Floating, Sinking & Force Chains | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Floating, Sinking & Force Chains

Upthrust rarely appears alone in Paper 2. It sits inside a force balance with weight, drag, tension or a hinge.

The first decision is whether the object is in equilibrium or accelerating. Draw the forces, write the correct resultant-force equation, then solve for the missing piece.

Float or sink

Rises or floats	Neutral	Sinks
mean density below the fluid density	mean density equal to the fluid density	mean density above the fluid density
settles partly submerged with U = W	can remain fully submerged with U = W	has a downward resultant before drag grows

A floating object pushes aside just enough fluid for the upthrust to equal its weight. It then sits in equilibrium: $U = W$ .
Here density means the object's mean density: total mass divided by total external volume. A hollow steel ship can float because the steel and enclosed air together have a mean density below that of water.

For a floating object, upthrust equals weight. Replacing each force by density × volume × g gives a useful relationship:

\frac{V_{\text{submerged}}}{V_{\text{object}}} = \frac{\rho_{\text{object, mean}}}{\rho_{\text{fluid}}}

A larger mean density therefore means that a larger fraction must be below the surface. If both the object and fluid densities change, the new floating fraction cannot be predicted from only one change.

A light jar floating high in a tank and an identical heavier jar resting fully submerged on the floor. — Same external volume, different mean density: the lighter jar floats where upthrust equals weight; the denser jar sinks.

Slide the block's density and watch the balance between weight and upthrust decide how it floats:

Float or sink

Water has density 1000 kg m⁻³. Slide the block's density and watch how deep it sits. Upthrust always equals the weight of the water pushed aside.

Block density600 kg m⁻³

Weight

5.9 N

Upthrust

5.9 N

Result

floats, 60% under

Worked example

How deep does a floating cylinder sit?

A cylinder of cross-section area $\pi (0.39)^2 \text{ m}^2$ floats upright in liquid of density 990 kg m⁻³. The total downward force on it (its weight plus a load) is 1400 N. How deep is its base below the surface?

Floating equilibrium: upthrust = 1400 N.
The submerged volume is a cylinder of depth y: $V = Ay$ , so $\rho g A y = 1400$ .
$y = \dfrac{1400}{990 \times 9.81 \times \pi (0.39)^2} = 0.30 \text{ m}$ .

Answer

y = 0.30 \text{ m}

The equilibrium chains

Falling through a fluid: weight down, upthrust and drag up. At terminal (constant) speed the resultant force is zero, so $W = U + D$ . The full explanation is in the Drag & Terminal Velocity lesson.
Held under by a vertical string: for wood tied under water in equilibrium, up = down gives $U = W + T$ , so the tension is $T = \rho g V - mg$ .
Hanging at rest from a newton meter in liquid: with no other force, the reading is $W-U$ , so it is lower than the reading in air by the upthrust. A denser liquid produces a greater drop if the submerged volume is unchanged.
Rope cut at one instant: if a vertical rope was holding an object in equilibrium and the other forces have not yet changed, removing the tension leaves an initial resultant equal in magnitude to the old tension, in the opposite direction.

Your turn9702/22/M/J/22 Q2

A sphere is tied to a riverbed by a wire at 68° to the horizontal. It is fully submerged and in equilibrium. Its weight is 32 N and the upthrust is 280 N. Water density = 1.0 × 10³ kg m⁻³. Find the wire tension, the sphere's volume, and the sphere's density. (9702/22/M/J/22 Q2)

Show worked answer

Vertical balance: $T\sin 68^\circ + 32 = 280$ , so $T = \dfrac{248}{\sin 68^\circ} = 270 \text{ N}$ .
$V = \dfrac{U}{\rho g} = \dfrac{280}{1.0 \times 10^{3} \times 9.81} = 0.029 \text{ m}^3$ .
$\rho_{\text{sphere}} = \dfrac{m}{V} = \dfrac{32/9.81}{0.029} = 110 \text{ kg m}^{-3}$ .

Answer: T = 270 N, V = 0.029 m³, density ≈ 110 kg m⁻³.

The hybrid to expect: one 2024 question found the upthrust on a cylinder by taking moments on the beam holding it, then turned $U = \rho g h A$ around to find the area. Show-that questions fix the route: read which quantity must come from which law.

Draw the forces, write up = down (and, if needed, moments), then use $U = \rho g V$ to move between forces and volumes. Before calculating, check whether the object is in equilibrium and whether the whole object or only part of it is submerged.

Floating, Sinking & Force Chains

Upthrust rarely appears alone in Paper 2. It sits inside a force balance with weight, drag, tension or a hinge.

The first decision is whether the object is in equilibrium or accelerating. Draw the forces, write the correct resultant-force equation, then solve for the missing piece.

Float or sink

Rises or floats	Neutral	Sinks
mean density below the fluid density	mean density equal to the fluid density	mean density above the fluid density
settles partly submerged with U = W	can remain fully submerged with U = W	has a downward resultant before drag grows

A floating object pushes aside just enough fluid for the upthrust to equal its weight. It then sits in equilibrium: $U = W$ .
Here density means the object's mean density: total mass divided by total external volume. A hollow steel ship can float because the steel and enclosed air together have a mean density below that of water.

For a floating object, upthrust equals weight. Replacing each force by density × volume × g gives a useful relationship:

\frac{V_{\text{submerged}}}{V_{\text{object}}} = \frac{\rho_{\text{object, mean}}}{\rho_{\text{fluid}}}

Slide the block's density and watch the balance between weight and upthrust decide how it floats:

Float or sink

Water has density 1000 kg m⁻³. Slide the block's density and watch how deep it sits. Upthrust always equals the weight of the water pushed aside.

Block density600 kg m⁻³

Weight

5.9 N

Upthrust

5.9 N

Result

floats, 60% under

Worked example

How deep does a floating cylinder sit?

Floating equilibrium: upthrust = 1400 N.
The submerged volume is a cylinder of depth y: $V = Ay$ , so $\rho g A y = 1400$ .
$y = \dfrac{1400}{990 \times 9.81 \times \pi (0.39)^2} = 0.30 \text{ m}$ .

Answer

y = 0.30 \text{ m}

The equilibrium chains

Falling through a fluid: weight down, upthrust and drag up. At terminal (constant) speed the resultant force is zero, so $W = U + D$ . The full explanation is in the Drag & Terminal Velocity lesson.
Held under by a vertical string: for wood tied under water in equilibrium, up = down gives $U = W + T$ , so the tension is $T = \rho g V - mg$ .
Hanging at rest from a newton meter in liquid: with no other force, the reading is $W-U$ , so it is lower than the reading in air by the upthrust. A denser liquid produces a greater drop if the submerged volume is unchanged.
Rope cut at one instant: if a vertical rope was holding an object in equilibrium and the other forces have not yet changed, removing the tension leaves an initial resultant equal in magnitude to the old tension, in the opposite direction.

Your turn9702/22/M/J/22 Q2

Show worked answer

Vertical balance: $T\sin 68^\circ + 32 = 280$ , so $T = \dfrac{248}{\sin 68^\circ} = 270 \text{ N}$ .
$V = \dfrac{U}{\rho g} = \dfrac{280}{1.0 \times 10^{3} \times 9.81} = 0.029 \text{ m}^3$ .
$\rho_{\text{sphere}} = \dfrac{m}{V} = \dfrac{32/9.81}{0.029} = 110 \text{ kg m}^{-3}$ .

Answer: T = 270 N, V = 0.029 m³, density ≈ 110 kg m⁻³.

The hybrid to expect: one 2024 question found the upthrust on a cylinder by taking moments on the beam holding it, then turned $U = \rho g h A$ around to find the area. Show-that questions fix the route: read which quantity must come from which law.