Buoyancy Force Practice Problems- Physics Concepts Explained

Buoyancy Force: What You Actually Need to Understand

Buoyancy force is the upward push a fluid exerts on any object placed in it. It's not magic—it's pressure. The deeper parts of an object experience higher pressure than the shallower parts, and that difference creates a net upward force.

The formula is simple:

F_b = ρ × g × V

Where:

• F_b = buoyant force (in Newtons)
• ρ (rho) = density of the fluid (kg/m³)
• g = gravitational acceleration (9.8 m/s² on Earth)
• V = volume of displaced fluid (m³)

That's it. Memorize this. Everything else is just plugging numbers in.

Archimedes' Principle: The Core Concept

Archimedes' principle states that the buoyant force equals the weight of the fluid displaced by the object. This is useful because sometimes it's easier to find the weight of displaced fluid than to calculate pressure differences directly.

F_b = weight of displaced fluid = m_fluid × g

To find the mass of displaced fluid:

m_fluid = ρ_fluid × V_displaced

Object Behavior in Fluids

Objects behave differently depending on their density compared to the fluid:

• If ρ_object < ρ_fluid → the object floats
• If ρ_object = ρ_fluid → the object suspends (neutral buoyancy)
• If ρ_object > ρ_fluid → the object sinks

Salt water has a density of about 1025 kg/m³, which is why objects float slightly better in the ocean than in fresh water (997 kg/m³).

Practice Problem 1: Basic Buoyant Force Calculation

Problem: A stone with a volume of 0.002 m³ is completely submerged in water. Calculate the buoyant force acting on it.

Given:

• V = 0.002 m³
• ρ_water = 1000 kg/m³
• g = 9.8 m/s²

Solution:

F_b = ρ × g × V

F_b = 1000 × 9.8 × 0.002

F_b = 19.6 N

The stone experiences an upward buoyant force of 19.6 Newtons. Whether it sinks or floats depends on the stone's weight, which brings us to the next concept.

Practice Problem 2: Will It Float or Sink?

Problem: A wooden block with dimensions 0.1 m × 0.1 m × 0.2 m has a mass of 1.2 kg. Will it float in water?

Step 1: Find the volume

V = 0.1 × 0.1 × 0.2 = 0.002 m³

Step 2: Find the density of the wood

ρ_wood = mass / volume = 1.2 / 0.002 = 600 kg/m³

Step 3: Compare densities

ρ_wood (600 kg/m³) < ρ_water (1000 kg/m³)

Answer: Yes, it floats.

The buoyant force when floating equals the object's weight:

F_b = mg = 1.2 × 9.8 = 11.76 N

Only 0.0012 m³ of water needs to be displaced to support this block—that's 60% of its total volume.

Practice Problem 3: Partially Submerged Object

Problem: A plastic ball with radius 0.15 m and density 400 kg/m³ floats in water. Find the submerged volume.

Step 1: Find total volume

V_total = (4/3)πr³ = (4/3) × π × (0.15)³ = 0.0141 m³

Step 2: Find mass of the ball

m = ρ × V = 400 × 0.0141 = 5.64 kg

Step 3: Find weight (equals buoyant force at equilibrium)

W = mg = 5.64 × 9.8 = 55.27 N

Step 4: Find submerged volume from buoyant force equation

F_b = ρ_water × g × V_submerged

55.27 = 1000 × 9.8 × V_submerged

V_submerged = 55.27 / 9800 = 0.00564 m³

This is about 40% of the total volume—matching the density ratio (400/1000 = 0.4). That shortcut works when an object floats.

Practice Problem 4: Apparent Weight in Fluid

Problem: A metal cube weighing 50 N in air appears to weigh 30 N when submerged in water. Find the volume of the cube.

Step 1: Find the buoyant force

F_b = Weight_air - Weight_apparent = 50 - 30 = 20 N

Step 2: Solve for volume

F_b = ρ_water × g × V

20 = 1000 × 9.8 × V

V = 20 / 9800 = 0.00204 m³

This is how divers measure volume underwater—the difference in weight tells you exactly how much fluid is displaced.

Practice Problem 5: Floating in Salt Water vs Fresh Water

Problem: A boat with mass 500 kg floats in both fresh water and ocean water. Compare the volume of water displaced in each case.

For fresh water (ρ = 1000 kg/m³):

V_displaced = m / ρ = 500 / 1000 = 0.5 m³

For salt water (ρ = 1025 kg/m³):

V_displaced = m / ρ = 500 / 1025 = 0.488 m³

The boat sits slightly higher in salt water because the denser fluid provides more buoyant force per unit volume displaced. The difference is small but measurable—about 2.4% less submerged volume.

Comparison: Buoyancy in Different Fluids

How to Solve Any Buoyancy Problem

Follow this sequence every time:

Step 1: Identify What You Know

Write down mass, volume, density, or weight. Circle what you're solving for.

Step 2: Choose the Right Equation

• Need buoyant force? → F_b = ρ × g × V
• Comparing floating vs sinking? → Compare densities
• Object is floating? → Buoyant force = weight of object
• Object is submerged? → V_displaced = V_object

Step 3: Check Unit Consistency

Convert everything to kg, m³, m/s², and Newtons before plugging in. Mixing units is the fastest way to get wrong answers.

Step 4: Verify Your Answer

Ask: Does this make physical sense? A floating object can't have buoyant force exceeding its weight. A sinking object must have weight exceeding buoyant force.

Common Mistakes to Avoid

• Using object density instead of fluid density in the buoyant force formula. ρ is always the fluid's density.
• Forgetting that floating objects only displace part of their volume. The displaced volume equals the volume below the surface, not the total volume.
• Confusing mass and weight. Weight is force (Newtons); mass is not. Use mg to convert.
• Using wrong density values. Water at room temperature is 1000 kg/m³. Don't use 1 g/cm³ in calculations expecting SI units—convert first.

Quick Reference Formulas

• Buoyant force: F_b = ρ_fluid × g × V_displaced
• Density: ρ = m / V
• Weight: W = mg
• Floating equilibrium: F_b = W_object
• Apparent weight: W_apparent = W_actual - F_b

Work through each practice problem until the steps become automatic. The formula structure never changes—only the numbers do.