Pressure

01

Pressure at a surface

Pressure describes the perpendicular force acting per unit area of a surface. The same force produces a greater pressure when spread over a smaller area.

Force is measured in newtons; area is measured in square units. Review mass and weight and square-unit conversions if needed. Area covers a surface; volume measures the space occupied by an object.

Pressure may also be written P. When P means pressure, its unit is Pa; when P means power, its unit is W. Use the named quantity and unit to identify the meaning.

The formula gives the average pressure over the area. If the force is spread uniformly, the pressure is the same across that area. Identify the surface and the force perpendicular to it before calculating.

The same block resting on two faces

A block weighing 30 N rests on a horizontal floor. Only its weight and the floor's upward support act vertically on the block. In either stationary arrangement below, it presses on the floor with a perpendicular force of 30 N.

Use the area of the face touching the floor, not the sum of all the block's faces. Dividing 30 N directly by 150 cm² gives 0.20 N/cm², which is a valid pressure unit, but it is not 0.20 Pa. Convert the area or the final pressure unit correctly.

Determine pressure from measurements

Use a rigid block resting on its flat rectangular face on a horizontal surface. Assume that only its weight and the surface's upward support act vertically.

1. Measure the block's mass: choose a balance whose range includes the mass and whose resolution is suitable for the reading. Check its zero before adding the block and wait for a steady reading.
2. Measure the actual contact face: use a ruler or calipers with suitable range and resolution for both perpendicular side lengths. Align the instrument with each side; check the caliper zero or subtract the ruler's endpoint readings. Read a ruler with the line of sight perpendicular to its scale.
3. Determine the perpendicular force: use the measured mass and supplied gravitational field strength in F = mg. Because the block is stationary, weight and upward support balance; the block presses on the surface with the same force magnitude.
4. Calculate the pressure: multiply the measured side lengths to obtain the contact area, convert it to m², then divide the force in N by this area.

For the block above, a mass reading of 3.0 kg with supplied g = 10 N/kg gives F = 30 N. A measured 15 cm x 10 cm face gives 0.015 m², so the average pressure is the calculated 2000 Pa.

The chosen area must actually support the block. If it rests only on small feet, use their contact areas instead of the whole rectangular footprint. Overestimating the area makes the calculated pressure too small; repeating the length reading does not fix a wrong choice of contact area.

Compare the force and area together

• Same force, larger area: lower pressure. Snowshoes spread the same person's weight over a larger contact area than ordinary shoes.
• Same area, larger perpendicular force: greater pressure.
• Both force and area change: compare F / A. If both double, the pressure stays the same.

The condition matters. A smaller contact area does not always mean a greater pressure if the force also becomes smaller. Calculate the ratio or compare the factors rather than using area alone.

Pressure is not an extra force. Force is measured in N; pressure describes force per area and is measured in Pa. Do not add a separate "pressure force" alongside the same contact force and count the interaction twice.

02

Density from measurements

Density is mass per unit volume. It tells us how much mass occupies a given amount of space.

Compare two samples that each occupy 10 cm³. If one has a mass of 27 g and the other 80 g, the second is denser: it has more mass in the same volume. A larger total mass alone does not show greater density when the volumes are different.

Mass is measured in g or kg; volume is measured in cubic units such as cm³ or m³. Review prefixes and volume conversions if needed. Density uses mass, not the weight in newtons.

Density is a scalar quantity: it has no spatial direction. For a uniform material under the same conditions, doubling the volume doubles the mass, leaving the ratio m / V unchanged.

An irregular solid: measure the displaced volume

Use a balance to find the solid's mass. To find the volume of a suitable irregular solid, record the water volume in a measuring cylinder before and after fully submerging the solid. The increase is the solid's volume.

The solid must be fully submerged, insoluble and non-absorbing, with no water lost. Remove trapped air bubbles: they displace extra water, making the calculated volume too large and the calculated density too small. An object that floats is only partly under water, so the rise in water level gives only part of its volume. Use this method for a floating object only if it is held completely under the water.

Choose a cylinder that fits the object and final level while still having useful divisions. Keep it upright and read the bottom of the water meniscus at eye level. The measurement methods page explains the reading technique.

Convert both parts of a density unit

Changing g/cm³ to kg/m³ changes both the mass unit and the volume unit.

Therefore 2.7 g/cm³ = 2700 kg/m³. Dividing 2.7 by 1000 would convert the grams but leave the cubic centimetres unchanged. That would give kg/cm³, not kg/m³.

A regular solid: calculate volume from its shape

For a cuboid, measure its mass and its three perpendicular dimensions. Its volume is length x width x height. Use an instrument with a suitable range and resolution for each dimension.

V = 4.0 x 3.0 x 2.0 = 24 cm³, so ρ = 64.8 / 24 = 2.7 g/cm³. Here geometry provides the volume that displacement provided for the irregular sample.

Measure the actual solid, not a surrounding container or gaps around it. Repeat dimension readings at different positions to judge small variations. If the shape is substantially irregular, assuming a cuboid can give a misleading volume; choose a suitable displacement method instead.

A liquid: subtract the container's mass

1. Weigh a clean, dry, empty measuring cylinder, or zero the balance with the empty cylinder on it.
2. Add the liquid and record its volume at eye level using the appropriate meniscus reading. Keep the outside of the cylinder dry.
3. Weigh the cylinder and liquid. Subtract the empty cylinder's mass, unless it was correctly tared.
4. Divide the liquid's mass by the volume of that same liquid.

For example, an empty cylinder has a mass of 42.6 g. With 50 cm³ of liquid inside, its mass is 82.6 g. The liquid's mass is 82.6 - 42.6 = 40.0 g, giving a density of 40.0 / 50 = 0.80 g/cm³.

Keep the mass and volume measurements matched. If liquid spills after the volume is recorded but before weighing, the smaller measured mass divided by the old volume gives too small a density. Liquid on the outside adds mass without adding to the measured inside volume, so it can make the density too large.

Keep observations separate from calculations. Record the balance and cylinder readings with their units, then show the calculated liquid or solid volume and density. This makes a mistaken subtraction or unit conversion easier to find.

03

Pressure in a liquid

In a stationary liquid, pressure increases with vertical depth. The increase depends on the liquid's density and gravitational field strength.

Pressure is force per unit area, measured in Pa. Density tells us the mass per unit volume. Use density in kg/m³ and depth in m in the calculation below.

Pressure is a scalar quantity. The force that a liquid exerts on a surface acts perpendicular to that surface, so liquid can push on a container's side walls as well as its bottom.

This relationship assumes a stationary liquid of uniform density in a uniform gravitational field. The height h is measured vertically from the liquid surface, not along a sloping wall or tube.

Why width is not in the relationship

Consider a vertical liquid column of area A and height h. Its volume is Ah, its mass is ρAh, and its weight is ρAhg. Dividing that weight by area A gives the pressure contribution ρgh. A wider column has more weight, but that weight is spread over a larger area.

At the same horizontal level in the same connected stationary liquid, the pressure is equal. Different widths or shapes of the connected regions do not change that equality.

Extra pressure and total pressure are different

The liquid contribution ρgh is an increase below the surface. There may already be pressure acting on that surface. For an open container, the surrounding atmosphere supplies that surface pressure.

For the same liquid and g, doubling depth doubles the liquid contribution. It does not double the total pressure when the surface pressure stays unchanged. At a fixed depth, a denser liquid gives a larger pressure increase.

Choose and measure the vertical height

Use a vertical scale and identify the two levels being compared. Read both levels from the same reference and subtract. The distance along an inclined tube is longer than the vertical separation and would give too large a pressure difference if substituted for h.

Identify the requested pressure before calculating. ρgh gives a pressure difference. Add the stated surface pressure when the question asks for total pressure; do not silently assume an open surface has zero pressure.

04

Hydraulic systems

An applied increase in pressure is transmitted through a confined liquid. Acting over a larger piston area, that increase can produce a larger force.

Use p = F / A and convert square units carefully. An increase in pressure is different from the total pressure already present in the liquid.

A hydraulic press connects a small input piston to a larger output piston through liquid. Pushing the input piston increases the pressure in the liquid. The pressure increase is transmitted throughout the confined liquid, including to the output piston.

For the simple model here, the piston faces are at the same height, the same atmospheric pressure acts above both pistons, and piston weights and friction are neglected. Compare the additional forces caused by the applied pressure increase.

The atmosphere also exerts pressure above the pistons. In this model its equal contribution is accounted for on both sides, leaving the extra force due to the applied increase. The 4000 N is the force available to balance an additional load at the output, with the piston in equilibrium.

The larger force comes with a smaller movement

Treat the liquid as incompressible, with no leaks. The volume pushed in by the small piston must be accommodated at the other piston. The large-area piston therefore rises a shorter distance for that displaced volume.

The press trades movement for force; it does not create energy. The force comparison above does not mean the large piston can move the same distance as the small one while providing the larger force in this ideal system.

Equal pressure increase does not mean equal force. Multiply that increase by each piston area. Check that the piston levels and the stated assumptions justify the simple force comparison.

05

Measuring atmospheric pressure

A liquid-column barometer measures atmospheric pressure from the vertical height of a supported liquid column.

In a stationary liquid, a vertical height difference gives a pressure difference of ρgh. A pressure reference is needed to turn that difference into an atmospheric-pressure reading.

The supplied barometer diagram shows a closed-top tube connected at its lower end to a reservoir of mercury. The reservoir is open to the atmosphere. The space above the mercury column has such a small pressure that we treat it as a vacuum in this model.

Follow the two pressure references

1. At the open reservoir surface, the pressure is atmospheric pressure.
2. At that same horizontal level inside the connected mercury, the pressure is equal.
3. Inside the tube, that pressure is the pressure above the column plus the mercury contribution ρgh.
4. With the top pressure approximately zero, atmospheric pressure is approximately ρgh.

Why the liquid's density matters

For the same pressure, a less dense liquid needs a taller column. If water of density 1000 kg/m³ were used to represent the same 102000 Pa, then:

For a given liquid with unchanged density, a greater atmospheric pressure supports a greater vertical height. Use the current reservoir level as the lower reference when reading h.

The space above the column is part of the explanation

If appreciable gas is trapped above the column, p_top is not zero. Then atmospheric pressure is p_top + ρgh. Using ρgh alone would underestimate it.

Read the pressure unit mmHg

Millimetres of mercury, mmHg (also written mm Hg), is a conventional pressure unit based on a mercury column under fixed reference conditions. A measurement in mm alone is a length; a reading in mmHg is a pressure.

This conversion uses the conventional pressure unit. It does not use the earlier column model's approximation g = 10 N/kg. When calculating pressure from a measured column height using ρgh, use that problem's stated density, g and vertical height.

Identify the top reference and the vertical height. The atmosphere acts on the reservoir and supports the column. The pressure above the column cannot be ignored unless the stated model makes it negligible.

06

Manometers and pressure differences

A manometer uses the level difference in a connected liquid to compare two pressures. First identify which side's liquid surface is lower.

At the same horizontal level in a connected stationary liquid, pressure is equal. A vertical difference h gives a pressure difference of ρgh.

An open-end manometer has one arm connected to a gas and the other open to the atmosphere. The greater pressure pushes its side's liquid surface lower. The two surface pressures are therefore the gas pressure and the atmospheric pressure.

Gas-side surface lower: add the difference

Choose the horizontal reference at the lower, gas-side surface. Its pressure is the gas pressure. At that same level in the open arm, the point is h below the atmosphere-exposed surface, so its pressure is atmospheric pressure plus ρgh.

Gas-side surface higher: subtract the difference

Now the open-side surface is lower. At that horizontal level, atmospheric pressure equals the gas pressure plus the pressure of the liquid column below the higher gas-side surface:

With the same supplied water density, g and 0.120 m separation, the difference is still 1200 Pa. Gas pressure is now 101000 - 1200 = 99800 Pa. The gas pressure is positive but lower than atmospheric pressure.

If both levels are equal, the pressure difference is zero. That means the gas and atmosphere have equal pressures, not that their pressures are both zero.

Measure the full vertical separation

Read both water menisci from the same vertical scale and use the bottom of each meniscus at eye level. Wait for a steady liquid. These example readings are consistent with the higher-gas-pressure case:

The separation is 254 - 134 = 120 mm = 0.120 m. Use the difference between the two surfaces, not the length around the U-bend or the movement of just one surface from an earlier position.

If h is overestimated, the magnitude of the pressure difference is overestimated. In the higher-gas-pressure case this makes the calculated gas pressure too high; in the lower-gas-pressure case it makes it too low because too much is subtracted. A vertical scale and eye-level readings address a specific measurement cause.

A difference needs a reference before it gives a gas pressure. The height and liquid density give ρgh. To find gas pressure relative to a vacuum, also use the atmospheric pressure or another supplied reference. Do not assume that ρgh alone is the absolute gas pressure.