A horse weighs roughly 5 000 newtons. Spread that weight across four hooves and the ground handles it fine. Now imagine the same force through a 1 cm² stiletto heel. The floor dents — and even soft ground can collapse. This lesson is about pressure: how much force acts on each tiny piece of area.
Key formula:
$$P = \frac{F}{A}$$
$P$ = pressure (N/cm² or N/m² — the unit N/m² is called a Pascal, Pa)
$F$ = force in newtons (N) — on Earth, weight in kg × 10 ≈ force in N
$A$ = contact area (cm² or m²)
Rearranged: $F = P \times A$ · $A = F \div P$
Math connection — Lesson 12 (Proportionality): Pressure and area are inversely proportional: double the area and pressure halves; halve the area and pressure doubles. That is $P \propto \tfrac{1}{A}$. It is the same idea as the speed–time relationship when distance is fixed.
Worked Example
Worked Example — One hoof
A horse has weight 5 000 N. It stands on one hoof with contact area 80 cm². Find the pressure.
$$P = \frac{F}{A} = \frac{5\,000}{80} = 62.5 \text{ N/cm}^2$$
If the horse wears a shoe that increases the area to 100 cm²: $P = 5\,000 \div 100 = 50$ N/cm². The shoe reduces pressure by 20%.
The surprising fact: A woman in stiletto heels (area ≈ 1 cm², weight ≈ 600 N) exerts around 600 N/cm² on the floor. A horse on one hoof exerts ≈ 63 N/cm². The stilettos win by almost 10×. This is why high heels mark wooden floors and horses don't.
Warm-Up — formula given
-
Problem 1
A force of 900 N acts on an area of 45 cm². Calculate the pressure.
$P = 900 \div 45$
divide →
-
Problem 2
A horse of mass 500 kg stands on all four hooves. Each hoof has contact area 80 cm².
(a) Calculate the horse's weight in newtons. (Weight ≈ mass × 10.)
(b) Assuming weight is shared equally, find the pressure on each hoof.
$F = 500 \times 10$ · $P = F \div (4 \times 80)$
weight → force per hoof → pressure →
-
Problem 3
A person of mass 60 kg stands on one stiletto heel with contact area 1 cm². Calculate the pressure. Then compare it to Problem 2 — how many times larger is it?
weight → pressure → ratio to P2 →
Core Problems — write the equation yourself
-
Problem 4
A horseshoe increases a hoof's contact area from 75 cm² to 125 cm². The horse's total weight is 4 800 N and it stands on four hooves.
(a) Find the pressure per hoof without shoes.
(b) Find the pressure per hoof with shoes.
(c) By what percentage does the shoe reduce pressure?
force per hoof = total ÷ 4 → calculate both pressures → percentage change →
-
Problem 5
A pressure of 30 N/cm² acts over an area of 240 cm². What is the total force?
$F = P \times A$
rearrange formula → multiply →
-
Problem 6
A rectangular ice-skate blade is 30 cm long. A skater weighs 600 N and balances on one skate. The pressure on the ice must not exceed 12 N/cm² or the blade sinks too far.
How wide (in cm) must the blade be at minimum? Round up to 1 decimal place.
A = F ÷ P → area of rectangle = length × width → solve for width →
-
Problem 7 Challenge
An African elephant of mass 6 000 kg stands on four feet, each with area 1 500 cm². A racehorse of mass 520 kg gallops on one hoof (contact area 75 cm²).
(a) Calculate the pressure for each animal.
(b) Which exerts more pressure, and by how many times?
(c) The elephant then lies down, distributing its weight over 6 000 cm² of body surface. Does this help or make it worse? Explain.
elephant: P = F ÷ A → horse: same → compare → lying down pressure →
-
Problem 8 Open
Design a simple experiment using a block of soft foam or a tray of sand to demonstrate pressure. List the objects you would press in, predict the order from highest to lowest pressure (without calculating), then describe how you would measure which left the deepest mark. What variables must you keep the same to make it a fair test?
materials → prediction (ranked) → method → fair test variables →
Show answers
Problem 1
$P = 900 \div 45 = 20$ N/cm²
Problem 2
(a) $F = 500 \times 10 = 5\,000$ N · (b) $P = 5\,000 \div (4 \times 80) = 5\,000 \div 320 = 15.625$ N/cm²
Problem 3
$F = 60 \times 10 = 600$ N · $P = 600 \div 1 = 600$ N/cm² · Ratio: $600 \div 15.625 \approx 38.4\times$ larger.
Problem 4
Force per hoof $= 4\,800 \div 4 = 1\,200$ N · (a) $P = 1\,200 \div 75 = 16$ N/cm² · (b) $P = 1\,200 \div 125 = 9.6$ N/cm² · (c) Reduction $= (16-9.6) \div 16 = 40\%$
Problem 5
$F = 30 \times 240 = 7\,200$ N
Problem 6
$A = 600 \div 12 = 50$ cm² · Width $= 50 \div 30 = 1.7$ cm (rounded up)
Problem 7
(a) Elephant: $F = 60\,000$ N, $A = 4 \times 1\,500 = 6\,000$ cm², $P = 10$ N/cm² · Horse (gallop): $F = 5\,200$ N, $P = 5\,200 \div 75 \approx 69.3$ N/cm² · (b) Horse: ≈ 6.9× more pressure · (c) Lying: $P = 60\,000 \div 6\,000 = 10$ N/cm² — same as standing. Still much less than a galloping horse.
Coming up next → s07: Density and Floating
Can a 500 kg horse float? Spoiler: yes! You'll learn why objects float or sink, calculate the density of everyday objects, and discover what fraction of an iceberg is hidden underwater.