K326 / K327 / 2027
Force and pressure overview

Topic 3 of 3

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 = perpendicular force / contact area
p = F / A
p = pressure in pascals (Pa); F = perpendicular force in N; A = area in m2. 1 Pa = 1 N/m2.

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.

Same block, different contact face

Block dimensions: 15 cm x 10 cm x 5 cm. Its weight is 30 N in both stationary arrangements.

On the 15 cm x 10 cm face

A 30 N block on a 150 square centimetre contact faceThe same rectangular block rests on a horizontal floor. A side view shows its height of 5 centimetres in this orientation. The green lower edge touches the floor. A separate view from below shows the actual contact face, 15 centimetres by 10 centimetres, with an area of 150 square centimetres. The block exerts a 30 N contact force downwards on the floor, perpendicular to the contact face. Dividing this force by 0.015 square metres gives an average pressure of 2000 pascals. Both orientations use the same force-arrow length and the same scale for dimensions.Side viewFloor5 cm30 Non floorContact face, viewed from below150 cm215 cm10 cm

Contact area = 15 x 10 = 150 cm2

150 cm2 = 0.015 m2

Average pressure = 30 / 0.015 = 2000 Pa.

On the 10 cm x 5 cm face

A 30 N block on a 50 square centimetre contact faceThe same rectangular block rests on a horizontal floor. A side view shows its height of 15 centimetres in this orientation. The green lower edge touches the floor. A separate view from below shows the actual contact face, 10 centimetres by 5 centimetres, with an area of 50 square centimetres. The block exerts a 30 N contact force downwards on the floor, perpendicular to the contact face. Dividing this force by 0.0050 square metres gives an average pressure of 6000 pascals. Both orientations use the same force-arrow length and the same scale for dimensions.Side viewFloor15 cm30 Non floorContact face, viewed from below50 cm210 cm5 cm

Contact area = 10 x 5 = 50 cm2

50 cm2 = 0.0050 m2

Average pressure = 30 / 0.0050 = 6000 Pa.

The perpendicular force stays the same. One third of the contact area gives three times the average pressure.

The same block rests first on its 15 cm x 10 cm face, then on its 10 cm x 5 cm face. The highlighted contact areas are 150 cm2 and 50 cm2. The force on the floor remains 30 N.

Worked comparison

Change the area while keeping the force fixed

Convert the contact areas to square metres before calculating in pascals. Since 1 cm2 = 0.0001 m2:

  • 150 cm2 = 150 x 0.0001 = 0.015 m2.
  • 50 cm2 = 50 x 0.0001 = 0.0050 m2.
The perpendicular force is 30 N in both cases
Contact areaPressure calculation
0.015 m2p = 30 / 0.015 = 2000 Pa
0.0050 m2p = 30 / 0.0050 = 6000 Pa

The contact area falls to one third, so the pressure becomes three times as large. The block's weight, mass, volume and material density are unchanged. Only its orientation and contact area have changed.

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.

Worked example

Find the force from a pressure

A pad exerts a uniform pressure of 1500 Pa over a contact area of 0.020 m2. Find the perpendicular force it exerts.

  1. Rearrange p = F / A: F = pA.
  2. Substitute: F = 1500 N/m2 x 0.020 m2 = 30 N.

The area units cancel, leaving newtons. The force acts perpendicular to the surface. If area were the unknown, rearrange to A = F / p.

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.

Optional check A pad's perpendicular force doubles while its contact area becomes four times as large. How does its average pressure change?
A pad's perpendicular force doubles while its contact area becomes four times as large. How does its average pressure change?