fizikzauglom

GRADE 7 · BUOYANT FORCE AND FLOATING

Buoyant force

The origin and measurement of buoyant force.

22 minutes6 theory cards
Start interactive lesson

Read the full lesson theory here without an account. In the interactive lesson, the same theory is presented as cards, followed by questions with feedback and practice problems. Your progress is saved.

Understanding step by step

A stone hanging from a force meter gives a lower reading when immersed in water. Its mass has not disappeared. Which forces have changed, and why does water push the stone upwards at all?

In airIn waterhigher readinglower
The same stone hangs stationary from identical force meters, first in air and then fully in water, without touching the vessel. In water, the spring stretches less and the reading is lower: water balances part of the gravitational force. No numerical readings are specified here.

A stronger push from below

Imagine a rectangular block immersed in water. Water pushes down on its top face and up on its bottom face. The bottom is deeper, so its pressure and force are greater for equal face areas. Sideways forces at equal depths cancel. Adding all the forces leaves an upward force: buoyant force, FА. Water does not act only from below; the lift comes from pressure acting over the entire immersed surface.

Column forcesTheir resultantFАThe resultant force points upwards
The left diagram shows forces due to the weight of the water column; equal pressure above the water cancels. The right shows their resultant FА. Vertical arrows are shifted sideways for clarity. The bottom force exceeds the top force, leaving an upward resultant; side forces are equal. FА is not added again to the original forces.

Imagine replacing the object with water

How can we avoid calculating pressure on every face of a complicated object? Imagine replacing it with the same liquid, filling the space it occupied. In a stationary medium, this portion would remain in equilibrium: external pressure forces would balance its gravitational force. The surrounding liquid acts on the same boundary in the same way when the object replaces that portion. Therefore, buoyant force equals the weight of the displaced liquid. This is Archimedes' principle.

Object in waterEquivalent waterFАdisplaces volume VFАmg of waterat rest: forces equalThe object gets the same buoyant force
An equal volume of water could remain at rest in the object's place. The surrounding water would support it, balancing its gravitational force, and it acts on the object with that same force.

Count only the immersed part

We need the volume from which the object excludes the medium, Vпогр, measured in m³. For a solid or sealed object fully immersed, it is the external volume; for a floating object, it is the underwater part. A cavity freely filled by water does not itself displace water. Imagine an open cup and a sealed box of the same external size: once water enters the cup, their displaced volumes can differ.

Underwater partSame VпогрFАmgVпогр < VтелаDisplaced water
Only the underwater part of the partially immersed block is highlighted. The right shows a portion of water of exactly the same volume and shape. FА equals its gravitational force in magnitude, and hence its weight at rest: ρжgVпогр. Do not substitute the entire volume of the block sticking out above the surface.

Archimedes' principle

The mass of the displaced uniform medium is ρсредыVпогр, and its weight is this product multiplied by g. Thus FА = ρсреды·g·Vпогр. Doubling the immersed volume doubles the force; a denser medium pushes more strongly on the same volume. Using kg/m³, N/kg and m³ gives N. The object's own density does not appear in this formula: two fully immersed blocks of equal volume experience equal buoyant forces, although their gravitational forces may differ.

FА = ρсреды · g · Vпогрρ of mediumкг/м³gН/кгVпогрм³result in newtonsobject density is not in the formula
Multiply the medium's density by gravitational field strength and displaced volume to obtain buoyant force. The object's own density does not enter this formula.

When the principle applies

Lower a rigid, fully immersed block deeper into a uniform incompressible liquid. Pressure above and below increases equally, so the difference stays the same; Vпогр and FА also remain unchanged. Here the medium is stationary and surrounds the object, including underneath. If the object is pressed tightly against the bottom so water cannot reach beneath it, the set of pressure forces changes: the usual formula needs a separate analysis. A compressible object may also change its displaced volume.

One object at different depthsFАFАSame displaced volume, same force
The same rigid object, fully immersed at two depths without touching the bottom, displaces the same volume of uniform liquid. Its buoyant force is therefore unchanged, although pressure itself increases with depth.

Measuring with a force meter

Hang an object from a force meter and hold it stationary first in air, then in water, without touching the sides or bottom. In water, both the suspension and buoyancy act upwards, so the suspension needs less force for equilibrium. Neglecting air buoyancy gives FА = Pвозд − Pжидк, where P denotes force-meter readings. The difference measures the support provided by water. The gravitational force on the object changes negligibly when immersed.

Н036Н0365 Н3 НAirWaterFА = 5 − 3 = 2 Н
The force meter reads 5 N in air and 3 N in water. Under otherwise identical conditions, water's buoyant force is 5 − 3 = 2 Н. Air buoyancy is neglected.

Not just water

The same reasoning applies to a gas: its pressure in a gravitational field increases downwards, and the difference in pressure forces produces an upward buoyant force. Use the gas density in the formula, not water density. For a small dense object, air buoyancy is weak because air has low density; a large balloon displaces a substantial mass of air. To predict lift, we must also compare buoyancy with the total gravitational force on the balloon. That comes next.

Balloon in airFАmgUse air density, not water density
Gas pressure also increases downwards, so air exerts buoyancy on objects within it. For a balloon, use air density. Here FА exceeds mg, so the resultant points upwards. The system's gravitational force is shown schematically inside the envelope; its exact centre of mass is not specified.

An immersed block

A block is fully immersed in water; V = 0,002 м³, ρводы = 1000 кг/м³, g = 10 Н/кг. Find FА.

Fully immersedV = 0,002 м³Waterρ = 1000 кг/м³g = 10 Н/кгBuoyant force — ?
A block of volume 0.002 m³ is fully immersed in water of density 1000 kg/m³ with g = 10 Н/кг. It touches neither the bottom nor the sides. Find its buoyant force; the data and apparatus remain the same throughout the steps.

Let's look at an example · 1

We need the volume of displaced water. For full immersion, it equals the block's volume.

The whole block is underwaterV = 0,002 м³Waterρ = 1000 кг/м³g = 10 Н/кгVпогр = V = 0,002 м³
The original block remains unchanged. All of it lies below the surface, so Vпогр = V = 0,002 м³. Partial immersion does not apply to this problem.

Let's look at an example · 2

FА = 1000·10·0,002 = 20 Н.

Calculate buoyant forceV = 0,002 м³Waterρ = 1000 кг/м³g = 10 Н/кгFАFА = 1000 · 10 · 0,002 = 20 Н
For the same apparatus, use water density and the block's entire volume. The product 1000 · 10 · 0,002 is 20 Н. Buoyant force points upwards.

Let's look at an example · 3

This is the weight of two kilograms of displaced water. The block's own mass is not needed for this calculation.

What 20 newtons meansV = 0,002 м³Waterρ = 1000 кг/м³g = 10 Н/кг2 kg of water displacedWeight of this water: 2 · 10 = 20 НThe block's mass is not needed
The original block and data remain unchanged. The displaced water has mass 1000 · 0,002 = 2 кг; its weight at rest is 2 · 10 = 20 Н. The block's own mass is unnecessary for calculating buoyant force.

Let's check the result

The buoyant force is 20 N upwards.

Key idea · 1

Buoyancy comes from the difference in pressure forces, not from gravity disappearing.

Key idea · 2

FА equals the weight of the displaced medium: use its density and the volume actually displaced.

Key idea · 3

Greater depth alone does not increase buoyancy on a rigid object in a uniform liquid; how the medium contacts the object matters.

Where do we go next?

To predict floating, compare the buoyant force we found with the gravitational force on the object.

Basic Concepts

3 concepts

Buoyant force

The upward resultant of the pressure forces exerted by a liquid or gas on an immersed object.

The bottom of an object is usually deeper and experiences higher pressure, producing a net upward force.

Archimedes' principle

Buoyant force equals the weight of the liquid or gas occupying the volume of the immersed part of an object.

The force depends on the medium's density, gravitational acceleration and displaced volume, but not on depth in a uniform medium.

Measuring buoyant force

Buoyant force is found from the difference between force-meter readings in air and with the object immersed.

The experiment shows a reduction in apparent weight and lets us compare the result with the weight of displaced liquid.

Physical quantities

2 quantities

Buoyant force

F_A

The buoyant force exerted by a liquid or gas on an immersed object.

Unit: newton · \mathrm{Н}

Displaced volume

V_{\text{погр}}

The volume of the immersed part of an object and of the medium it displaces.

Unit: cubic meter · \mathrm{м}^3

Lesson formulas

1 formula

Buoyant force

F_A = \rho_{\text{ж}}gV_{\text{погр}}

Buoyant force equals the weight of the medium occupying the volume of the object's immersed part.

Test the relationship

Make a prediction first. The button changes the stated quantity; then compare the result and explore other values.

Floating and density

Fₐ = ρжgVпогр; mg = ρтgV; Vпогр/V = min(1, ρт/ρж)

Uniform object in water of density 1000 kg/m³, g = 10 m/s². No surface tension. A sinking object is shown fully submerged before touching the bottom, not in equilibrium.

The volume of an object with the same density increases from 2 to 4 L. How will the submerged fraction change?

Your prediction

Initial values

Show vectors
Fₐmg

Calculated result

Vпогр/V
60 %
Fₐ
12 Н
mg
12 Н

Floats: Fₐ = mg

Explaining the result

For floating, Fₐ = mg: volume cancels from the density ratio. At ρт > ρж, even the maximum buoyant force is smaller than gravity.

Related formulas

The same formulas in the reference, with symbols and units.

Problems on this topic

Statements, answer options and the walkthrough.