A Thin Tube and a Wide Drum Push Equally Hard on Their Bases
Learn to separate thrust from pressure and calculate each, derive the expression for pressure at a depth, use the laws of liquid pressure on dams and tanks, and see why shape and area make no difference.
Why does the shape of the vessel not change the pressure at its base?
Fill a narrow glass tube with water to a height of m. Fill a wide drum with water to the same height of m. The drum holds hundreds of times as much water.
Now measure the pressure at the base of each. The readings are identical:
That looks wrong. The drum is vastly heavier, so surely it presses harder?
It presses with a far greater total force — but over a far greater area, and pressure is force per unit area. Divide the bigger force by the bigger area and the two cancel out exactly.
What is left in the formula tells you what actually matters: the depth , the density , and . The width, the shape and the volume of liquid have all disappeared.
That is why a dam holding back a shallow lake needs no more strength than one holding back a deep pond of the same depth, and why the wall must be thickest at its base rather than where the water is widest.
This page covers the first part of the ICSE Class 9 Physics chapter on pressure in fluids — thrust and pressure, the expression for pressure at a depth, and the laws of liquid pressure.
Now measure the pressure at the base of each. The readings are identical:
That looks wrong. The drum is vastly heavier, so surely it presses harder?
It presses with a far greater total force — but over a far greater area, and pressure is force per unit area. Divide the bigger force by the bigger area and the two cancel out exactly.
What is left in the formula tells you what actually matters: the depth , the density , and . The width, the shape and the volume of liquid have all disappeared.
That is why a dam holding back a shallow lake needs no more strength than one holding back a deep pond of the same depth, and why the wall must be thickest at its base rather than where the water is widest.
This page covers the first part of the ICSE Class 9 Physics chapter on pressure in fluids — thrust and pressure, the expression for pressure at a depth, and the laws of liquid pressure.
Formula
What is the difference between thrust and pressure?
Thrust is the force acting normally on a surface; pressure is that thrust spread over each unit of area.
- Thrust — a force, so a vector, measured in newtons (N)
- Pressure — a scalar, measured in , which is given the name pascal (Pa)
The word normally matters: only the component perpendicular to the surface counts as thrust.
Worked example 1 — same thrust, different pressure. A block of weight N rests first on a face of area and then on a face of area :
**The thrust was N both times. Only the area changed, and the pressure went up five times.
Worked example 2 — standing on one foot.** A person of mass kg has a weight of
Standing on both feet with a total contact area of :
Lifting one foot halves the area and doubles the pressure to Pa, with no change in weight at all.
Worked example 3 — why a sharp knife cuts. A sharpened blade has a very small contact area, so a modest force produces an enormous pressure. A blunt blade spreads the same force over a wider edge and cuts nothing. The force your hand supplies is the same; the area is what the sharpening changed.
Worked example 4 — finding the area. A thrust of N produces a pressure of Pa:
Worked example 5 — a nail and a drawing pin. The same push on a drawing pin's flat head and on its point gives an almost unnoticeable pressure at the head, where the area is wide, and a piercing pressure at the tip. One object, one thrust, two very different pressures — which is the whole design of the pin.
Thrust and pressure are not two words for the same thing. A camel walks on soft sand on broad feet because its weight cannot be changed but the pressure can. A tractor uses wide tyres for the same reason. So a question asking why something sinks or does not sink is a question about area, not about force.
The unit tells them apart. Thrust is in newtons and pressure in pascals, and an answer in the wrong unit is a wrong answer even when the number is right.
- Thrust — a force, so a vector, measured in newtons (N)
- Pressure — a scalar, measured in , which is given the name pascal (Pa)
The word normally matters: only the component perpendicular to the surface counts as thrust.
Worked example 1 — same thrust, different pressure. A block of weight N rests first on a face of area and then on a face of area :
**The thrust was N both times. Only the area changed, and the pressure went up five times.
Worked example 2 — standing on one foot.** A person of mass kg has a weight of
Standing on both feet with a total contact area of :
Lifting one foot halves the area and doubles the pressure to Pa, with no change in weight at all.
Worked example 3 — why a sharp knife cuts. A sharpened blade has a very small contact area, so a modest force produces an enormous pressure. A blunt blade spreads the same force over a wider edge and cuts nothing. The force your hand supplies is the same; the area is what the sharpening changed.
Worked example 4 — finding the area. A thrust of N produces a pressure of Pa:
Worked example 5 — a nail and a drawing pin. The same push on a drawing pin's flat head and on its point gives an almost unnoticeable pressure at the head, where the area is wide, and a piercing pressure at the tip. One object, one thrust, two very different pressures — which is the whole design of the pin.
Thrust and pressure are not two words for the same thing. A camel walks on soft sand on broad feet because its weight cannot be changed but the pressure can. A tractor uses wide tyres for the same reason. So a question asking why something sinks or does not sink is a question about area, not about force.
The unit tells them apart. Thrust is in newtons and pressure in pascals, and an answer in the wrong unit is a wrong answer even when the number is right.
How do you derive the pressure at a given depth in a liquid?
Work out the weight of the liquid column standing above the point, then divide by the area it rests on — and watch the area cancel.
The derivation. Consider a point at depth below the surface of a liquid of density . Imagine a column of that liquid standing on a small horizontal area at that point.
- Volume of the column:
- Mass of the column:
- Weight of the column, which is the thrust on :
So the pressure is
**The area cancelled, and with it went every trace of the vessel's width and shape. That cancellation is the derivation's whole point — it is the reason the thin tube and the wide drum agreed in the opening section.
Worked example 1 — water at m.** With :
**Worked example 2 — water at m.** Twice the depth:
Double the depth, double the pressure. The relationship is a straight proportion, unlike the inverse square of the gravitation chapter.
Worked example 3 — a denser liquid. Mercury has . At a depth of m:
That number is standard atmospheric pressure, which is why a mercury barometer stands at about cm — the subject of the next part of this chapter.
Worked example 4 — finding the depth. At what depth in water is the pressure Pa?
Worked example 5 — total pressure at a depth. A diver m down in water experiences the water's pressure plus the atmosphere pressing on the surface above:
roughly twice atmospheric pressure. "Pressure due to the liquid" and "total pressure" are different answers, and the question's wording decides which is wanted.
Worked example 6 — a tank's base. A tank m deep is filled with a liquid of density . Pressure at the base due to the liquid:
And the thrust on a base of area is
Here the area returns. Pressure at the base is independent of area, but the total thrust on the base is pressure times area — so the wide drum of the opening section does experience a far greater thrust, exactly as it should. Pressure and thrust answer different questions about the same tank.
The derivation. Consider a point at depth below the surface of a liquid of density . Imagine a column of that liquid standing on a small horizontal area at that point.
- Volume of the column:
- Mass of the column:
- Weight of the column, which is the thrust on :
So the pressure is
**The area cancelled, and with it went every trace of the vessel's width and shape. That cancellation is the derivation's whole point — it is the reason the thin tube and the wide drum agreed in the opening section.
Worked example 1 — water at m.** With :
**Worked example 2 — water at m.** Twice the depth:
Double the depth, double the pressure. The relationship is a straight proportion, unlike the inverse square of the gravitation chapter.
Worked example 3 — a denser liquid. Mercury has . At a depth of m:
That number is standard atmospheric pressure, which is why a mercury barometer stands at about cm — the subject of the next part of this chapter.
Worked example 4 — finding the depth. At what depth in water is the pressure Pa?
Worked example 5 — total pressure at a depth. A diver m down in water experiences the water's pressure plus the atmosphere pressing on the surface above:
roughly twice atmospheric pressure. "Pressure due to the liquid" and "total pressure" are different answers, and the question's wording decides which is wanted.
Worked example 6 — a tank's base. A tank m deep is filled with a liquid of density . Pressure at the base due to the liquid:
And the thrust on a base of area is
Here the area returns. Pressure at the base is independent of area, but the total thrust on the base is pressure times area — so the wide drum of the opening section does experience a far greater thrust, exactly as it should. Pressure and thrust answer different questions about the same tank.
What are the laws of liquid pressure, and what do they explain?
**Five statements, all of which follow from , and each one explains something you can see.
Law 1 — pressure increases with depth.** From , doubling doubles .
Law 2 — pressure is the same at all points on the same horizontal level in a liquid at rest, because those points share the same .
Law 3 — at a given depth, pressure acts equally in all directions, not only downward. A liquid pushes sideways on the walls of its container and upward on a body immersed in it.
Law 4 — pressure depends on the density of the liquid. At the same depth, mercury exerts times the pressure of water.
Law 5 — pressure at a given depth is independent of the area and the shape of the vessel, because cancelled in the derivation.
Worked application 1 — a dam wall is thicker at its base. By law 1 the water pressure grows steadily with depth, so the lowest part of the wall has the greatest push to resist. At the surface the pressure is zero; at a depth of m in water it is
The wall is therefore built to taper — thin at the top, massive at the bottom. It is thicker where the pressure is greater, not where the water is widest, which is law 5 doing the work.
Worked application 2 — water spouts farther from a lower hole. Punch two holes in the side of a tall tank of water. The lower hole is at a greater depth, so by law 1 the pressure there is greater, the water leaves faster, and it lands farther from the tank. The upper hole dribbles; the lower one jets.
Worked application 3 — sideways thrust on the walls. By law 3 the water pushes outward on the tank's sides, which is why a tank needs strong walls and not merely a strong floor, and why a rubber hose bulges when the tap is opened.
Worked application 4 — the level in a kettle's spout. By law 2 the water in the spout stands at the same height as the water in the body, since the two are connected at the same levels. That is why the spout tells you how full the kettle is, and it is the principle of every water-level indicator.
Worked application 5 — mercury against water. At a depth of m,
- water: Pa
- mercury: Pa
Law 4 in numbers, and the ratio is just the ratio of the densities.
Law 3 is the one that is hardest to believe and most useful. A liquid at rest pushes upward at a point as hard as it pushes downward — which is why a bucket's bottom is pushed out and why an immersed body is pushed up. That upward push is upthrust, and it is the subject of the last chapter of this group.
Law 5 is the hydrostatic paradox. The thin tube and the wide drum have the same base pressure and wildly different weights of liquid, and it stops being paradoxical the moment you notice that pressure is a force per unit area. **The liquid's total weight never appears in **, and looking for it is what makes the result feel strange.
Law 1 — pressure increases with depth.** From , doubling doubles .
Law 2 — pressure is the same at all points on the same horizontal level in a liquid at rest, because those points share the same .
Law 3 — at a given depth, pressure acts equally in all directions, not only downward. A liquid pushes sideways on the walls of its container and upward on a body immersed in it.
Law 4 — pressure depends on the density of the liquid. At the same depth, mercury exerts times the pressure of water.
Law 5 — pressure at a given depth is independent of the area and the shape of the vessel, because cancelled in the derivation.
Worked application 1 — a dam wall is thicker at its base. By law 1 the water pressure grows steadily with depth, so the lowest part of the wall has the greatest push to resist. At the surface the pressure is zero; at a depth of m in water it is
The wall is therefore built to taper — thin at the top, massive at the bottom. It is thicker where the pressure is greater, not where the water is widest, which is law 5 doing the work.
Worked application 2 — water spouts farther from a lower hole. Punch two holes in the side of a tall tank of water. The lower hole is at a greater depth, so by law 1 the pressure there is greater, the water leaves faster, and it lands farther from the tank. The upper hole dribbles; the lower one jets.
Worked application 3 — sideways thrust on the walls. By law 3 the water pushes outward on the tank's sides, which is why a tank needs strong walls and not merely a strong floor, and why a rubber hose bulges when the tap is opened.
Worked application 4 — the level in a kettle's spout. By law 2 the water in the spout stands at the same height as the water in the body, since the two are connected at the same levels. That is why the spout tells you how full the kettle is, and it is the principle of every water-level indicator.
Worked application 5 — mercury against water. At a depth of m,
- water: Pa
- mercury: Pa
Law 4 in numbers, and the ratio is just the ratio of the densities.
Law 3 is the one that is hardest to believe and most useful. A liquid at rest pushes upward at a point as hard as it pushes downward — which is why a bucket's bottom is pushed out and why an immersed body is pushed up. That upward push is upthrust, and it is the subject of the last chapter of this group.
Law 5 is the hydrostatic paradox. The thin tube and the wide drum have the same base pressure and wildly different weights of liquid, and it stops being paradoxical the moment you notice that pressure is a force per unit area. **The liquid's total weight never appears in **, and looking for it is what makes the result feel strange.
Why does the weight of the liquid not appear in the formula?
Because the weight and the area grow together, so their ratio — the pressure — depends only on the height of liquid above the point.
The derivation showed it symbolically. Here it is in numbers.
Worked example 1 — two vessels, same depth. Both hold water to a height of m.
- A narrow tube of base area : volume , mass kg, weight N, and
- A wide drum of base area : volume , mass kg, weight N, and
A thousand times the weight, a thousand times the area, the same pressure. And gives Pa for both without either calculation.
Worked example 2 — depth is what changes the answer. Take the same narrow tube and fill it to m instead:
Twice the pressure from twice the depth, even though this tube still holds far less water than the drum did.
Worked example 3 — density is the other thing that matters. Fill the narrow tube to m with a liquid of density :
Worked example 4 — an oddly shaped vessel. A vessel that flares out near the top, so that most of its water sits above a narrow neck, still has
at its base, with measured from the free surface straight down. The liquid sitting off to the sides is supported by the sloping walls, not by the base, which is why it makes no contribution.
Worked example 5 — total thrust, where the area does matter. A swimming pool m deep has a base of . The pressure at the base is
the same as in the narrow tube. But the thrust on the base is
nearly five million newtons. Same pressure, hugely different thrust — and this pair of answers is the clearest statement of what the two quantities mean.
**Measure from the free surface, vertically. Not along a sloping wall, and not from the top of the vessel if the vessel is not full. An empty cm of tube above the water contributes nothing, and using the vessel's height instead of the liquid's depth is the standard numerical error here.
And gives the pressure due to the LIQUID only.** The atmosphere pressing on the free surface adds its own Pa, so the total pressure at the base of that swimming pool is
Read the question for which one is wanted — pressure due to the water and total pressure differ by a whole atmosphere, and that difference is bigger than the water's own contribution in a shallow pool.
The derivation showed it symbolically. Here it is in numbers.
Worked example 1 — two vessels, same depth. Both hold water to a height of m.
- A narrow tube of base area : volume , mass kg, weight N, and
- A wide drum of base area : volume , mass kg, weight N, and
A thousand times the weight, a thousand times the area, the same pressure. And gives Pa for both without either calculation.
Worked example 2 — depth is what changes the answer. Take the same narrow tube and fill it to m instead:
Twice the pressure from twice the depth, even though this tube still holds far less water than the drum did.
Worked example 3 — density is the other thing that matters. Fill the narrow tube to m with a liquid of density :
Worked example 4 — an oddly shaped vessel. A vessel that flares out near the top, so that most of its water sits above a narrow neck, still has
at its base, with measured from the free surface straight down. The liquid sitting off to the sides is supported by the sloping walls, not by the base, which is why it makes no contribution.
Worked example 5 — total thrust, where the area does matter. A swimming pool m deep has a base of . The pressure at the base is
the same as in the narrow tube. But the thrust on the base is
nearly five million newtons. Same pressure, hugely different thrust — and this pair of answers is the clearest statement of what the two quantities mean.
**Measure from the free surface, vertically. Not along a sloping wall, and not from the top of the vessel if the vessel is not full. An empty cm of tube above the water contributes nothing, and using the vessel's height instead of the liquid's depth is the standard numerical error here.
And gives the pressure due to the LIQUID only.** The atmosphere pressing on the free surface adds its own Pa, so the total pressure at the base of that swimming pool is
Read the question for which one is wanted — pressure due to the water and total pressure differ by a whole atmosphere, and that difference is bigger than the water's own contribution in a shallow pool.
Exam tip
Exam tip: measure the depth from the free surface, and mind the unit
Thrust is in newtons; pressure is in pascals. Getting the unit wrong loses the mark even with the right number.
** needs the area in .** Convert first — remember .
**Use with measured vertically from the FREE SURFACE, never from the top of the vessel and never along a slope.
Take and **, with .
Read whether the question wants the liquid's pressure or the TOTAL pressure — the total adds atmospheric pressure, about Pa.
Pressure is independent of area; thrust is not. Base pressure is , and base thrust is .
**Show the cancelling** when asked to derive : volume , mass , weight , then divide by . That cancellation is the marked step.
Quote the relevant law by name when explaining: pressure increases with depth for a dam wall, pressure acts in all directions for sideways thrust.
Pressure grows in direct proportion to depth — double the depth, double the pressure.
For a dam, say thicker where the pressure is greater, which is at the base, not where the reservoir is widest.
And state which liquid you used the density of; a question with two liquids is testing law 4.
** needs the area in .** Convert first — remember .
**Use with measured vertically from the FREE SURFACE, never from the top of the vessel and never along a slope.
Take and **, with .
Read whether the question wants the liquid's pressure or the TOTAL pressure — the total adds atmospheric pressure, about Pa.
Pressure is independent of area; thrust is not. Base pressure is , and base thrust is .
**Show the cancelling** when asked to derive : volume , mass , weight , then divide by . That cancellation is the marked step.
Quote the relevant law by name when explaining: pressure increases with depth for a dam wall, pressure acts in all directions for sideways thrust.
Pressure grows in direct proportion to depth — double the depth, double the pressure.
For a dam, say thicker where the pressure is greater, which is at the base, not where the reservoir is widest.
And state which liquid you used the density of; a question with two liquids is testing law 4.
Did you know
Why a diver ten metres down carries a second atmosphere
Stand at the seaside and the air above you is already pressing on every part of your body at about Pa. You never notice it, because it has been there all your life and it pushes equally from every direction.
Now swim down. Every m of water adds
which is almost exactly one more atmosphere. So at m the total is about two atmospheres, at m about three, at m about four.
Put another way: a column of water just m tall weighs about as much per square metre as the entire atmosphere stacked above it. The air goes up for many kilometres and the water goes down for ten metres, and they press equally — because water is roughly eight hundred times denser than air near the ground.
That single comparison explains why a mercury barometer is a sensible instrument and a water barometer is not. To balance the atmosphere with water you would need a column of
a tube three storeys high. Mercury is times denser, so it does the same job in
which fits on a laboratory bench. **Same pressure, same , and the density chooses the height for you.
It also explains why a diver's ears hurt after a few metres and not after a few centimetres. The ear responds to a difference** in pressure across the eardrum, and a few metres of water is enough to build a difference comparable to the whole atmosphere — while a whole day of walking about at sea level builds none at all.
Now swim down. Every m of water adds
which is almost exactly one more atmosphere. So at m the total is about two atmospheres, at m about three, at m about four.
Put another way: a column of water just m tall weighs about as much per square metre as the entire atmosphere stacked above it. The air goes up for many kilometres and the water goes down for ten metres, and they press equally — because water is roughly eight hundred times denser than air near the ground.
That single comparison explains why a mercury barometer is a sensible instrument and a water barometer is not. To balance the atmosphere with water you would need a column of
a tube three storeys high. Mercury is times denser, so it does the same job in
which fits on a laboratory bench. **Same pressure, same , and the density chooses the height for you.
It also explains why a diver's ears hurt after a few metres and not after a few centimetres. The ear responds to a difference** in pressure across the eardrum, and a few metres of water is enough to build a difference comparable to the whole atmosphere — while a whole day of walking about at sea level builds none at all.
Exam relevance
How is fluid pressure tested in JEE Main and NEET?
Because is the starting point of an entire Class 11 chapter, and the law that pressure acts in all directions is what makes buoyancy possible.
This is the foundation for Class 11 Physics Mechanical Properties of Fluids, examined in JEE Main and NEET. The expression derived on this page is used there in the form
with the pressure at the surface — the total pressure distinction drawn in the fourth section, now written as part of the formula. Every manometer, barometer and U-tube problem in that chapter is this equation applied at two points and set equal, using law 2 to pick points on the same horizontal level.
U-tube problems are the standard question type. Two immiscible liquids in the arms of a tube balance when their columns exert equal pressure at the common level:
which is laws 2 and 4 combined. Finding an unknown density from two measured heights is a recurring JEE Main numerical, and the mercury-and-water comparison in the previous section is its simplest case.
Law 3 is what upthrust is made of. Class 11 derives Archimedes' principle by noting that the pressure on the bottom face of a submerged body exceeds that on the top face, and the difference times the area gives an upward force — which is the last chapter of this ICSE group and the direct consequence of pressure acting in every direction.
Where the shape-independence reappears. The hydrostatic paradox is examined as an assertion-reason item in both exams, usually with two vessels of different shapes and the same liquid height. The answer always turns on pressure being a force per unit area, exactly as on this page.
Beyond statics. Class 11 adds flowing fluids, where Bernoulli's equation contains as its potential-energy term alongside a kinetic term — so the spouting-water application here becomes quantitative there, giving the speed of efflux as . That result explains why the lower hole jets farther, and it is a standard derivation.
What the questions look like. For board work, expect distinguish thrust from pressure with units, calculate pressure from a thrust and area, **derive showing the area cancel, state the laws of liquid pressure, explain the dam wall and the spouting tank, and find a total pressure at a stated depth. For JEE Main and NEET, expect U-tube and manometer numericals, speed of efflux, buoyancy problems, and Pascal's law applications from the next part of this chapter.
How board and competitive emphasis differ. A board paper rewards the derivation with the cancellation shown and the named law in each explanation. A competitive paper assumes the formula and tests whether you can equate pressures at a chosen level — the skill is picking the right level, not substituting.
The single trap that costs the most marks.** Measuring from the top of the vessel rather than from the liquid's free surface. A tank m tall filled to m has a base pressure of Pa and not Pa, and the wrong answer looks perfectly reasonable. **The defence is to mark the free surface on the diagram and draw the vertical arrow for ** before writing anything — and to check whether the atmosphere's contribution has been asked for as well.
This is the foundation for Class 11 Physics Mechanical Properties of Fluids, examined in JEE Main and NEET. The expression derived on this page is used there in the form
with the pressure at the surface — the total pressure distinction drawn in the fourth section, now written as part of the formula. Every manometer, barometer and U-tube problem in that chapter is this equation applied at two points and set equal, using law 2 to pick points on the same horizontal level.
U-tube problems are the standard question type. Two immiscible liquids in the arms of a tube balance when their columns exert equal pressure at the common level:
which is laws 2 and 4 combined. Finding an unknown density from two measured heights is a recurring JEE Main numerical, and the mercury-and-water comparison in the previous section is its simplest case.
Law 3 is what upthrust is made of. Class 11 derives Archimedes' principle by noting that the pressure on the bottom face of a submerged body exceeds that on the top face, and the difference times the area gives an upward force — which is the last chapter of this ICSE group and the direct consequence of pressure acting in every direction.
Where the shape-independence reappears. The hydrostatic paradox is examined as an assertion-reason item in both exams, usually with two vessels of different shapes and the same liquid height. The answer always turns on pressure being a force per unit area, exactly as on this page.
Beyond statics. Class 11 adds flowing fluids, where Bernoulli's equation contains as its potential-energy term alongside a kinetic term — so the spouting-water application here becomes quantitative there, giving the speed of efflux as . That result explains why the lower hole jets farther, and it is a standard derivation.
What the questions look like. For board work, expect distinguish thrust from pressure with units, calculate pressure from a thrust and area, **derive showing the area cancel, state the laws of liquid pressure, explain the dam wall and the spouting tank, and find a total pressure at a stated depth. For JEE Main and NEET, expect U-tube and manometer numericals, speed of efflux, buoyancy problems, and Pascal's law applications from the next part of this chapter.
How board and competitive emphasis differ. A board paper rewards the derivation with the cancellation shown and the named law in each explanation. A competitive paper assumes the formula and tests whether you can equate pressures at a chosen level — the skill is picking the right level, not substituting.
The single trap that costs the most marks.** Measuring from the top of the vessel rather than from the liquid's free surface. A tank m tall filled to m has a base pressure of Pa and not Pa, and the wrong answer looks perfectly reasonable. **The defence is to mark the free surface on the diagram and draw the vertical arrow for ** before writing anything — and to check whether the atmosphere's contribution has been asked for as well.
Key takeaways
Thrust, pressure and the laws of liquid pressure: quick revision
- Thrust is the force acting normally on a surface: a vector, in newtons.
- Pressure is thrust per unit area: , a scalar, in or pascals.
- A N block on gives Pa; on it gives Pa. Same thrust, five times the pressure.
- A kg person ( N) on of feet gives Pa, and Pa on one foot.
- A thrust of N at Pa acts over .
- A sharp knife, a drawing pin, a camel's foot and a tractor tyre are all area arguments, not force arguments.
- **Derivation of **: column volume , mass , weight , divided by — and the area cancels.
- Water at m: Pa. At m: Pa. Double the depth, double the pressure.
- Mercury at m: Pa — standard atmospheric pressure.
- Pressure Pa in water means a depth of m; a m tank of density gives Pa.
- Total pressure adds the atmosphere: a diver at m has Pa.
- The five laws: pressure increases with depth; is equal at the same horizontal level; acts equally in all directions; depends on density; and is independent of the area and shape of the vessel.
- Dam wall thicker at the base — greatest pressure at the greatest depth; at m it is Pa.
- Water spouts farther from a lower hole, since greater depth means greater pressure and greater speed.
- Sideways thrust on walls and upward push on an immersed body both follow from law 3; that upward push is upthrust.
- A kettle's spout shows the level by law 2.
- At m depth: water Pa, mercury Pa — the ratio is , the density ratio.
- The hydrostatic paradox: a narrow tube (base , weight N) and a wide drum (base , weight N) filled to m both give Pa.
- Thrust does depend on area: a m pool with a base has Pa of pressure and N of thrust.
- **Measure vertically from the free surface** — an empty part of the vessel contributes nothing.
Fill a tall bottle with water, punch two small holes at different heights and watch where each jet lands — then work out the pressure at both holes and see whether the order matches.
- Pressure is thrust per unit area: , a scalar, in or pascals.
- A N block on gives Pa; on it gives Pa. Same thrust, five times the pressure.
- A kg person ( N) on of feet gives Pa, and Pa on one foot.
- A thrust of N at Pa acts over .
- A sharp knife, a drawing pin, a camel's foot and a tractor tyre are all area arguments, not force arguments.
- **Derivation of **: column volume , mass , weight , divided by — and the area cancels.
- Water at m: Pa. At m: Pa. Double the depth, double the pressure.
- Mercury at m: Pa — standard atmospheric pressure.
- Pressure Pa in water means a depth of m; a m tank of density gives Pa.
- Total pressure adds the atmosphere: a diver at m has Pa.
- The five laws: pressure increases with depth; is equal at the same horizontal level; acts equally in all directions; depends on density; and is independent of the area and shape of the vessel.
- Dam wall thicker at the base — greatest pressure at the greatest depth; at m it is Pa.
- Water spouts farther from a lower hole, since greater depth means greater pressure and greater speed.
- Sideways thrust on walls and upward push on an immersed body both follow from law 3; that upward push is upthrust.
- A kettle's spout shows the level by law 2.
- At m depth: water Pa, mercury Pa — the ratio is , the density ratio.
- The hydrostatic paradox: a narrow tube (base , weight N) and a wide drum (base , weight N) filled to m both give Pa.
- Thrust does depend on area: a m pool with a base has Pa of pressure and N of thrust.
- **Measure vertically from the free surface** — an empty part of the vessel contributes nothing.
Fill a tall bottle with water, punch two small holes at different heights and watch where each jet lands — then work out the pressure at both holes and see whether the order matches.