8.2 Pressure

1. A3.0 × 10¹ Pa
   A student who converts the area by dividing 4000 cm² by 100, as if it were a length, gets 40 m² and 1200/40 = 30 Pa. Since 1 cm = 10⁻² m, 1 cm² = 10⁻⁴ m², so 4000 cm² = 0.40 m².
2. B3.0 × 10³ Pa Correct
   The crate presses on the floor with a force equal to its weight, mg = 1200 N, spread over A = 0.80 m × 0.50 m = 0.40 m². P = F_⊥/A = 1200/0.40 = 3.0 × 10³ Pa.
3. C3.0 × 10² Pa
   A student who divides the mass instead of the weight by the area gets 120/0.40 = 300. That quantity has unit kg/m², not Pa; the force is the weight, mg = 1200 N.
4. D1.2 × 10³ Pa
   A student who treats pressure as the same thing as force gives the crate's weight, 1200 N, as the pressure. Pressure is force per unit area, so the weight must be divided by 0.40 m².

Working The crate is at rest on a level floor, so it presses on the floor with a perpendicular force equal to its weight: F_⊥ = mg = (120 kg)(10 m/s²) = 1200 N. A = (0.80 m)(0.50 m) = 0.40 m². P = F_⊥/A = 1200 N / 0.40 m² = 3.0 × 10³ Pa.

CED 8.2.A.1 · Read this in Fix

1. AThe claim is right: pressure is simply another name for the force that a fluid exerts, and force is a vector.
   A student who treats pressure and force as the same quantity picks this. Pressure is the magnitude of the perpendicular force per unit area, P = F_⊥/A; the force is a vector, but the pressure is not.
2. BThe claim is right: fluid pressure points straight downward, in the direction of the weight of the fluid above it.
   A student who thinks fluid pressure acts only downward picks this. A fluid pushes sideways on the walls of its container and upward on the underside of submerged objects, with the same pressure in every direction at a given point.
3. CThe claim holds only for a moving fluid, whose pressure points along the flow; still fluid has none.
   A student who links pressure to the flow of a fluid picks this. A fluid at rest exerts pressure on everything it touches, as the still air in a room does; pressure comes from the random motion of the particles, not from bulk flow.
4. DThe claim is wrong: at a point the fluid pushes equally on a surface facing any way; only the force has a direction. Correct
   Pressure is a scalar. At a point in a fluid at rest, the fluid pushes with the same pressure on a small surface facing up, down or sideways; the force on each surface is perpendicular to it. The direction belongs to the force, and is set by the surface, not by the pressure.

CED 8.2.A.2 · Read this in Fix

1. A3P, since the liquid's density is the same at every depth. Correct
   The gauge pressure is ρgh. An ideal fluid is incompressible, so its density does not change with the pressure on it and ρg is the same at every depth; tripling the depth triples the gauge pressure to 3P.
2. BMore than 3P, since the deeper liquid is squeezed and so denser.
   A student who thinks a liquid is compressed by the liquid above it picks this. An ideal fluid is incompressible: its volume and density stay the same however great the pressure, so the gauge pressure grows in proportion to the depth.
3. CLess than 3P, since atmospheric pressure does not grow with depth.
   A student who mixes up gauge and absolute pressure picks this. The atmospheric pressure is part of the absolute pressure, not of the gauge pressure; the gauge pressure, ρgh, is exactly proportional to the depth.
4. DP, since a fluid at rest has the same pressure at all points in it.
   A student who thinks the pressure is the same throughout a fluid at rest picks this. Pressure is the same in all directions at one point, but it increases with depth: at 3d the gauge pressure is 3P.

Working P_gauge = ρgh. An ideal fluid is incompressible, so ρ is the same at every depth, whatever the pressure. So P_gauge ∝ h: at 3d, P_gauge = ρg(3d) = 3P.

CED 8.2.A.3 · Read this in Fix

1. AThe weight of the air in the container pushes down on the floor of the container beneath it.
   A student who thinks fluid pressure is a downward push from weight picks this. The air presses on the sides and top of the container as well as the bottom, and for air near atmospheric pressure the weight of a container's worth of air is far too small to account for its pressure.
2. BEach air molecule swells until the molecules fill the container and press outward on its walls.
   A student who gives the molecules the properties of the bulk gas picks this. The molecules stay the same size and are far apart; the air presses on the walls because its moving molecules keep colliding with them.
3. CAir molecules keep striking the walls, and each collision exerts a tiny force perpendicular to the wall. Correct
   The molecules of the air are in constant random motion. Enormous numbers of them strike every part of the walls each second, and each interaction exerts a tiny force on the wall; the total force per unit area is the pressure. This happens even though the air as a whole is at rest.
4. DThe air is one continuous substance, with no separate particles in it, that is squashed against the walls.
   A student who pictures air as a continuous substance picks this. Air is made of separate molecules with empty space between them, and its pressure is the combined effect of their collisions with the walls.

CED 8.2.B.1 · Read this in Fix

1. A1.0 × 10⁵ N
   A student who uses the absolute pressure of the seawater and forgets the air inside gets (5.0 × 10⁵)(0.20) = 1.0 × 10⁵ N. That is the force of the water alone; the inside air pushes the other way with (1.0 × 10⁵)(0.20) = 2.0 × 10⁴ N.
2. B4.0 × 10⁵ N
   A student who treats pressure as a force gives the gauge pressure, 4.0 × 10⁵ Pa, as the answer. To find the force, the pressure difference must be multiplied by the window's area, 0.20 m².
3. C2.0 × 10⁶ N
   A student who rearranges P = F/A as F = P/A gets 4.0 × 10⁵ / 0.20 = 2.0 × 10⁶ N. Multiplying both sides of P = F/A by A gives F = PA = 8.0 × 10⁴ N.
4. D8.0 × 10⁴ N Correct
   The seawater pushes on the window with an absolute pressure of 1.0 × 10⁵ + 4.0 × 10⁵ = 5.0 × 10⁵ Pa, and the inside air pushes back with 1.0 × 10⁵ Pa. The atmospheric part cancels, leaving the gauge pressure ρgd = 4.0 × 10⁵ Pa, so the net force is (4.0 × 10⁵)(0.20) = 8.0 × 10⁴ N.

Working Outside: P = P₀ + ρgd = 1.0 × 10⁵ Pa + (1.0 × 10³ kg/m³)(10 m/s²)(40 m) = 1.0 × 10⁵ Pa + 4.0 × 10⁵ Pa = 5.0 × 10⁵ Pa, pushing inward. Inside: 1.0 × 10⁵ Pa, pushing outward. Net pressure difference = ρgd = 4.0 × 10⁵ Pa. Net force = (4.0 × 10⁵ Pa)(0.20 m²) = 8.0 × 10⁴ N, into the submarine.

CED 8.2.B.2 · Read this in Fix

1. A1.5 × 10⁴ Pa
   A student who uses the density of water for the whole 1.5 m depth gets 1000 × 10 × 1.5 = 15 000 Pa. The top 0.50 m is oil, which is less dense, so it adds only 4000 Pa, not 5000 Pa.
2. B1.0 × 10⁴ Pa
   A student who counts only the water around point X gets 1000 × 10 × 1.0 = 10 000 Pa. The oil pushes down on the water, adding 4000 Pa to the pressure everywhere in the water layer.
3. C1.1 × 10⁵ Pa
   A student who adds atmospheric pressure gets 1.0 × 10⁵ + 1.4 × 10⁴ ≈ 1.1 × 10⁵ Pa. That is the absolute pressure at X; the gauge pressure is the part above atmospheric pressure, 1.4 × 10⁴ Pa.
4. D1.4 × 10⁴ Pa Correct
   Each layer adds its own ρgh. The oil adds 800 × 10 × 0.50 = 4000 Pa at the boundary, and the water adds 1000 × 10 × 1.0 = 10 000 Pa more, so the gauge pressure at X is 1.4 × 10⁴ Pa.

Working Gauge pressure at the oil–water boundary: ρ_oil g h_oil = (800 kg/m³)(10 m/s²)(0.50 m) = 4000 Pa. Gauge pressure at X: 4000 Pa + ρ_water g h_water = 4000 Pa + (1000 kg/m³)(10 m/s²)(1.0 m) = 4000 Pa + 10 000 Pa = 1.4 × 10⁴ Pa.

CED 8.2.B.3 · Read this in Fix

1. A5.5 × 10² Pa
   A student who adds the whole 50 N push to the weight gets 110/0.20 = 550 Pa. Only the component of the push perpendicular to the floor, 50 sin 37° = 30 N, adds to the force on the floor.
2. B5.0 × 10² Pa
   A student who takes the component as 50 cos 37° = 40 N gets 100/0.20 = 500 Pa. The angle is measured from the horizontal, so 40 N is the horizontal (parallel) component; the vertical component is 50 sin 37° = 30 N.
3. C4.5 × 10² Pa Correct
   The push has a downward component of 50 × sin 37° = 30 N, so the floor must support 60 N + 30 N = 90 N and the box presses on it with 90 N. The 40 N horizontal component is parallel to the floor and adds nothing. P = 90/0.20 = 4.5 × 10² Pa.
4. D3.0 × 10² Pa
   A student who assumes the box always presses on the floor with its weight uses 60 N and gets 300 Pa. The student's push has a downward component, so the floor must push up with 90 N, and the box presses on the floor with 90 N.

Working Vertical component of the push, downward: (50 N) sin 37° = 30 N; horizontal component (40 N) is parallel to the floor and is balanced by static friction. The box is at rest, so vertically F_N = F_g + 30 N = 60 N + 30 N = 90 N. By Newton's third law the box presses on the floor with a perpendicular force of 90 N. P = F_⊥/A = 90 N / 0.20 m² = 4.5 × 10² Pa.

CED 8.2.A.1 · Read this in Fix

1. A×4 Correct
   From P = F_⊥/A, the force is F = PA. Doubling the radius multiplies the area πr² by 2² = 4, so with the same pressure the force is 4 times as large.
2. B×2
   A student who thinks the area doubles when the radius doubles picks this. Area depends on r², so the area, and therefore the force, is 4 times as large.
3. C×1
   A student who treats pressure and force as the same thing reasons that equal pressures give equal forces. The same pressure over a larger area produces a larger force: F = PA.
4. D×¼
   A student who rearranges P = F/A as F = P/A concludes that a 4 times larger area gives a quarter of the force. Multiplying both sides by A gives F = PA, so the force grows with the area.

Working F = PA with A = πr². Second piston: A' = π(2r)² = 4πr² = 4A. Same P, so F' = P(4A) = 4F.

CED 8.2.A.1 · Read this in Fix

1. AThe low pressure inside the can pulls the walls inward, just as sucking on a straw pulls a drink upward.
   A student who thinks low pressure pulls picks this. Pressure only pushes: the nearly empty can exerts almost no outward push, and it is the push of the outside air, no longer balanced, that crushes it. (The drink in a straw is likewise pushed up by the atmosphere.)
2. BFew gas particles now strike the inside of the walls, so the outside air pushes in harder than the gas inside pushes out. Correct
   When the steam condenses, very few gas particles are left inside, so collisions with the inside of the walls become rare and the pressure inside falls far below atmospheric. Air molecules still strike the outside as often as before, so the net force on each part of the wall points inward and the can is crushed.
3. CThe weight of the atmosphere above the can presses down on the top of the can until the can is flattened.
   A student who thinks air pressure acts only downward picks this. The outside air pushes perpendicular to every part of the can's surface, sides and bottom included, and the can is crushed because the inside push has almost vanished, not because air presses only on its top.
4. DThe steam particles shrink as they cool, and as they shrink they drag the walls of the can inward with them.
   A student who gives particles the properties of the bulk gas picks this. The water molecules do not change size; they condense into a small amount of liquid, leaving few gas particles to strike the inside of the walls.

CED 8.2.B.1 · Read this in Fix

1. Ah = 2P₀/(ρg)
   A student who thinks the pressure in the water comes only from the water above sets ρgh = 2P₀. The atmosphere already supplies P₀ at every depth, so the water needs to add only another P₀: h = P₀/(ρg).
2. Bh = P₀/ρ
   A student who writes the pressure due to the water as ρh, leaving out g, picks this. Its unit is Pa/(kg/m³) = m²/s², not m; the water's contribution is ρgh.
3. Ch = P₀/(ρg) Correct
   The absolute pressure is P = P₀ + ρgh. Setting P₀ + ρgh = 2P₀ gives ρgh = P₀, so h = P₀/(ρg): the gauge pressure must equal one atmosphere.
4. Dh = ρg/P₀
   A student who rearranges ρgh = P₀ with the quantities upside down picks this. Its unit is 1/m, not m; dividing both sides by ρg gives h = P₀/(ρg).

Working Absolute pressure P = P₀ + ρgh. Set P = 2P₀: P₀ + ρgh = 2P₀ → ρgh = P₀ → h = P₀/(ρg). Units: Pa/((kg/m³)(m/s²)) = (N/m²)/(N/m³) = m.

CED 8.2.B.2 · Read this in Fix

1. AThe pressure is proportional to the depth, because all of the data points lie on a single straight line.
   A student who thinks every straight-line graph shows proportionality picks this. A proportional relationship gives a line through the origin; this line starts at 100 kPa, so the pressure increases linearly with depth but is not proportional to it.
2. BDoubling the depth does not double the pressure, because the line does not pass through the origin. Correct
   The line has an intercept of 100 kPa, the atmospheric pressure at the surface: P = P₀ + ρgh. At 2.0 m the pressure is 124 kPa and at 4.0 m it is 148 kPa, so doubling the depth increases the pressure by only about 19%. Only the gauge pressure, P − P₀, doubles.
3. CThe liquid gets denser as the depth increases, because the pressure increases as the depth increases.
   A student who thinks a liquid is squashed denser at depth picks this. The slope of the graph is ρg; it is constant (12 kPa per meter all the way down), so the density is the same at every depth, as the ideal-fluid model predicts.
4. DA wider tank filled with this liquid would give a steeper line, because it would then hold more liquid.
   A student who thinks the pressure depends on how much liquid there is picks this. The slope is ρg, which depends only on the liquid, not on the width of the tank or the amount of liquid in it, and the graph gives no evidence about other tanks.

Working Intercept 100 kPa = P₀ (atmosphere at the surface). Slope (148 − 100) kPa / 4.0 m = 12 kPa/m, constant, so ρg is constant (ρ = 1.2 × 10³ kg/m³ with g = 10 m/s²). P(2.0 m) = 124 kPa, P(4.0 m) = 148 kPa: doubling the depth multiplies P by 1.19, not 2, because P = P₀ + ρgh has a nonzero intercept.

CED 8.2.B.2 · Read this in Fix

1. AP₁ = P₂ = P₃ Correct
   The gauge pressure at the bottom is ρgh. All three contain water to the same depth h, so the pressures are equal whatever the shape, width or volume of the container. In container 2 the sloping walls hold up the extra water.
2. BP₂ > P₁ > P₃
   A student who ranks the pressures by the amount of water (2V, V, V/2) picks this. The pressure at the bottom depends only on the depth and the density of the water, ρgh, and all three depths are equal.
3. CP₂ > P₁ = P₃
   A student who divides the total weight of water by the base area gets ρgV/A for containers 1 and 3 but 2ρgV/A for container 2. That method works only for vertical walls: container 2's sloping walls support part of the water's weight, so its base carries only ρghA and P₂ = ρgh.
4. DP₃ > P₁ > P₂
   A student who thinks water in a narrower column is squeezed to a higher pressure ranks the narrow container 3 first and the widening container 2 last. For water at rest the width does not matter: the pressure at the bottom is ρgh in each.

Working P_gauge = ρgh at the bottom of each. Same liquid (ρ), same depth (h): P₁ = P₂ = P₃. (Check with cylinders: container 1, weight ρgV over area A with V = Ah gives ρgh; container 3, ρg(V/2) over A/2 gives ρgh. In container 2 the outward-sloping walls support the extra water's weight, so the base still carries only ρghA.)

CED 8.2.B.3 · Read this in Fix

1. Amg/A
   A student who thinks an object always presses on the surface under it with its full weight picks mg/A. On a ramp only the component of the weight perpendicular to the surface, mg cos θ, is balanced by the ramp's normal force, so the perpendicular force on the ramp is less than mg.
2. Bm cos θ/A
   A student who uses the mass as if it were the force picks this. Pressure is force per unit area, and the force here is the perpendicular component of the weight, mg cos θ, in newtons; m cos θ/A has the wrong unit.
3. Cmg/(A cos θ)
   A student who sets the vertical component of the normal force equal to the weight, F_N cos θ = mg, gets F_N = mg/cos θ, more than the weight. Static friction also pushes up the ramp and has a vertical component, so that equation is wrong; balancing forces perpendicular to the ramp gives F_N = mg cos θ.
4. Dmg cos θ/A Correct
   Only the force perpendicular to the ramp produces pressure. The block is at rest, so the perpendicular forces balance and the block presses on the ramp with mg cos θ; friction acts along the surface. P = F_⊥/A = mg cos θ/A.

Working The block is at rest, so the forces perpendicular to the ramp balance: F_N = mg cos θ (friction acts along the ramp and has no perpendicular component). By Newton's third law the block presses on the ramp with a perpendicular force mg cos θ. P = F_⊥/A = mg cos θ/A.

CED 8.2.A.1 · Read this in Fix