10 N/kg = 0,054 N
Because | F_A | > | G | a resultant force appears, which acts on the body vertically, upwards, called the ascending force (F_a) which causes the partial body to come out of the liquid.
| F_a | = | F_A - G | = 0,27 – 0,054 = 0,216 N

c)The submerged portion displaces a volume of liquid equal to the weight of the body:
| F_A | = | G |

5.15. A body weighs 800 g in the air, and when immersed in glycerin it weighs 600 g. Knowing the density of glycerin of 1260 kg/m³, find out:

a) Body volume.

b) The resulting force. What does the body do immersed in this liquid?

Solution:

We write down the problem data and turn it into SI:
m = 800 g = 0,8 kg
m_apparent = 600 g = 0,6 kg
ρ_glycerin = 1260 kg/m³
a) V _body = ?
b) R = ?

We calculate body weight and its apparent weight:
G = m ∙ g = 0,8 kg ∙ 10 N/kg = 8 N
G_ap = m_ap ∙ g = 0,6 kg ∙ 10 N/kg = 6 N

We calculate the Archimedean force with its two formulas:
F_A = G – G_ap = 8 N – 6 N = 2 N
F_A = ρ_glycerin ∙ V_body ∙ g = 1260 ∙ V_body ∙ 10
2 = 1260 ∙ V_body ∙ 10
V_body = 2/12600 = 0,0001 m³

b)Because | F_A | < | G | a resultant force appears, which acts on the body vertically, downwards, called the apparent weight (G_ap) which causes the body to sink into the liquid.
| G_ap | = | G – F_A | = 8 N – 2 N = 6 N

5.16. A steel sphere of 500 cm³ and a density of 7800 kg/m³ is suspended by a dynamometer in the air and then immersed in water (with a density of 1000 kg/m³). What does the dynamometer indicate when the sphere is in the air or in the water?

Solution:

We write down the problem data and turn it into SI:
V_c = 500 cm³ = 0,0005 m³
ρ_c = 7800 kg/m³
ρ_water = 1000 kg/m³
G, G_ap = ?

We calculate the mass of the sphere:
m = ρ_c ∙ V_c = 7800 kg/m³ ∙ 0,0005 m³ = 3,9 kg

We calculate the weight of the sphere:
G = m ∙ g = 3,9 kg ∙ 10 N/kg = 39 N

We calculate the Archimedean force:
F_A = ρ_water ∙ V_c ∙ g = 1000 kg/m³ ∙ 0,0005 m³ ∙ 10 N/kg = 5 N

We calculate the apparent weight (when the sphere is submerged in water):
| F_A | = | G – G_ap |
| G_ap | = | G – F_A | = 39 N – 5 N = 34 N

5.17. An iceberg floating in ocean water (with a density of ρ = 1010 kg/m³), has the upper part, the one above water level, with a volume V₀ = 600 m³. Knowing the density of ice ρ₀ = 920 kg/m³, find the total volume, V, of the iceberg.

Solution:

We write down the problem data:
V₀ = 600 m³
ρ = 1010 kg/m³
ρ₀ = 920 kg/m³
V = ?

a) We calculate the volume of the iceberg submerged in water:
V_submerged = V - V₀

The submerged portion displaces a volume of liquid equal to the weight of the body:
| F_A | = | G |
ρ ∙ V_submerged ∙ g = ρ₀ ∙ V ∙ g
ρ ∙ V_submerged = ρ₀ ∙ V
ρ ∙ (V-V₀) = ρ₀ ∙ V
ρ ∙ V - ρ ∙ V₀ = ρ₀ ∙ V
V ∙ (ρ - ρ₀) = ρ ∙ V₀

Exercises.

🔐 Exercises - Law of Arhimedes.

5.18. What happens to a ship when it crosses the Danube into the Black Sea?

🔐 Recapitulative exercises - Pressure.

5.19. Complete the following statements.

a) The pressure is a physical parameter equal to the ratio between the pressing force and …………… the surface on which the pressing force is distributed.

b) Atmospheric pressure is the pressure ……………………… due ………………………

c) The static pressure inside a liquid is called the ……………………………… pressure, being due to ……………………………… the liquid.

d) In two or more communicating vessels the liquid rises to …………………………………….

e) The external pressure exerted on a fluid is transmitted throughout the whole …………………………………… fluid.

f) On a body immersed in a fluid, a force is exerted on the …………………………………… direction oriented ………………, called the force …………, equal to …………… the volume of fluid displaced by the body.

5.20. Recognizes what the following applications are based on:

a) Densimeter

b) Level indicator of a water (oil) tank

c) Car foot brake

d) Thinner footprints on snow with skis than those left with boots

e) Hot air balloons

f) Vacuum pump

g) Car lift

h) The altimeter

i) The sprinkler

j) The dental chair

k) The submarine

5.21. Complete the following statements regarding the archimedical force, F_A:

a) It has …………………… direction, oriented …………………… and the point of application in the weight center of the …………………… liquid volume.

b) Equal in module with displaced fluid ………………………….

c) …… depends on the weight and shape of the body, the depth of immersion, the height of the liquid in the vessel.

d) Its value depends on the ……………………… of the liquid in which we immerse the body and ……………… of the body.

5.22. What are the two factors on which atmospheric pressure depends?

5.23. Draw a hydraulic press and name its components.

5.24. How did Torricelli measure atmospheric pressure?

5.25. The pressure exerted on an area of 1 m² is 10 Pa. What is the angle that the direction of the force makes with the normal to the respective surface, knowing that the pressing force is 20 N?

5.26. A steel cylinder of 400 cm³, with a density of 7800 kg/m³ has a diameter of 5 cm. What pressure does it exert on a surface?

5.27. Glycerin with a density of 1260 kg/m³ is poured into a U tube and then water, with a density of 1000 kg/m³ is added. The glycerin column is 15 cm long. How much is the difference between the two liquids in the branches of the tube?

5.28. The diameters of the pistons of a hydraulic press are 20 mm and 8 cm, respectively. A force of 40 N is exerted on the small piston, with a stroke of 30 cm. Find out the force that lifts the big piston and its stroke.

5.29. A 60 g egg with a volume of 60 cm³ is placed in a brine (salt solution). The sunken volume of the egg is 50 cm³. What is the density of brine?

5.30. A wooden cube with a side of 4 cm and a density of 600 kg/m³ is placed in water, which has a density of 1000 kg/m³. Find out:

a) The value of the Archimedean force.

b) The value of the resulting force acting on the body in water. What is this force called?

c) What is the height of the portion of the cube that is under water?

5.31. How tall should Torricelli's tube be if we used oil with a density of 800 kg/m³ instead of mercury, knowing the normal atmospheric pressure of 100.000 Pa?

Self-assessment test.

📝 TEST1: Self-assessment test - Pressure

5.32. Complete the following statements: -1p

a) Atmospheric pressure is the pressure exerted by .....................................

b) The external pressure exerted on a fluid is transmitted throughout the whole ........ of the fluid.

c) Hydrostatic pressure is due ...................................................................

d) An ............................ oriented force, called archimedean force is exerted on a body immersed in a fluid.

5.33. What principle or law are based on the following applications: -1p

a) The excavator

b) Sink siphon

c) Floating ships

d) Cutting sharp objects.

5.34. Draw a hydraulic press and name its components. -1p

5.35. How did physicist Torricelli measure atmospheric pressure? -1p

5.36. How do we experimentally demonstrate the pressure of the air from the bottom up? - 1p

5.37. What is the depth inside a vessel of alcohol, with a density of 800 kg/m³, where a pressure of 200 Pa is exerted? -1p

5.38. A body weighs 120 g in the air, and when immersed in alcohol it weighs 110 g. Knowing the alcohol density of 800 kg/m³, find out:

a) Body volume. -1p

b) The resulting force. What does the body do immersed in this liquid? Represents the forces acting on the body. -1p

Ex officio - 2p