Physics 9702 · Paper 2

Floating, Sinking & Force Chains

Upthrust rarely appears alone in Paper 2. It sits inside a force balance with weight, drag, tension or a hinge.
Every one of those questions is the same skill: write the equilibrium equation, then solve for the missing piece.

Float or sink

Floats	Sinks
less dense than the fluid	denser than the fluid
upthrust can grow to equal its weight	upthrust stays below its weight even fully under
settles part-submerged, U = W	rests on the bottom

A floating object pushes aside just enough fluid for the upthrust to equal its weight. It then sits in equilibrium: $U = W$ .
A denser-than-the-fluid object cannot reach that balance even fully submerged, so it sinks.

A light jar floating high in a tank; an identical heavier jar resting fully submerged on the floor. — Same size, different density: the lighter jar floats where U = W; the denser one 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? (9702/22/F/M/25 Q2(b))

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 one-mark equation is $W = U + D$ (9702/22/O/N/24 Q1(b)). The full terminal-velocity explanation is in lesson 2.7.
Held under by a string: for wood tied under water, up = down gives $U = W + T$ , so the tension is $T = \rho g V - mg$ (9702/12/M/J/25 Q14).
Hanging from a newton meter in liquid: the reading drops by exactly the upthrust. A denser liquid drops it further (9702/11/M/J/25 Q14).

Worked example

Chain 1: a balloon on a rope

A hot-air balloon of weight 3.39 × 10⁴ N is held just off the ground by a vertical rope of tension 4.00 × 10² N. The air density is 1.23 kg m⁻³. Find the balloon's volume, then its acceleration just after the rope is released. (9702/21/O/N/23 Q2)

Equilibrium: up = down, so $U = W + T = 3.39 \times 10^{4} + 4.00 \times 10^{2} = 3.43 \times 10^{4} \text{ N}$ .
$V = \dfrac{U}{\rho g} = \dfrac{3.43 \times 10^{4}}{1.23 \times 9.81} = 2.84 \times 10^{3} \text{ m}^3$ .
Rope gone: resultant = old tension = $4.00 \times 10^{2} \text{ N}$ upward. Mass = $3.39 \times 10^{4}/9.81 = 3.46 \times 10^{3} \text{ kg}$ .
$a = \dfrac{4.00 \times 10^{2}}{3.46 \times 10^{3}} = 0.12 \text{ m s}^{-2}$ upwards.

Answer

V = 2.84 \times 10^{3} \text{ m}^3, \; a = 0.12 \text{ m s}^{-2}

Your turn— tap to reveal the worked answer (9702/22/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)

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 (9702/23/O/N/24 Q3).

Draw the forces, write up = down (and, if needed, moments), then use $U = \rho g V$ to move between forces and volumes. Every upthrust question in 2024/25 fits this pattern.

PreviousUpthrust NextWork & Energy Transfer

Floating, Sinking & Force Chains

Upthrust rarely appears alone in Paper 2. It sits inside a force balance with weight, drag, tension or a hinge.
Every one of those questions is the same skill: write the equilibrium equation, then solve for the missing piece.

Float or sink

Floats	Sinks
less dense than the fluid	denser than the fluid
upthrust can grow to equal its weight	upthrust stays below its weight even fully under
settles part-submerged, U = W	rests on the bottom

A floating object pushes aside just enough fluid for the upthrust to equal its weight. It then sits in equilibrium: $U = W$ .
A denser-than-the-fluid object cannot reach that balance even fully submerged, so it 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?

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 one-mark equation is $W = U + D$ (9702/22/O/N/24 Q1(b)). The full terminal-velocity explanation is in lesson 2.7.
Held under by a string: for wood tied under water, up = down gives $U = W + T$ , so the tension is $T = \rho g V - mg$ (9702/12/M/J/25 Q14).
Hanging from a newton meter in liquid: the reading drops by exactly the upthrust. A denser liquid drops it further (9702/11/M/J/25 Q14).

Worked example

Chain 1: a balloon on a rope

Equilibrium: up = down, so $U = W + T = 3.39 \times 10^{4} + 4.00 \times 10^{2} = 3.43 \times 10^{4} \text{ N}$ .
$V = \dfrac{U}{\rho g} = \dfrac{3.43 \times 10^{4}}{1.23 \times 9.81} = 2.84 \times 10^{3} \text{ m}^3$ .
Rope gone: resultant = old tension = $4.00 \times 10^{2} \text{ N}$ upward. Mass = $3.39 \times 10^{4}/9.81 = 3.46 \times 10^{3} \text{ kg}$ .
$a = \dfrac{4.00 \times 10^{2}}{3.46 \times 10^{3}} = 0.12 \text{ m s}^{-2}$ upwards.

Answer

V = 2.84 \times 10^{3} \text{ m}^3, \; a = 0.12 \text{ m s}^{-2}

Your turn— tap to reveal the worked answer (9702/22/M/J/22 Q2)

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 (9702/23/O/N/24 Q3).

Draw the forces, write up = down (and, if needed, moments), then use $U = \rho g V$ to move between forces and volumes. Every upthrust question in 2024/25 fits this pattern.