GRADE 7 · BUOYANT FORCE AND FLOATING
Buoyant force
The origin and measurement of buoyant force.
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?
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.
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.
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.
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.
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.
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.
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.
An immersed block
A block is fully immersed in water; V = 0,002 м³, ρводы = 1000 кг/м³, g = 10 Н/кг. Find FА.
Let's look at an example · 1
We need the volume of displaced water. For full immersion, it equals the block's volume.
Let's look at an example · 2
FА = 1000·10·0,002 = 20 Н.
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.
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
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
Buoyant force
F_AThe 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
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?
Initial values
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.