Free AP Physics 1 Fluids

AP Physics 1 · Fluids · free lesson

Pressure

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Pressure

A sharp knife cuts and a blunt one does not, though you push on both with the same force. A person on snowshoes stays on top of the snow, and the same person in boots sinks. What matters here is not only the force but the area over which the force is spread. Pressure measures this.

Definition.

Pressure is the magnitude of the force component perpendicular to a surface, per unit of the area of the surface: P=F⊥AP = \dfrac{F_\perp}{A}. Its SI unit is the pascal (Pa): 1 Pa = 1 N/m². Pressure is a scalar quantity: it has a size and no direction.

A pascal is a very small pressure (about the pressure of a sheet of paper lying on a table), so pressures are often given in kilopascals: 1 kPa = 1000 Pa. The next picture shows one brick, of mass 2.0 kg, resting on a table in three ways. Its weight is the same each time, and the table pushes back with the same force, but the area in contact with the table is different.

The force on the table is the brick's weight, 19.6 N19.6\,\mathrm{N}, in every case. The pressure on the table is that force divided by the area in contact: 0.980 kPa0.980\,\mathrm{kPa} when the brick lies on its largest face, 1.96 kPa1.96\,\mathrm{kPa} on its middle face, and 3.92 kPa3.92\,\mathrm{kPa} when it stands on end. The same force on one quarter of the area gives four times the pressure.

Note.

Pressure is not a vector, so it has no direction. The force that a pressure exerts on a surface does have a direction: it is always perpendicular to the surface, and its size is F=PAF = PA.

Only the perpendicular part counts

If a force pushes on a surface at a slant, only its component perpendicular to the surface pushes on the surface. The component along the surface slides the object along it and does not add to the pressure. Drag the slider to change the angle between the force and the line perpendicular to the surface. The force is 50 N, and the area of the surface it pushes on is 0.0020 m².

The red arrow is the force, the blue dashed line its perpendicular component and the green dashed line its component along the surface. At 0 degrees the whole force is perpendicular and the pressure is the greatest, FA\frac{F}{A} = 25 kPa. As the angle grows, the perpendicular component Fcos⁡θF\cos\theta shrinks, and so does the pressure. At 60 degrees it is half as great. At 90 degrees the force would only slide along the surface, and the pressure would be zero.

Fperp=Fscos⁡(θ∘)=120 N×cos⁡(30∘)=104 NP=FperpAs=104 N0.004 m2=26.0 kPa\begin{aligned} F_{perp} &= F_s\cos(\theta^\circ) \\ &= 120\,\mathrm{N} \times \cos(30^\circ) \\ &= 104\,\mathrm{N} \\ P &= \frac{F_{perp}}{A_s} \\ &= \frac{104\,\mathrm{N}}{0.004\,\mathrm{m^{2}}} \\ &= 26.0\,\mathrm{kPa} \end{aligned}

A force of 120 N120\,\mathrm{N} at 3030 degrees to the perpendicular, on 0.00400 m20.00400\,\mathrm{m^{2}}, gives a perpendicular component of 104 N104\,\mathrm{N} and a pressure of 26.0 kPa26.0\,\mathrm{kPa}.

Fluids and pressure

Note.

An incompressible fluid. The volume and the density of a given amount of an incompressible fluid are constant, whatever the pressure exerted on it. A liquid in a syringe with the end sealed does not become denser when you push on the plunger. Gases do change: a gas squeezed into half the volume has twice the density.

A fluid presses on every surface it touches: the walls of its container, the bottom, and any object in it. This pressure is the result of all the interactions between the particles of the fluid and the surface. The particles are always moving, and each one that strikes the surface rebounds and pushes on it for an instant. There are enormous numbers of collisions every second, so the many tiny pushes add up to a steady force on the surface, and the force divided by the area is the pressure. Since a particle that rebounds from a surface pushes it straight away from the fluid, the force of the fluid on the surface is perpendicular to it.

The picture shows particles of a fluid striking a wall. Each one arrives (blue arrow) and rebounds (green arrow). Together they push the wall to the right with a force perpendicular to it (red arrow).

Absolute and gauge pressure

The air above us is a fluid too, and it presses on everything. At sea level its pressure, the atmospheric pressure, is PatmP_{atm} = 1.01 × 10⁵ Pa (101 kPa, or 1 atm = 101 325 Pa). Most pressure gauges, such as a tire gauge, read the pressure in excess of the atmosphere around them. Physicists therefore distinguish two quantities.

Definition.

The absolute pressure of a fluid at a point is the sum of a reference pressure P0P_0, such as the atmospheric pressure PatmP_{atm}, and the gauge pressure PgaugeP_{gauge}:

P=P0+PgaugeP = P_0 + P_{gauge}

A tire gauge that reads 220 kPa means a gauge pressure of 220 kPa, and the air in the tire has an absolute pressure of about 220 + 101 = 321 kPa. The gauge pressure of a fluid open to the air at its surface is zero at the surface.

Pressure at a depth

Take a liquid at rest in an open container. Imagine a vertical column of the liquid, with a horizontal cross section of area AA, reaching from the surface down to a depth hh. The column's volume is AhAh, so its mass is ρAh\rho Ah (the density is the same all the way down, for the fluid is incompressible) and its weight is ρAhg\rho Ahg. The liquid below holds this weight up, by a force on the bottom of the column, so the gauge pressure at that depth is the weight over the area.

Rule.

The gauge pressure at a depth hh below the surface of a liquid of density ρ\rho is Pgauge=ρghP_{gauge} = \rho g h. The absolute pressure there is P=P0+ρghP = P_0 + \rho g h, where P0P_0 is the pressure at the surface (for an open container, the atmospheric pressure).

Drag the slider to move the point down in a tank of water. The gauge pressure grows in proportion to the depth.

For water, ρg\rho g = 1000 kg/m³ × 9.8 m/s² = 9800 N/m³, so every meter of depth adds 9.8 kPa. At a depth of 10 m the water adds about 98 kPa, nearly one more atmosphere. The graph shows the absolute pressure in a tank of water against depth. It is a straight line, since P=P0+ρghP = P_0 + \rho g h has the form of a line: the intercept is the pressure at the surface, P0P_0, and the slope is ρg\rho g.

The line starts at 101 kPa at the surface and rises 98 kPa over each 10 m, a slope of 9.8 kPa per meter. A denser liquid gives a steeper line, and the same intercept if both are open to the air.

Pg=ρgh=1000 kg/m3×9.8 m/s2×6 m=58.8 kPaP=P0+Pg=101 kPa+58.8 kPa=160 kPa\begin{aligned} P_g &= \rho gh \\ &= 1000\,\mathrm{kg/m^{3}} \times 9.8\,\mathrm{m/s^{2}} \times 6\,\mathrm{m} \\ &= 58.8\,\mathrm{kPa} \\ P &= P_0 + P_g \\ &= 101\,\mathrm{kPa} + 58.8\,\mathrm{kPa} \\ &= 160\,\mathrm{kPa} \end{aligned}

At a depth of 6.00 m6.00\,\mathrm{m} in fresh water, the gauge pressure is 58.8 kPa58.8\,\mathrm{kPa} and the absolute pressure is 160 kPa160\,\mathrm{kPa}.

What the pressure depends on

The formula contains the density of the liquid, gg and the depth, and nothing else. It does not contain the width of the container, its shape or the total amount of liquid. In the picture below, three open containers are filled with water to the same level. One is wide, one is a narrow tube, and one is wider at the top. A point at the bottom of each is at the same depth, so the pressure there is the same in all three.

For a depth of 3.00 m3.00\,\mathrm{m} of water the gauge pressure is 29.4 kPa29.4\,\mathrm{kPa} at the bottom of each. The force on the bottom is F=PAF = PA, so it is larger for a larger bottom, but the pressure is the same. The wide container holds more water and pushes on a larger area, but the pressure at a point does not depend on either.

Warning.

Three traps. The depth hh in ρgh\rho g h is measured from the surface of the liquid, not from the bottom of the container. The formula gives the gauge pressure, so for the absolute pressure you must add P0P_0. And the pressure depends on the density and the depth, not on how much liquid there is: a lake and a bucket give the same pressure at the same depth.

The force of a fluid on a surface

The force that a fluid exerts on a surface is perpendicular to the surface and has the size F=PAF = PA. If the air is on both sides of a surface, its pressure pushes equally on both sides and the forces cancel, so the net force from the liquid on a submerged surface depends on the gauge pressure.

Pg=ρghw=1000 kg/m3×9.8 m/s2×2.5 m=24.5 kPaF=PgAw=24.5 kPa×0.02 m2=490 N\begin{aligned} P_g &= \rho gh_w \\ &= 1000\,\mathrm{kg/m^{3}} \times 9.8\,\mathrm{m/s^{2}} \times 2.5\,\mathrm{m} \\ &= 24.5\,\mathrm{kPa} \\ F &= P_gA_w \\ &= 24.5\,\mathrm{kPa} \times 0.02\,\mathrm{m^{2}} \\ &= 490\,\mathrm{N} \end{aligned}

A square hatch of area 0.0200 m20.0200\,\mathrm{m^{2}} on the bottom of a tank of water 2.50 m2.50\,\mathrm{m} deep has a gauge pressure of 24.5 kPa24.5\,\mathrm{kPa} on it. The water pushes down on the hatch with a force of 490 N490\,\mathrm{N}, and with air pressure on its other side, this is the net force.

Finding the density of a liquid: an experiment

A pressure sensor on a long tube is lowered into a liquid, and the absolute pressure is read at several depths. The dots in the graph are the readings and the line is the best straight line through them. The intercept is the pressure at the surface. The slope is ρg\rho g, so the density of the liquid is the slope divided by gg.

The slope of the best-fit line is 10.0 kPa/m10.0\,\mathrm{kPa/m} and the intercept is 101 kPa101\,\mathrm{kPa}, close to atmospheric pressure, as it should be. The density is the slope over gg, which is 1020 kg/m31020\,\mathrm{kg/m^{3}}: the liquid is probably water. Using the slope of the line through all the points reduces the effect of one poor reading.

Worked examples

Worked examples

Six examples of the kinds the AP exam asks on this topic: a pressure from a force and an area, a force at a slant, the pressure at a depth, a graph of pressure against depth, the force of a liquid on a surface, and a claim about pressure justified. Use g = 9.8 m/s², a density of 1000 kg/m³ for water, and an atmospheric pressure of 101 kPa.

Example.

Example 1. A force on an area

A student of mass 65.0 kg65.0\,\mathrm{kg} stands on a level floor. Each shoe is in contact with the floor over an area of 0.0200 m20.0200\,\mathrm{m^{2}}. What is the pressure on the floor when the student stands on both feet? What is it when the student stands on one foot?

Solution

The floor pushes up on the student with a force equal to the weight, and the student pushes down on the floor with the same force, perpendicular to the floor. The pressure is that force divided by the area.

F1=m1g=65 kg×9.8 m/s2=637 NP2=F1A2=637 N0.04 m2=15.9 kPaP1=F1A1=637 N0.02 m2=31.9 kPa\begin{aligned} F_1 &= m_1g \\ &= 65\,\mathrm{kg} \times 9.8\,\mathrm{m/s^{2}} \\ &= 637\,\mathrm{N} \\ P_2 &= \frac{F_1}{A_2} \\ &= \frac{637\,\mathrm{N}}{0.04\,\mathrm{m^{2}}} \\ &= 15.9\,\mathrm{kPa} \\ P_1 &= \frac{F_1}{A_1} \\ &= \frac{637\,\mathrm{N}}{0.02\,\mathrm{m^{2}}} \\ &= 31.9\,\mathrm{kPa} \end{aligned}
Result.

The weight is 637 N637\,\mathrm{N}, spread over 0.0400 m20.0400\,\mathrm{m^{2}} on two feet, so the pressure is 15.9 kPa15.9\,\mathrm{kPa}. On one foot the same force acts on half the area, so the pressure is 31.9 kPa31.9\,\mathrm{kPa}, twice as much. The force on the floor is the same in both cases. Only the area changes.

Example.

Example 2. A force at a slant

A force of 80.0 N80.0\,\mathrm{N} pushes on a flat plate of area 0.0100 m20.0100\,\mathrm{m^{2}}. The force makes an angle of 3030 degrees with the perpendicular to the plate. What is the pressure on the plate? What would the pressure be if the angle were 6060 degrees?

Solution

Only the component perpendicular to the plate counts, F⊥=Fcos⁡θF_\perp = F\cos\theta, with θ measured from the perpendicular.

Fp=F2cos⁡(θ∘)=80 N×cos⁡(30∘)=69.3 NP3=FpA3=69.3 N0.01 m2=6.93 kPa\begin{aligned} F_p &= F_2\cos(\theta^\circ) \\ &= 80\,\mathrm{N} \times \cos(30^\circ) \\ &= 69.3\,\mathrm{N} \\ P_3 &= \frac{F_p}{A_3} \\ &= \frac{69.3\,\mathrm{N}}{0.01\,\mathrm{m^{2}}} \\ &= 6.93\,\mathrm{kPa} \end{aligned}
Result.

The perpendicular component is 69.3 N69.3\,\mathrm{N} and the pressure is 6.93 kPa6.93\,\mathrm{kPa}. At 6060 degrees the perpendicular component is 40.0 N40.0\,\mathrm{N} and the pressure is 4.00 kPa4.00\,\mathrm{kPa}. Doubling the angle does not halve the pressure: it is 0.5770.577 of the first. A force of 80.0 N80.0\,\mathrm{N} divided by the area, with no regard to the angle, would give 8.00 kPa8.00\,\mathrm{kPa}, which is too large.

Example.

Example 3. A diver in a lake

A diver is 15.0 m15.0\,\mathrm{m} below the surface of a lake. What are the gauge pressure and the absolute pressure at that depth? How many times as great as the atmospheric pressure is the absolute pressure?

Solution

The gauge pressure of a liquid at depth hh is ρgh\rho g h, and the absolute pressure adds the pressure at the surface, here the atmosphere.

Pg=ρgh=1000 kg/m3×9.8 m/s2×15 m=147 kPaP=P0+Pg=101 kPa+147 kPa=248 kPa\begin{aligned} P_g &= \rho gh \\ &= 1000\,\mathrm{kg/m^{3}} \times 9.8\,\mathrm{m/s^{2}} \times 15\,\mathrm{m} \\ &= 147\,\mathrm{kPa} \\ P &= P_0 + P_g \\ &= 101\,\mathrm{kPa} + 147\,\mathrm{kPa} \\ &= 248\,\mathrm{kPa} \end{aligned}
Result.

The gauge pressure is 147 kPa147\,\mathrm{kPa} and the absolute pressure is 248 kPa248\,\mathrm{kPa}, which is 2.462.46 times the atmospheric pressure. The water's pressure alone is about one and a half atmospheres. Forgetting to add the atmospheric pressure would give only 147 kPa147\,\mathrm{kPa}, the gauge pressure.

Example.

Example 4. Reading a graph of pressure against depth

The graph shows the absolute pressure in a tank of an unknown liquid, open to the air, against the depth below the surface. What is the density of the liquid? What is the pressure at a depth of 8.00 m8.00\,\mathrm{m}?

Solution

The line is P=P0+ρghP = P_0 + \rho g h, so the intercept, 101 kPa101\,\mathrm{kPa}, is the pressure at the surface and the slope is ρg\rho g. Read the two marked points: at a depth of 0 the pressure is 101 kPa, and at a depth of 5.00 m5.00\,\mathrm{m} it is 140.2 kPa140.2\,\mathrm{kPa}.

s=P1−P0h1=140.2 kPa−101 kPa5 m=7.84 kPa/mρ=sg=7.84 kPa/m9.8 m/s2=800 kg/m3P2=P0+ρgh2=101 kPa+800 kg/m3×9.8 m/s2×8 m=164 kPa\begin{aligned} s &= \frac{P_1 - P_0}{h_1} \\ &= \frac{140.2\,\mathrm{kPa} - 101\,\mathrm{kPa}}{5\,\mathrm{m}} \\ &= 7.84\,\mathrm{kPa/m} \\ \rho &= \frac{s}{g} \\ &= \frac{7.84\,\mathrm{kPa/m}}{9.8\,\mathrm{m/s^{2}}} \\ &= 800\,\mathrm{kg/m^{3}} \\ P_2 &= P_0 + \rho gh_2 \\ &= 101\,\mathrm{kPa} + 800\,\mathrm{kg/m^{3}} \times 9.8\,\mathrm{m/s^{2}} \times 8\,\mathrm{m} \\ &= 164\,\mathrm{kPa} \end{aligned}
Result.

The slope is 7.84 kPa/m7.84\,\mathrm{kPa/m}, so the density of the liquid is 800 kg/m3800\,\mathrm{kg/m^{3}}, less than the density of water. At a depth of 8.00 m8.00\,\mathrm{m} the pressure is 164 kPa164\,\mathrm{kPa}. A liquid of greater density would give a steeper line with the same intercept.

Example.

Example 5. The force of water on a hatch

A rectangular hatch, 0.600 m0.600\,\mathrm{m} by 0.500 m0.500\,\mathrm{m}, is on the flat bottom of a tank. The tank is open to the air and is filled with water to a depth of 2.00 m2.00\,\mathrm{m}. The air outside the tank is at atmospheric pressure. What is the net force of the water and air on the hatch, and in what direction?

Solution

The air pushes on the outside of the hatch with the atmospheric pressure, and it pushes on the water's surface with the same pressure, which acts through the water to the hatch. These two forces are equal and opposite, so the net force is due to the gauge pressure at the depth of the hatch. It is perpendicular to the hatch, pointing down.

A=L5W5=0.6 m×0.5 m=0.300 m2Pg=ρgh=1000 kg/m3×9.8 m/s2×2 m=19.6 kPaF=PgA=19.6 kPa×0.3 m2=5880 N\begin{aligned} A &= L_5W_5 \\ &= 0.6\,\mathrm{m} \times 0.5\,\mathrm{m} \\ &= 0.300\,\mathrm{m^{2}} \\ P_g &= \rho gh \\ &= 1000\,\mathrm{kg/m^{3}} \times 9.8\,\mathrm{m/s^{2}} \times 2\,\mathrm{m} \\ &= 19.6\,\mathrm{kPa} \\ F &= P_gA \\ &= 19.6\,\mathrm{kPa} \times 0.3\,\mathrm{m^{2}} \\ &= 5880\,\mathrm{N} \end{aligned}
Result.

The area is 0.300 m20.300\,\mathrm{m^{2}} and the gauge pressure at the bottom is 19.6 kPa19.6\,\mathrm{kPa}, so the net force of the water and air on the hatch is 5880 N5880\,\mathrm{N}, directed downward, out of the tank, perpendicular to the hatch. The absolute pressure 121 kPa121\,\mathrm{kPa} is not the one to use for the net force.

Example.

Example 6. A narrow tube and a wide tank

A narrow vertical tube with a cross-sectional area of 0.00100 m20.00100\,\mathrm{m^{2}} and a wide tank with a base area of 0.500 m20.500\,\mathrm{m^{2}} are both open to the air and filled with water to a depth of 2.00 m2.00\,\mathrm{m}. A student says that the pressure at the bottom of the wide tank must be greater, because there is much more water on it. Is the student right? Compare the force of the water on the bottoms.

Solution

The gauge pressure depends only on the density and the depth, which are the same in both.

Pg=ρgh6=1000 kg/m3×9.8 m/s2×2 m=19.6 kPaFt=PgAt=19.6 kPa×0.001 m2=19.6 NFw=PgAw=19.6 kPa×0.5 m2=9800 N\begin{aligned} P_g &= \rho gh_6 \\ &= 1000\,\mathrm{kg/m^{3}} \times 9.8\,\mathrm{m/s^{2}} \times 2\,\mathrm{m} \\ &= 19.6\,\mathrm{kPa} \\ F_t &= P_gA_t \\ &= 19.6\,\mathrm{kPa} \times 0.001\,\mathrm{m^{2}} \\ &= 19.6\,\mathrm{N} \\ F_w &= P_gA_w \\ &= 19.6\,\mathrm{kPa} \times 0.5\,\mathrm{m^{2}} \\ &= 9800\,\mathrm{N} \end{aligned}
Result.

The student is not right. The gauge pressure is 19.6 kPa19.6\,\mathrm{kPa} at the bottom of both. The force on the bottom of the tube is 19.6 N19.6\,\mathrm{N} and on the bottom of the tank is 9800 N9800\,\mathrm{N}, 500500 times as much, because the area is that many times as great. More water gives a larger force on a larger area, and not a larger pressure.

Practice

Answer AP-style questions on pressure. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise pressure with new numbers