Pressure and hydrostatic pressure

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 5-1Core

   A skier's weight is $720\,\text{N}$. It is spread evenly over the bases of the skis, whose total area is $0.30\,\text{m}^2$.

   How large is the pressure that the skis exert on the snow?

   PaCheck
   Answer

   $2400\,\text{Pa}$

   Show the solution

   Given

   the skier's weight, that is, the force on the snow, $F = 720\,\text{N}$, total area of the skis $A = 0.30\,\text{m}^2$

   Find

   the pressure $p$ on the snow in pascals

   Principle

   Pressure is the ratio of the force acting perpendicular to a surface to the area of that surface. The weight is spread evenly over the bases of the skis, so the whole force $720\,\text{N}$ is spread over the whole area $0.30\,\text{m}^2$.

   Formula

   $p = \dfrac{F}{A}$; the unknown $p$ is already alone on the left-hand side

   Conversions

   No conversions are needed, because the force is in newtons and the area in square metres: $1\,\text{N}/\text{m}^2 = 1\,\text{Pa}$.

   Substitution

   $p$$= \dfrac{720\,\text{N}}{0.30\,\text{m}^2}$

   $= 2400\,\tfrac{\text{N}}{\text{m}^2}$

   $= 2400\,\text{Pa}$

   Answer

   $2400\,\text{Pa}$

   Two significant figures like the area $0.30\,\text{m}^2$ (the zeros are only place holders).

   Reasonableness

   Estimate: $700\,\text{N} / 0.3\,\text{m}^2 \approx 2300\,\text{Pa}$. If the same person stood in boots whose soles have about a tenth of the area of the skis, the pressure would be ten times greater, about $24\,\text{kPa}$. That is why skis carry you on deep snow although boots sink in. The result is only about $2\,\%$ of atmospheric pressure ($101\,\text{kPa}$).

2. Exercise 5-2Core

   The deepest point of Lake Päijänne is at a depth of about $95\,\text{m}$.

   How large is the hydrostatic pressure there? The density of water is $1000\,\text{kg/m}^3$. Give the answer in kilopascals.

   kPaCheck
   Answer

   $930\,\text{kPa}$

   Show the solution

   Given

   depth $h = 95\,\text{m}$, density of water $\rho = 1000\,\text{kg/m}^3$, $g = 9.81\,\text{m/s}^2$

   Find

   the hydrostatic pressure $\Delta p$ in kilopascals

   Principle

   In a liquid at rest the pressure grows with depth, because deeper down there is more liquid above. Water is practically incompressible, so $\rho$ is constant at all depths and the pressure grows in direct proportion to the depth. The hydrostatic pressure asked for is only the part caused by the water; atmospheric pressure is not included.

   Formula

   $\Delta p = \rho g h$; the unknown $\Delta p$ is already alone on the left-hand side

   Conversions

   All quantities are in SI units, so the result comes in pascals. At the end $1\,\text{kPa} = 1000\,\text{Pa}$.

   Substitution

   $\Delta p$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 95\,\text{m}$

   $= 931\,950\,\text{Pa}$

   $= 931.95\,\text{kPa}$

   Answer

   $930\,\text{kPa}$

   Two significant figures like the depth $95\,\text{m}$.

   Reasonableness

   Water adds about $10\,\text{kPa}$ per metre, so $95\,\text{m}$ gives about $950\,\text{kPa}$. The exact value is slightly smaller, because $g < 10\,\text{m/s}^2$. The result is about $930/101 \approx 9.2$ atmospheres.

3. Exercise 5-3

   A pressure gauge has been installed on the bottom of a lake at a depth of $18.0\,\text{m}$. During a low-pressure weather system the atmospheric pressure at the surface of the lake is $98.5\,\text{kPa}$.

   What is the total pressure at the gauge? The density of water is $1000\,\text{kg/m}^3$.

   kPaCheck
   Answer

   $275\,\text{kPa}$

   Show the solution

   Given

   depth $h = 18.0\,\text{m}$, atmospheric pressure at the surface $p_0 = 98.5\,\text{kPa}$, $\rho = 1000\,\text{kg/m}^3$, $g = 9.81\,\text{m/s}^2$

   Find

   the total pressure $p$ at the gauge in kilopascals

   Principle

   The gauge feels both the weight of the water and the weight of the air above the water surface. The surface of the liquid transmits atmospheric pressure to the whole liquid unchanged, and the hydrostatic pressure $\rho g h$ is added on top of it, so the pressures are added.

   Formula

   $p = p_0 + \rho g h$; the unknown $p$ is already alone on the left-hand side

   Conversions

   Atmospheric pressure in pascals: $98.5\,\text{kPa} = 98\,500\,\text{Pa}$. At the end the result is converted back: $1\,\text{kPa} = 1000\,\text{Pa}$.

   Substitution

   $\rho g h$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 18.0\,\text{m}$

   $= 176\,580\,\text{Pa}$

   $p$$= 98\,500\,\text{Pa} + 176\,580\,\text{Pa}$

   $= 275\,080\,\text{Pa}$

   $= 275.08\,\text{kPa}$

   Answer

   $275\,\text{kPa}$

   Three significant figures like the depth $18.0\,\text{m}$ and the atmospheric pressure $98.5\,\text{kPa}$.

   Reasonableness

   Estimate: $18\,\text{m}$ of water is about $1.8$ atmospheres, that is, about $180\,\text{kPa}$, and atmospheric pressure of about $100\,\text{kPa}$ is added to it, giving about $280\,\text{kPa}$ in all. The result is greater than the hydrostatic pressure alone, $177\,\text{kPa}$, and almost three atmospheres.

4. Exercise 5-4

   The weight of a sauna stove is $750\,\text{N}$. The stove stands on four legs whose total area against the floor is $0.0025\,\text{m}^2$.

   How large is the pressure that the legs exert on the floor? Give the answer in kilopascals.

   kPaCheck
   Answer

   $300\,\text{kPa}$

   Show the solution

   Given

   the weight of the stove, that is, the force on the floor, $F = 750\,\text{N}$, total area of the legs $A = 0.0025\,\text{m}^2$

   Find

   the pressure $p$ on the floor in kilopascals

   Principle

   Pressure is the ratio of force to area. The stove stands still, so the force on the floor is equal to the weight of the stove, and it is transmitted to the floor only through the small area of the legs.

   Formula

   $p = \dfrac{F}{A}$; the unknown $p$ is already alone on the left-hand side

   Conversions

   The force is in newtons and the area in square metres, so the result comes in pascals. At the end $1\,\text{kPa} = 1000\,\text{Pa}$.

   Substitution

   $p$$= \dfrac{750\,\text{N}}{0.0025\,\text{m}^2}$

   $= 300\,000\,\text{Pa}$

   $= 300\,\text{kPa}$

   Answer

   $300\,\text{kPa}$

   Two significant figures like the area $0.0025\,\text{m}^2$ (the zeros are place holders).

   Reasonableness

   The area of one leg is $0.0025\,\text{m}^2 / 4 \approx 6\,\text{cm}^2$, for example $2.5\,\text{cm} \cdot 2.5\,\text{cm}$, which is a believable size for a leg. The pressure is about three atmospheres, so the legs of a heavy stove can sink into a soft floor. If the area were ten times greater, the pressure would be only a tenth.

5. Exercise 5-5

   The depth of the water in a sauna bucket is $0.45\,\text{m}$.

   How large is the hydrostatic pressure at the bottom of the bucket? The density of water is $1000\,\text{kg/m}^3$.

   PaCheck
   Answer

   $4400\,\text{Pa}$

   Show the solution

   Given

   depth $h = 0.45\,\text{m}$, $\rho = 1000\,\text{kg/m}^3$, $g = 9.81\,\text{m/s}^2$

   Find

   the hydrostatic pressure $\Delta p$ in pascals

   Principle

   In the bucket the water is at rest and its density is constant, so the pressure grows in direct proportion to the depth. At the bottom the hydrostatic pressure depends only on the depth, not on the shape of the bucket or the amount of water.

   Formula

   $\Delta p = \rho g h$; the unknown $\Delta p$ is already alone on the left-hand side

   Conversions

   No conversions are needed, because all quantities are in SI units and the result is asked for in pascals.

   Substitution

   $\Delta p$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 0.45\,\text{m}$

   $= 4414.5\,\text{Pa}$

   Answer

   $4400\,\text{Pa}$

   Two significant figures like the depth $0.45\,\text{m}$.

   Reasonableness

   Water adds about $10\,\text{kPa}$ per metre, so in less than half a metre the pressure is about $4.5\,\text{kPa}$. The result is only about $4\,\%$ of atmospheric pressure, so the bottom does not feel a large gauge pressure.

6. Exercise 5-6

   A nail goes into wood point first when it is pressed with a thumb, but not head first. The force of the thumb is equally large in both cases.

   What explains the difference?

   Choose an option.

   • The pressure is equally large at the point and at the head, so the difference is caused only by the weight of the nail itself.
   • The area of the point is small, so the same force causes a greater pressure at the point.
   • The force on the point is greater than on the head, because the nail amplifies the force of the thumb.
   • The area of the head is large, so the pressure on the head is greater than on the point.

   Answer
   I have done my solution.
   Answer

   The area of the point is small, so the same force causes a greater pressure at the point.

   Show the solution

   Find

   why a nail goes into wood point first but not head first, when the force of the thumb is the same.

   Principle

   Pressure is the ratio of the force perpendicular to a surface to the area of the surface, $p = F/A$. The nail is in equilibrium between the thumb and the wood, so the force is equally large at the point, at the head and at the thumb. The pressure therefore depends only on how small an area the same force is concentrated on.

   Model answer

   The area of the point is small, so the same force causes a greater pressure at the point, and the fibres of the wood give way. The area of the head is many times greater, so the pressure there is small and the nail does not penetrate the wood. The nail does not amplify the force: the force is equally large in both cases, but the pressure is not. The weight of the nail itself does not matter, because it is small compared with the force of the thumb.

7. Exercise 5-7

   The water of the Baltic Sea is brackish water with a density of $1005\,\text{kg/m}^3$.

   How large is the hydrostatic pressure in the water of the Baltic Sea at a depth of $40\,\text{m}$? Give the answer in kilopascals.

   kPaCheck
   Answer

   $390\,\text{kPa}$

   Show the solution

   Given

   depth $h = 40\,\text{m}$, density of Baltic Sea water $\rho = 1005\,\text{kg/m}^3$, $g = 9.81\,\text{m/s}^2$

   Find

   the hydrostatic pressure $\Delta p$ in kilopascals

   Principle

   In a liquid at rest the pressure grows with depth, because deeper down there is more liquid above. Brackish water, like fresh water, is practically incompressible, so $\rho$ is constant. The hydrostatic pressure asked for is only the part caused by the water; atmospheric pressure is not included.

   Formula

   $\Delta p = \rho g h$; the unknown $\Delta p$ is already alone on the left-hand side

   Conversions

   All quantities are in SI units, so the result comes in pascals. At the end $1\,\text{kPa} = 1000\,\text{Pa}$.

   Substitution

   $\Delta p$$= 1005\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 40\,\text{m}$

   $= 394\,362\,\text{Pa}$

   $= 394.362\,\text{kPa}$

   Answer

   $390\,\text{kPa}$

   Two significant figures like the depth $40\,\text{m}$.

   Reasonableness

   Water adds about $10\,\text{kPa}$ per metre, so $40\,\text{m}$ gives about $400\,\text{kPa}$. The exact value is slightly smaller, because $g < 10\,\text{m/s}^2$. With the density of fresh water, $1000\,\text{kg/m}^3$, the result would be $392\,\text{kPa}$, so the greater density of brackish water changes the result by less than $1\,\%$.

8. Exercise 5-8

   A student claims: “In a lake, the total pressure at a depth of $10\,\text{m}$ is twice as large as at a depth of $5\,\text{m}$, because the depth is twice as great.” What is the error in the reasoning?

   Choose an option.

   • Deeper down the water is lighter, so the pressure grows more slowly than the depth.
   • The pressure grows with the square of the depth, so it is more than double at double the depth.
   • Only the share of the water doubles, because the atmospheric pressure is the same at both depths.
   • The pressure does not depend on depth, because the pressure of a liquid is determined only by its density and the atmospheric pressure.

   Answer
   I have done my solution.
   Answer

   Only the share of the water doubles, because the atmospheric pressure is the same at both depths.

   Show the solution

   Find

   what the error is in the claim that at double the depth the total pressure is also double.

   Principle

   In a liquid the total pressure is $p = p_0 + \rho g h$. The hydrostatic part $\rho g h$ is directly proportional to the depth, but the pressure $p_0$ at the surface is a constant term, which is the same at every depth.

   Model answer

   Only the share of the water doubles, because the atmospheric pressure is the same at both depths. For example, with atmospheric pressure $101\,\text{kPa}$, the pressure at a depth of $5\,\text{m}$ is $101\,\text{kPa} + 49\,\text{kPa} = 150\,\text{kPa}$ and at a depth of $10\,\text{m}$ it is $101\,\text{kPa} + 98\,\text{kPa} = 199\,\text{kPa}$, that is, only about $1.3$ times as great. The pressure grows with depth, but not with the square of the depth, and the density of water is the same at every depth.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 5-9Core

   A person's weight is $600\,\text{N}$. At the start of a step the person presses on a parquet floor for a moment with only the heel of a high-heeled shoe, whose contact area with the floor is $1.5\,\text{cm}^2$.

   How large is the pressure that the heel exerts on the floor? Give the answer in megapascals (MPa); the prefix mega means a million.

   MPaCheck
   Answer

   $4.0\,\text{MPa}$

   Show the solution

   Given

   force $F = 600\,\text{N}$, contact area of the heel $A = 1.5\,\text{cm}^2$

   Find

   the pressure $p$ in megapascals

   Principle

   Pressure is the ratio of force to area. At the start of a step the whole weight $600\,\text{N}$ is transmitted to the floor through the small area of the heel, so the force is spread only over the area $1.5\,\text{cm}^2$.

   Formula

   $p = \dfrac{F}{A}$; the unknown $p$ is already alone on the left-hand side

   Conversions

   $1\,\text{cm}^2$$= (0.01\,\text{m})^2$

   $= 10^{-4}\,\text{m}^2$

   so $A = 1.5 \cdot 10^{-4}\,\text{m}^2$. At the end $1\,\text{MPa} = 10^6\,\text{Pa}$.

   Substitution

   $p$$= \dfrac{600\,\text{N}}{1.5 \cdot 10^{-4}\,\text{m}^2}$

   $= 4\,000\,000\,\text{Pa}$

   $= 4.0\,\text{MPa}$

   Answer

   $4.0\,\text{MPa}$

   Two significant figures like the area $1.5\,\text{cm}^2$.

   Reasonableness

   The pressure is about $4.0\,\text{MPa} / 0.101\,\text{MPa} \approx 40$ atmospheres. A bare sole with an area of about $200\,\text{cm}^2$ would give only about $600\,\text{N} / 0.02\,\text{m}^2 = 30\,\text{kPa}$, so the pressure under the heel is more than a hundred times greater. That is why a heel leaves a dent in a parquet floor.

2. Exercise 5-10Core

   A house gets its water from a water tower. When the taps are closed, a pressure gauge at the tap in the cellar shows a gauge pressure of $350\,\text{kPa}$.

   How high above the tap is the water surface in the tower? The density of water is $1000\,\text{kg/m}^3$.

   mCheck
   Answer

   $35.7\,\text{m}$

   Show the solution

   Given

   gauge pressure $\Delta p = 350\,\text{kPa}$, $\rho = 1000\,\text{kg/m}^3$, $g = 9.81\,\text{m/s}^2$

   Find

   the height $h$ of the water surface above the tap in metres

   Principle

   When the taps are closed, the water is at rest, so the gauge pressure at the tap is the hydrostatic pressure of the column of water, $\Delta p = \rho g h$. The water between the tower and the tap is continuous, so only the difference in height matters, not the length of the pipes.

   Formula

   $\Delta p = \rho g h$, so $h = \dfrac{\Delta p}{\rho g}$

   Conversions

   $350\,\text{kPa} = 350\,000\,\text{Pa}$, so that the units $\text{Pa} = \text{kg}/(\text{m}\cdot\text{s}^2)$ cancel to metres.

   Substitution

   $h$$= \dfrac{350\,000\,\text{Pa}}{1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2}}$

   $= 35.678\,\text{m}$

   Answer

   $35.7\,\text{m}$

   Three significant figures: the gauge pressure $350\,\text{kPa} = 3.50\cdot10^{2}\,\text{kPa}$ and the density $1000\,\text{kg/m}^3$ are given here to three figures.

   Reasonableness

   Water gives about $10\,\text{kPa}$ per metre, so $350\,\text{kPa}$ corresponds to about $35\,\text{m}$. The exact value is slightly greater, because $g < 10\,\text{m/s}^2$. Checking backwards,

   $\rho g h$$= 1000 \cdot 9.81 \cdot 35.7\,\text{Pa}$

   $\approx 350\,\text{kPa}$

   Water towers are about this tall.

3. Exercise 5-11

   The graph shows the pressures of two liquids, A and B, as a function of depth. At the surface the pressure is the same in both.

   At what depth in liquid A is the pressure equal to the pressure in liquid B at a depth of $4.0\,\text{m}$?

   mCheck
   Answer

   $h_\text{A} = 6.0\,\text{m}$

   Show the solution

   Given

   from the graph, in both liquids $p_0 = 100\,\text{kPa}$; line A passes through the point $(10\,\text{m}$; $200\,\text{kPa})$ and line B through the point $(10\,\text{m}$; $250\,\text{kPa})$; $h_\text{B} = 4.0\,\text{m}$

   Find

   the depth $h_\text{A}$ at which the pressure in liquid A equals the pressure in liquid B at depth $h_\text{B}$

   Principle

   In a liquid $p = p_0 + \rho g h$, so the $(h, p)$ graph is a straight line whose slope $k = \rho g$ is greater for the denser liquid. The densities do not have to be calculated; the slopes are read from the graph.

   Reading the graph

   The lines start from the same point $(0\,\text{m}$; $100\,\text{kPa})$ and end at crossings of the grid. So

   $k_\text{A}$$= \dfrac{(200 - 100)\,\text{kPa}}{10\,\text{m}}$

   $= 10\,\tfrac{\text{kPa}}{\text{m}}$

   $k_\text{B}$$= \dfrac{(250 - 100)\,\text{kPa}}{10\,\text{m}}$

   $= 15\,\tfrac{\text{kPa}}{\text{m}}$

   Formula

   $p_0 + k_\text{A} h_\text{A} = p_0 + k_\text{B} h_\text{B}$, so

   $h_\text{A}$$= \dfrac{k_\text{B}}{k_\text{A}}\, h_\text{B}$

   Conversions

   No conversions are needed: the kilopascals cancel in the ratio $k_\text{B}/k_\text{A}$.

   Substitution

   $h_\text{A}$$= \dfrac{15\,\text{kPa/m}}{10\,\text{kPa/m}} \cdot 4.0\,\text{m}$

   $= 6.0\,\text{m}$

   Answer

   $h_\text{A} = 6.0\,\text{m}$

   Two significant figures like the depth $4.0\,\text{m}$. A value read by eye in the range $5.8$–$6.2\,\text{m}$ is accepted.

   Reasonableness

   Liquid A is less dense, so a greater depth than $4.0\,\text{m}$ is needed to reach the same pressure. Check:

   $100\,\text{kPa} + 10\,\tfrac{\text{kPa}}{\text{m}} \cdot 6.0\,\text{m}$

   $= 160\,\text{kPa}$

   $= 100\,\text{kPa} + 15\,\tfrac{\text{kPa}}{\text{m}} \cdot 4.0\,\text{m}$
4. Exercise 5-12

   Inside a pressure cooker there is a gauge pressure of $0.85\,\text{bar}$ ($1\,\text{bar} = 100\,\text{kPa}$). The area of the lid is $380\,\text{cm}^2$.

   How large a force does the gauge pressure exert on the lid? Give the answer in kilonewtons.

   kNCheck
   Answer

   $3.2\,\text{kN}$

   Show the solution

   Given

   gauge pressure $\Delta p = 0.85\,\text{bar}$ (according to the problem $1\,\text{bar} = 100\,\text{kPa}$), area of the lid $A = 380\,\text{cm}^2$

   Find

   the force $F$ on the lid in kilonewtons

   Principle

   The gauge pressure pushes the lid from the inside at every point, and pressure is force divided by area. The total force is therefore the gauge pressure multiplied by the area of the lid. The pressure of the outside air is not part of this, because only the force caused by the gauge pressure is asked for.

   Formula

   $\Delta p = \dfrac{F}{A}$, so $F = \Delta p\,A$

   Conversions

   $0.85\,\text{bar}$$= 85\,\text{kPa}$

   $= 85\,000\,\text{Pa}$

   $380\,\text{cm}^2$$= 380 \cdot 10^{-4}\,\text{m}^2$

   $= 0.0380\,\text{m}^2$

   At the end $1\,\text{kN} = 1000\,\text{N}$.

   Substitution

   $F$$= 85\,000\,\text{Pa} \cdot 0.0380\,\text{m}^2$

   $= 3230\,\text{N}$

   $= 3.23\,\text{kN}$

   Answer

   $3.2\,\text{kN}$

   Two significant figures like the gauge pressure $0.85\,\text{bar}$.

   Reasonableness

   The force equals the weight of a mass of $3230\,\text{N} / 9.81\,\text{m/s}^2 \approx 330\,\text{kg}$, which is a lot, although the gauge pressure is less than one atmosphere. Estimate: $80\,\text{kPa} \cdot 0.04\,\text{m}^2 = 3200\,\text{N}$. That is why the lid must not be opened before the gauge pressure has been released.

5. Exercise 5-13

   Example data: a diver measured the pressure in a lake with a pressure gauge at different depths. The slope of the line fitted to the points is, calculated with a computer, $9.78\,\tfrac{\text{kPa}}{\text{m}}$, and the line passes through the mean point $(h_{\text{mean}}, p_{\text{mean}})$ of the points. The uncertainty of the gauge reading is $\pm 0.5\,\text{kPa}$.

   a) Calculate the density $\rho$ of the water from the slope.

   b) Calculate the mean point from the table and use it to find the constant term of the line, that is, the atmospheric pressure $p_0$ at the surface.

   The answer field takes the atmospheric pressure of part b) in kilopascals.

   Pressure in a lake as a function of depth (example data)

   $h\,(\text{m})$	$p\,(\text{kPa})$
   0.0	99.6
   2.0	119.5
   4.0	138.6
   6.0	158.7
   8.0	178.1
   10.0	197.3

   kPaCheck
   Answer

   $p_0 = 99.7\,\text{kPa}$

   Show the solution

   Given

   six points $(h, p)$ from the table; slope of the fitted line $k = 9.78\,\tfrac{\text{kPa}}{\text{m}}$; the line passes through the mean point $(h_{\text{mean}}, p_{\text{mean}})$; $g = 9.81\,\text{m/s}^2$

   Find

   a) the density $\rho$ of the water; b) the atmospheric pressure $p_0$ in kilopascals

   Principle

   The pressure in a liquid is $p = p_0 + \rho g h$, so the graph $(h, p)$ is a straight line whose slope is $\rho g$ and whose constant term is $p_0$. In real series of measurements the data points are usually not exactly on one line because of measurement uncertainty, so the fitted line passes through the middle of the set of points, and therefore through the mean point.

   Reading the table

   The constant term is the pressure of the line at $h = 0$, and it is obtained from the mean point. The sum of the depths is $30.0\,\text{m}$ and the sum of the pressures $891.8\,\text{kPa}$.

   Formula

   a) $k = \rho g$, so $\rho = \dfrac{k}{g}$. b) $p_{\text{mean}} = p_0 + k h_{\text{mean}}$, so $p_0 = p_{\text{mean}} - k h_{\text{mean}}$

   Conversions

   $k$$= 9.78\,\text{kPa/m}$

   $= 9780\,\text{Pa/m}$

   $= 9780\,\text{N/m}^3$

   so that the density comes in the unit $\text{kg/m}^3$.

   Substitution

   a)

   $\rho$$= \dfrac{9780\,\text{N/m}^3}{9.81\,\text{m/s}^2}$

   $= 997.0\,\tfrac{\text{kg}}{\text{m}^3}$

   b)

   $h_{\text{mean}}$$= 30.0\,\text{m} : 6$

   $= 5.0\,\text{m}$

   $p_{\text{mean}}$$= 891.8\,\text{kPa} : 6$

   $= 148.63\,\text{kPa}$

   so

   $$\begin{aligned} p_0 &= 148.63\,\text{kPa} - 9.78\,\tfrac{\text{kPa}}{\text{m}} \cdot 5.0\,\text{m} \\ &= 148.63\,\text{kPa} - 48.9\,\text{kPa} \\ &= 99.73\,\text{kPa} \end{aligned}$$

   Answer

   $p_0 = 99.7\,\text{kPa}$

   Three significant figures like the slope and the pressures. The density is $997\,\text{kg/m}^3$.

   Reasonableness

   At the point $h = 10.0\,\text{m}$ the line gives the pressure

   $99.73\,\text{kPa} + 9.78\,\tfrac{\text{kPa}}{\text{m}} \cdot 10.0\,\text{m}$

   $= 197.5\,\text{kPa}$

   which differs from the measured $197.3\,\text{kPa}$ by only $0.2\,\text{kPa}$, less than the uncertainty $0.5\,\text{kPa}$. The density is close to the density of water, $1000\,\text{kg/m}^3$, and the atmospheric pressure is close to the value $101\,\text{kPa}$.

6. Exercise 5-14

   Example data: four glass vessels of different shapes were filled with water to a height of $0.300\,\text{m}$, and the gauge pressure was measured with a pressure gauge at the bottom of each vessel. The uncertainty of the gauge reading is $\pm 0.1\,\text{kPa}$. The density of water is $1000\,\text{kg/m}^3$.

   a) Calculate the hydrostatic pressure at a depth of $0.300\,\text{m}$ and check whether the result agrees with the measurements in the table.

   b) Explain why the pressure at the bottom is almost the same in all the vessels although the amount of water is different.

   The answer field takes the result of part a) in kilopascals.

   Gauge pressure at the bottom of different vessels (example data)

   Vessel	$V\,(\text{l})$	$\Delta p\,(\text{kPa})$
   straight	1.5	2.9
   widening downwards	2.4	3.0
   widening upwards	0.8	3.0
   bottle	1.2	2.9

   kPaCheck
   Answer

   $2.94\,\text{kPa}$

   Show the solution

   Given

   height of the water column $h = 0.300\,\text{m}$, $\rho = 1000\,\text{kg/m}^3$, $g = 9.81\,\text{m/s}^2$; measured gauge pressures $2.9\ldots 3.0\,\text{kPa}$, uncertainty of the reading $\pm 0.1\,\text{kPa}$

   Find

   a) the hydrostatic pressure $\Delta p$ in kilopascals; b) why the pressure at the bottom is almost the same in the different vessels

   Principle

   In a liquid at rest the hydrostatic pressure depends only on the depth and the density of the liquid, $\Delta p = \rho g h$. The shape of the vessel and the amount of water do not appear in the formula, and the gauge measures the gauge pressure relative to atmospheric pressure.

   Reading the table

   The four gauge pressures in the table are $2.9\,\text{kPa}$, $3.0\,\text{kPa}$, $3.0\,\text{kPa}$ and $2.9\,\text{kPa}$, although the amount of water is $0.8\ldots 2.4\,\textsf{l}$. They differ from each other only by the uncertainty of the gauge.

   Formula

   $\Delta p = \rho g h$; the unknown $\Delta p$ is already alone on the left-hand side

   Conversions

   All quantities are in SI units, so the result comes in pascals. At the end $1\,\text{kPa} = 1000\,\text{Pa}$.

   Substitution

   $\Delta p$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 0.300\,\text{m}$

   $= 2943\,\text{Pa}$

   $= 2.94\,\text{kPa}$

   Answer

   $2.94\,\text{kPa}$

   Three significant figures like the height $0.300\,\text{m}$. Explanation for part b): the pressure at the bottom is determined only by the depth and the density, so the shape of the vessel and the amount of water do not affect it.

   Reasonableness

   Every measured value is within $0.1\,\text{kPa}$ of the calculation: the differences are $0.04\,\text{kPa}$ and $0.06\,\text{kPa}$. The largest vessel contains three times as much water as the smallest ($2.4\,\textsf{l} : 0.8\,\textsf{l} = 3$), but the pressure is the same, so the pressure cannot depend on the amount of water.

7. Exercise 5-15

   Example data: one arm of a U-tube manometer is open to the atmosphere and the other is connected to a closed gas tank. The tube contains water with a density of $1000\,\text{kg/m}^3$. The difference in height between the liquid surfaces was read three times; the water surface on the gas side was lower than the surface in the open arm. Atmospheric pressure is $101\,\text{kPa}$.

   a) Calculate the mean $\Delta h_{\text{mean}}$ of the height difference.

   b) Calculate the gauge pressure of the gas relative to atmospheric pressure.

   c) Calculate the total pressure of the gas and explain why the surface on the gas side is lower.

   The answer field takes the result of part b) in kilopascals.

   Height difference between the liquid surfaces (example data)

   Measurement	$\Delta h\,(\text{cm})$
   1	14.2
   2	14.8
   3	14.5

   kPaCheck
   Answer

   $1.42\,\text{kPa}$

   Show the solution

   Given

   differences in height of the liquid surfaces $14.2\,\text{cm}$, $14.8\,\text{cm}$ and $14.5\,\text{cm}$; $\rho = 1000\,\text{kg/m}^3$, $g = 9.81\,\text{m/s}^2$, atmospheric pressure $p_0 = 101\,\text{kPa}$

   Find

   a) the mean $\Delta h_{\text{mean}}$; b) the gauge pressure $\Delta p$ of the gas in kilopascals; c) the total pressure $p$ of the gas

   Principle

   The open arm of the tube is at atmospheric pressure. At the same horizontal level in the same liquid the pressure is the same, so the hydrostatic pressure $\rho g \Delta h$ caused by the height difference $\Delta h$ is the difference between the pressure of the gas and atmospheric pressure. The mean of repeated measurements reduces the random reading error.

   Formula

   $\Delta h_{\text{mean}}$$= \dfrac{\text{sum}}{3}$

   $\Delta p = \rho g \Delta h_{\text{mean}}$; $p = p_0 + \Delta p$

   Conversions

   Height from centimetres to metres:

   $14.5\,\text{cm}$$= 14.5 : 100\,\text{m}$

   $= 0.145\,\text{m}$

   At the end $1\,\text{kPa} = 1000\,\text{Pa}$.

   Substitution

   a) The sum is $14.2\,\text{cm} + 14.8\,\text{cm} + 14.5\,\text{cm} = 43.5\,\text{cm}$, so

   $\Delta h_{\text{mean}}$$= 43.5\,\text{cm} : 3$

   $= 14.5\,\text{cm}$

   b)

   $\Delta p$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 0.145\,\text{m}$

   $= 1422.45\,\text{Pa}$

   $= 1.42\,\text{kPa}$

   c)

   $p$$= 101\,\text{kPa} + 1.42\,\text{kPa}$

   $= 102.42\,\text{kPa}$

   Answer

   b) $1.42\,\text{kPa}$, three significant figures like the height difference $14.5\,\text{cm}$. In part c) the total pressure is $102\,\text{kPa}$, because the atmospheric pressure $101\,\text{kPa}$ is given to three figures. The surface on the gas side is lower because the pressure of the gas is greater than atmospheric pressure and it pushes the liquid down in its own arm, until the water column of the height difference balances the gauge pressure.

   Reasonableness

   A water column of $10\,\text{cm}$ corresponds to about $1\,\text{kPa}$, so $14.5\,\text{cm}$ corresponds to about $1.4\,\text{kPa}$. The gauge pressure is only about $1.4\,\%$ of atmospheric pressure ($1.42/101 = 0.014$), which fits a small height difference.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 5-16

   A hiker's mass with a rucksack is $85\,\text{kg}$. Soft snow supports the hiker if the pressure on it is at most $6.0\,\text{kPa}$. When walking, the hiker is for a moment on only one foot.

   How large must the area of one snowshoe be at least, so that the hiker does not sink?

   m²Check
   Answer

   $0.14\,\text{m}^2$

   Show the solution

   Given

   mass with the rucksack $m = 85\,\text{kg}$, greatest permitted pressure $p_\text{max} = 6.0\,\text{kPa}$, $g = 9.81\,\text{m/s}^2$; the whole weight on one foot

   Find

   the smallest area $A$ of one snowshoe in square metres

   Principle

   The hiker does not sink when the pressure on the snow is at most $p_\text{max}$. For a moment the whole weight is on one foot, so the force on this shoe is the whole weight $G = mg$, and the area is smallest when the pressure is exactly $p_\text{max}$.

   Formula

   $G = mg$ and $p_\text{max} = \dfrac{G}{A}$, so

   $A$$= \dfrac{G}{p_\text{max}}$

   $= \dfrac{mg}{p_\text{max}}$

   Conversions

   $6.0\,\text{kPa} = 6000\,\text{Pa}$. The mass is already in kilograms.

   Substitution

   $G$$= 85\,\text{kg} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2}$

   $= 833.85\,\text{N}$

   $A$$= \dfrac{833.85\,\text{N}}{6000\,\text{Pa}}$

   $= 0.13898\,\text{m}^2$

   Answer

   $0.14\,\text{m}^2$

   Two significant figures like the pressure $6.0\,\text{kPa}$ and the mass $85\,\text{kg}$.

   Reasonableness

   Checking backwards, $833.85\,\text{N} / 0.14\,\text{m}^2 \approx 5960\,\text{Pa}$, which is less than $6000\,\text{Pa}$. The area is, for example, $25\,\text{cm} \cdot 56\,\text{cm}$, the size of an ordinary snowshoe. The area of the sole of an ordinary shoe is about $0.03\,\text{m}^2$, and with it the pressure would be about $28\,\text{kPa}$, so the shoe would sink.

2. Exercise 5-17

   The rainwater barrel at a cottage is full of water, and its tight lid is against the water. The area of the lid is $0.25\,\text{m}^2$.

   A thin vertical hose is installed through the lid and filled with water to a height of $4.0\,\text{m}$ above the lid. The hose holds only $0.40\,\textsf{l}$ of water.

   By how much does the force with which the water in the barrel presses on the lid increase? Give the answer in kilonewtons.

   kNCheck
   Answer

   $9.8\,\text{kN}$

   Show the solution

   Given

   height of the hose above the lid $h = 4.0\,\text{m}$, area of the lid $A = 0.25\,\text{m}^2$, $\rho = 1000\,\text{kg/m}^3$, $g = 9.81\,\text{m/s}^2$; amount of water in the hose $0.40\,\textsf{l}$

   Find

   the increase $\Delta F$ in the force on the lid in kilonewtons

   Principle

   The water in the barrel and the water in the hose are one liquid at rest, so at the lid the pressure grows by the hydrostatic pressure $\rho g h$ of the column of water in the hose. The pressure depends only on the height, not on the amount of water, so a thin hose is enough. This increase in pressure acts on the whole area of the lid.

   Formula

   $\Delta p = \rho g h$ and $\Delta F = \Delta p \, A$, so $\Delta F = \rho g h A$

   Conversions

   All quantities are in SI units. At the end $1\,\text{kN} = 1000\,\text{N}$.

   Substitution

   $\Delta p$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 4.0\,\text{m}$

   $= 39\,240\,\text{Pa}$

   $\Delta F$$= 39\,240\,\text{Pa} \cdot 0.25\,\text{m}^2$

   $= 9810\,\text{N}$

   $= 9.81\,\text{kN}$

   Answer

   $9.8\,\text{kN}$

   Two significant figures like the height $4.0\,\text{m}$ and the area $0.25\,\text{m}^2$.

   Reasonableness

   The $0.40\,\textsf{l}$ of water in the hose weighs only $0.40\,\text{kg} \cdot 9.81\,\text{m/s}^2 \approx 3.9\,\text{N}$, but almost $10\,\text{kN}$ acts on the lid. The ratio is $9810 / 3.9 \approx 2500$, which is the same as the ratio of the cross-sectional areas of the lid and the hose ($1\,\text{cm}^2$). That is why a thin hose can break the barrel.

3. Exercise 5-18

   At the bottom of an oil separator well at a petrol station there is a $1.60\,\text{m}$ layer of water and on top of it a $0.40\,\text{m}$ layer of oil with a density of $850\,\text{kg/m}^3$.

   How large is the gauge pressure at the bottom of the well? The density of water is $1000\,\text{kg/m}^3$. Give the answer in kilopascals.

   kPaCheck
   Answer

   $19.0\,\text{kPa}$

   Show the solution

   Given

   layer of water $h_2 = 1.60\,\text{m}$, $\rho_2 = 1000\,\text{kg/m}^3$; on top of it a layer of oil $h_1 = 0.40\,\text{m}$, $\rho_1 = 850\,\text{kg/m}^3$; $g = 9.81\,\text{m/s}^2$

   Find

   the gauge pressure $\Delta p$ at the bottom of the well in kilopascals

   Principle

   The liquids are at rest, and the bottom carries the whole column of liquid above it. Because the density changes at the interface, the hydrostatic pressure of each layer is calculated with its own density and height, and the pressures are added. Gauge pressure means that atmospheric pressure is not included.

   Formula

   the pressures of the layers are added, so $\Delta p = \rho_1 g h_1 + \rho_2 g h_2$

   Conversions

   All quantities are in SI units, so the pressure comes in pascals. At the end $1\,\text{kPa} = 1000\,\text{Pa}$.

   Substitution

   $\rho_1 g h_1$$= 850\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 0.40\,\text{m}$

   $= 3335.4\,\text{Pa}$

   $\rho_2 g h_2$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 1.60\,\text{m}$

   $= 15\,696\,\text{Pa}$

   $\Delta p$$= 3335.4\,\text{Pa} + 15\,696\,\text{Pa}$

   $= 19\,031.4\,\text{Pa}$

   Answer

   $19.0\,\text{kPa}$

   In a sum the decimals are determined by the least precise term: the $3.3\,\text{kPa}$ of the oil layer (thickness $0.40\,\text{m}$, two figures) is precise to one decimal place, so $3.3 + 15.7 = 19.0\,\text{kPa}$.

   Reasonableness

   If the whole $2.00\,\text{m}$ were water, the gauge pressure would be $1000 \cdot 9.81 \cdot 2.00 = 19\,620\,\text{Pa}$. Oil is lighter, so the result is smaller: the difference is $(1000 - 850) \cdot 9.81 \cdot 0.40\,\text{Pa} = 588.6\,\text{Pa}$ and $19\,620 - 588.6 = 19\,031.4\,\text{Pa}$, as calculated.

4. Exercise 5-19

   Example data: in summer a lake has temperature layers, and the density of the water depends on the temperature. A group installed a pressure gauge in the lake at a depth of $20.0\,\text{m}$. The table shows the layers of water from the surface downwards. The uncertainty of the gauge reading is $\pm 0.5\,\text{kPa}$. Atmospheric pressure is ignored and $g = 9.81\,\text{m/s}^2$.

   a) Calculate the hydrostatic pressure at the gauge, taking the own density of each layer into account.

   b) Calculate the same pressure by assuming that the density of the whole water column is $1000\,\text{kg/m}^3$, and calculate how much greater this result is than the result of part a).

   c) Is the difference visible in the reading of the gauge? Also calculate how large an error there would be in the depth if the gauge pressure were converted to depth using the density $1000\,\text{kg/m}^3$, and compare the result with the depth corresponding to the uncertainty.

   The answer field takes the difference of part b) in kilopascals.

   Water layers of a lake (example data)

   Layer	$d\,(\text{m})$	$\theta\,({}^\circ\text{C})$	$\rho\,(\tfrac{\text{kg}}{\text{m}^3})$
   surface	5.0	20	998.2
   middle	10.0	10	999.7
   bottom	5.0	4	1000.0

   kPaCheck
   Answer

   $\delta p = 0.12\,\text{kPa}$

   Show the solution

   Given

   depth $20.0\,\text{m}$; layers: $5.0\,\text{m}$ ($998.2\,\text{kg/m}^3$), $10.0\,\text{m}$ ($999.7\,\text{kg/m}^3$) and $5.0\,\text{m}$ ($1000.0\,\text{kg/m}^3$); uncertainty of the gauge $\pm 0.5\,\text{kPa}$; $g = 9.81\,\text{m/s}^2$

   Find

   a) the hydrostatic pressure $\Delta p_{\text{a}}$ layer by layer; b) the pressure with a constant density and its difference $\delta p$ from part a) in kilopascals; c) whether the difference and the error in depth are visible

   Principle

   Each layer adds the pressure $\rho_i g d_i$, and the pressures of the layers are added, because the layers are on top of one another. If the density is taken to be constant, one density $\rho = 1000\,\text{kg/m}^3$ is used for the whole depth. The difference can be observed only if it is greater than the uncertainty of the gauge.

   Reading the table

   The thickness $d_i$ and the density $\rho_i$ of each layer are read from the table; the sum of the thicknesses is the depth of the gauge, $20.0\,\text{m}$.

   Formula

   a) $\Delta p_{\text{a}} = g\,(\rho_1 d_1 + \rho_2 d_2 + \rho_3 d_3)$. b) $\Delta p_{\text{b}} = \rho g h$ and $\delta p = \Delta p_{\text{b}} - \Delta p_{\text{a}}$. c) error in depth

   $\delta h$$= h - \dfrac{\Delta p_{\text{a}}}{\rho g}$

   Conversions

   All quantities are in SI units, so the pressures come in pascals. At the end $1\,\text{kPa} = 1000\,\text{Pa}$.

   Substitution

   a) The sum of the $\rho_i d_i$ is

   $998.2 \cdot 5.0 + 999.7 \cdot 10.0 + 1000.0 \cdot 5.0$

   $= 19\,988\,\tfrac{\text{kg}}{\text{m}^2}$

   so

   $\Delta p_{\text{a}}$$= 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 19\,988\,\tfrac{\text{kg}}{\text{m}^2}$

   $= 196\,082\,\text{Pa}$

   $= 196.08\,\text{kPa}$

   b)

   $\Delta p_{\text{b}}$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 20.0\,\text{m}$

   $= 196\,200\,\text{Pa}$

   so

   $\delta p$$= 196\,200\,\text{Pa} - 196\,082\,\text{Pa}$

   $= 118\,\text{Pa}$

   $= 0.118\,\text{kPa}$

   c) When the reading of the gauge $196\,082\,\text{Pa}$ is converted to depth with the density $1000\,\text{kg/m}^3$:

   $$\begin{aligned} h' &= \dfrac{196\,082\,\text{Pa}}{1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2}} \\ &= 19.988\,\text{m} \end{aligned}$$

   so the error in depth is

   $20.0\,\text{m} - 19.988\,\text{m}$

   $= 0.012\,\text{m}$

   $= 1.2\,\text{cm}$

   The uncertainty of the gauge, $0.5\,\text{kPa}$, corresponds to a depth of

   $500\,\text{Pa} : 9810\,\text{Pa/m}$

   $= 0.051\,\text{m}$

   $= 5.1\,\text{cm}$

   Answer

   $\delta p = 0.12\,\text{kPa}$

   Two significant figures like the difference in density $1000 - 998.2 = 1.8\,\text{kg/m}^3$. The difference in density between the layers therefore shows up in the pressure by only about $0.06\,\%$. It is smaller than the uncertainty of the gauge, $0.5\,\text{kPa}$, and the error in depth, $1.2\,\text{cm}$, is smaller than the $5.1\,\text{cm}$ corresponding to the uncertainty, so the difference is not visible in the reading of the gauge.

   Reasonableness

   The density of water differs from $1000\,\text{kg/m}^3$ most in the top layer, by less than $0.2\,\%$ (the ratio $1.8 : 1000 = 0.0018$). The deviation concerns the top $15\,\text{m}$ of the water column, and in the middle layer it is only $0.03\,\%$, so the change in the total pressure is even smaller: the calculated difference is about $0.06\,\%$.

5. Exercise 5-20

   Example data: a vertical concrete wall of a dam holds back water to a depth of $20.0\,\text{m}$. Consider a strip of the wall one metre wide, divided into four parts of equal height; the area of a part is its height times its width. Atmospheric pressure acts on the wall on both sides, so it is ignored. The pressure grows linearly with depth, so the force on a part is calculated as the mean of the pressures at the top and bottom edges of the part multiplied by the area of the part. The density of water is $1000\,\text{kg/m}^3$ and $g = 9.81\,\text{m/s}^2$.

   a) Calculate the pressure of the water at the top and bottom edge of each part.

   b) Calculate the force of the water on the top part and on the bottom part.

   c) Calculate the sum of the forces on all the parts and check it by calculating the force on the whole strip with the mean of the pressures at the top and bottom edges of the whole wall.

   d) Using the ratio of the forces, explain why the wall of a dam is made thicker at the bottom than at the top.

   The answer field takes the force on the bottom part from part b) in kilonewtons.

   Parts of a strip of the wall (example data)

   Part	$h_1\,(\text{m})$	$h_2\,(\text{m})$
   1	0.0	5.0
   2	5.0	10.0
   3	10.0	15.0
   4	15.0	20.0

   kNCheck
   Answer

   $F_4 = 858\,\text{kN}$

   Show the solution

   Given

   height of the wall $H = 20.0\,\text{m}$ in four parts, each of height $5.0\,\text{m}$; width of the strip $w = 1.0\,\text{m}$; $\rho = 1000\,\text{kg/m}^3$, $g = 9.81\,\text{m/s}^2$

   Find

   a) the pressures at the edges of the parts; b) the forces on the top and bottom parts, $F_1$ and $F_4$, in kilonewtons; c) the sum of the forces and a check; d) why the wall is thicker at the bottom

   Principle

   The water presses on the wall perpendicular to its surface, and the pressure is $\Delta p = \rho g h$. Atmospheric pressure acts on the wall on both sides and cancels itself, so only the gauge pressure is calculated. Because the pressure is a linear function of depth, the mean pressure of a part is the mean of the pressures at its top and bottom edges, and the force on the part is $F = p_{\text{mean}} A$.

   Reading the table

   The depths $h_1$ and $h_2$ of the top and bottom edges of each part are read from the table. The height of each part is $h_2 - h_1 = 5.0\,\text{m}$, so the area of a part is $A = 5.0\,\text{m} \cdot 1.0\,\text{m} = 5.0\,\text{m}^2$.

   Formula

   a) $\Delta p = \rho g h$. b)

   $p_{\text{mean}}$$= \dfrac{\rho g h_1 + \rho g h_2}{2}$

   $F = p_{\text{mean}} A$. c) the whole strip:

   $F_{\text{tot}}$$= \dfrac{0 + \rho g H}{2} \cdot H w$

   d) compare the forces $F_4$ and $F_1$

   Conversions

   All quantities are in SI units, so the pressures come in pascals and the forces in newtons. At the end $1\,\text{kN} = 1000\,\text{N}$.

   Substitution

   a) $\rho g = 9810\,\text{Pa/m}$, so the pressures at the edges are $0$, $49.05$, $98.1$, $147.15$ and $196.2\,\text{kPa}$ at the depths $0$, $5.0$, $10.0$, $15.0$ and $20.0\,\text{m}$. b) The mean pressure of part 1 is $(0 + 49.05\,\text{kPa}) : 2 = 24.525\,\text{kPa}$, so

   $F_1$$= 24.525\,\text{kPa} \cdot 5.0\,\text{m}^2$

   $= 122.6\,\text{kN}$

   The mean pressure of part 4 is $(147.15\,\text{kPa} + 196.2\,\text{kPa}) : 2 = 171.675\,\text{kPa}$, so

   $$\begin{aligned} F_4 &= 171.675\,\text{kPa} \cdot 5.0\,\text{m}^2 \\ &= 858.375\,\text{kN} \end{aligned}$$

   c) In the same way, parts 2 and 3 give $367.9\,\text{kN}$ and $613.1\,\text{kN}$. The sum is

   $122.6\,\text{kN} + 367.9\,\text{kN} + 613.1\,\text{kN} + 858.4\,\text{kN}$

   $= 1962\,\text{kN}$

   Check for the whole strip: $98.1\,\text{kPa} \cdot 20.0\,\text{m}^2 = 1962\,\text{kN}$. d) The ratio of the forces is

   $F_4 : F_1$$= 858.4\,\text{kN} : 122.6\,\text{kN}$

   $= 7.0$

   Answer

   $F_4 = 858\,\text{kN}$

   Three significant figures, because the depths $15.0\,\text{m}$ and $20.0\,\text{m}$ are given to three figures and the width of the strip is exactly one metre. The force on the bottom part is seven times the force on the top part ($F_1 = 123\,\text{kN}$), because the pressure grows with depth. That is why the wall is made thicker at the bottom: the lower part has to withstand a much greater force than the upper part.

   Reasonableness

   The forces on the parts grow in equal steps in the ratio $1 : 3 : 5 : 7$ ($122.6 \cdot 3 \approx 367.9$, $122.6 \cdot 5 \approx 613.1$, $122.6 \cdot 7 \approx 858.4$), which fits a linearly growing pressure.