Fluids and Newton's Laws

Lesson ~10 min read 8 MCQs

In simple terms: In simple terms, buoyancy is the upward push a fluid (like water or air) exerts on an object, making it feel lighter or even float. This push comes from pressure differences in the fluid.

Why this matters

This upward push from the water is a real force. It’s not magic; it’s physics. We call it the buoyant force. It’s the reason massive steel ships, weighing thousands of tons, can float on the ocean, and why a hot air balloon can rise into the sky over the Texas plains.

In this lesson, we'll break down exactly where this mysterious upward force comes from. We’ll connect it back to Newton's laws and learn how to calculate its strength using a principle discovered over two thousand years ago.

Diagram

This diagram shows the origin of buoyant force. A rectangular block is submerged in a fluid, with its top surface at depth h1 and its bottom at depth h2. A downward force F1 acts on the top, while a larger upward force F2 acts on the bottom, creating a net upward buoyant force.

Concept map

flowchart TD
    A[Object enters a fluid] --> B{Displaces a volume of fluid};
    B --> C{Fluid exerts pressure on object};
    C --> D[Pressure increases with depth];
    D --> E[Pressure on bottom > Pressure on top];
    E --> F[Creates a net upward force];
    F --> G[Buoyant Force, F_b];
    G --> H{Compare F_b to Weight F_g};
    H --> I[If F_b = F_g, it floats];
    H --> J[If F_b < F_g, it sinks];

Core explanation

From Tiny Collisions to a Giant Push

Let's start small. Imagine a glass of water. It looks perfectly still, but at the microscopic level, it's chaos. Trillions of water molecules are zipping around, constantly bumping into each other and the walls of the glass. Each tiny collision exerts a tiny force.

When you add up all these forces over a certain area, you get pressure. This is the core idea: the macroscopic behavior of a fluid (like its pressure) is the result of all the internal interactions between its particles.

Now, let's connect this to Newton's laws. For a fluid to stay put (i.e., not accelerate), the forces on any given "chunk" of it must be balanced. Gravity is always pulling the fluid down. What pushes back up? The pressure from the fluid below it. To counteract the weight of the fluid above it, the pressure must increase as you go deeper.

Think of it like a human pyramid. The person at the very bottom has to support the weight of everyone above them. The people in the middle row support less weight, and the person at the top supports no one. It’s the same in a fluid: the deeper you go, the more "weight" of the fluid is on top of you, and the greater the pressure.

The Origin of the Buoyant Force

So, pressure increases with depth. What happens when we submerge an object, like a rectangular block, in the water?

Let's look at the forces on the block from the water pressure.

The water pushes on the left side, and the water on the right side pushes back. These horizontal forces cancel each other out.
But what about the vertical forces? The top of the block is at a certain depth, h_1. The bottom of the block is deeper, at depth h_2.

Since pressure increases with depth, the pressure at the bottom (P_2) is greater than the pressure at the top (P_1).

Remember that Force = Pressure × Area.

The downward force on the top surface is F_1 = P_1 × A.
The upward force on the bottom surface is F_2 = P_2 × A.

Since P_2 > P_1, the upward force F_2 is stronger than the downward force F_1. This imbalance creates a net upward force.

This net upward force caused by the fluid is the buoyant force, F_b.

F_b = F_upward - F_downward = F_2 - F_1

This is a huge concept. Buoyancy isn't a new, magical force. It's simply the logical outcome of pressure increasing with depth.

Archimedes' Principle: The "Eureka!" Moment

The ancient Greek mathematician Archimedes figured out a brilliant way to calculate the exact magnitude of this buoyant force.

Archimedes' Principle states that the buoyant force on an object is equal to the weight of the fluid it displaces.

When you put a block in water, it shoves some water out of the way to make room for itself. The volume of the water it shoves aside is the "displaced volume," V_displaced.

The weight of that displaced fluid is mass_fluid × g. Since mass = density × volume, the weight of the displaced fluid is (ρ_fluid × V_displaced) × g.

And that gives us our key equation:

F_b = ρ_fluid * V_displaced * g

Where:

F_b is the buoyant force (in Newtons).
ρ_fluid (the Greek letter "rho") is the density of the fluid (in kg/m³).
V_displaced is the volume of the fluid displaced by the object (in m³). This is equal to the volume of the part of the object that is submerged.
g is the acceleration due to gravity (≈ 9.8 m/s²).

For the buoyant force, you always use the density of the fluid (ρ_fluid), not the object. The buoyant force is the water's reaction to being pushed aside.
The volume V in the equation is only the part of the object that's underwater. If an object is floating, V_displaced is less than the object's total volume. If it's fully submerged, V_displaced is equal to the object's total volume.

Putting It All Together: Sink or Float?

Now we can use this in a free-body diagram just like any other force. For an object in a fluid, there are two main vertical forces:

Its own weight, F_g = m_obj * g, pulling it down.
The buoyant force, F_b, pushing it up.

The object's motion depends on which force is bigger:

If F_g > F_b, the net force is downward. The object sinks.
If F_g < F_b, the net force is upward. The object rises to the surface.
If F_g = F_b, the net force is zero. The object is in equilibrium and floats (or hovers if fully submerged).

This simple comparison explains everything from why a rock sinks in a lake to why a cruise ship made of steel floats. The ship is shaped to displace a huge volume of water, making the buoyant force enormous—large enough to balance its incredible weight.

Worked examples

Example 1

Calculating Buoyant Force

Problem: A solid aluminum cube with side lengths of 0.5 meters is fully submerged in a tank of fresh water in Boston. The density of fresh water is 1000 kg/m³. What is the buoyant force acting on the cube?

Solution:

1
Identify the goal
We need to find the buoyant force, F_b. The key equation is F_b = ρ_fluid * V_displaced * g.
2
List your knowns
- Fluid: Fresh water, so ρ_fluid = 1000 kg/m³.
- Object shape: Cube with side s = 0.5 m.
- g ≈ 9.8 m/s².
3
Find the displaced volume (V_displaced). The problem states the cube is fully submerged. This means the volume of water displaced is equal to the total volume of the cube.
- V_cube = side³ = (0.5 m)³ = 0.125 m³.
- So, V_displaced = 0.125 m³.
- Why this step? The buoyant force depends on how much space the object takes up in the fluid, not what the object is made of.
4
Plug the values into the buoyant force equation
- F_b = (1000 kg/m³) * (0.125 m³) * (9.8 m/s²).
- F_b = 1225 N.
5
Final Answer
The buoyant force on the cube is 1225 Newtons. Notice we didn't even need to know the cube was made of aluminum. The buoyant force only cares about the fluid's density and the submerged volume.

Example 2

Why Do Things Float?

Problem: A block of wood has a total volume of 0.05 m³ and a density of 700 kg/m³. It is placed in a lake near Seattle (fresh water, ρ_fluid = 1000 kg/m³). Will it sink or float? If it floats, what percentage of its volume is submerged?

Solution:

1
Determine sink or float
We need to compare the block's weight (F_g) to the maximum possible buoyant force. The max buoyant force occurs when the block is fully submerged (V_displaced = V_obj).
- Calculate weight
  First, find the mass of the block.
  
  m_obj = ρ_obj * V_obj = (700 kg/m³) * (0.05 m³) = 35 kg.
  
  F_g = m_obj * g = (35 kg) * (9.8 m/s²) = 343 N.
- Calculate max buoyant force
  F_b,max = ρ_fluid * V_obj * g = (1000 kg/m³) * (0.05 m³) * (9.8 m/s²) = 490 N.
- Compare
  Since the maximum possible buoyant force (490 N) is greater than the block's weight (343 N), the block will float. The upward push of the water is strong enough to overcome the downward pull of gravity.
2
Find the submerged volume for a floating object
When an object floats, it's in equilibrium. This means the forces are balanced: F_net = 0.
- F_b - F_g = 0, which means F_b = F_g.
- We know F_g = 343 N. So, the buoyant force acting on the floating block must also be 343 N.
- Why this step? The block sinks just enough to displace a volume of water whose weight equals the block's own total weight.
3
Calculate the displaced volume
Now we use the buoyant force equation to solve for V_displaced.
- F_b = ρ_fluid * V_displaced * g
- 343 N = (1000 kg/m³) * V_displaced * (9.8 m/s²).
- V_displaced = 343 / (1000 * 9.8) = 0.035 m³.
4
Calculate the percentage submerged
- Percentage = (V_displaced / V_total) * 100%
- Percentage = (0.035 m³ / 0.05 m³) * 100% = 70%.

Final Answer: The block floats with 70% of its volume submerged. A neat shortcut: the ratio of the object's density to the fluid's density (700/1000) gives you the fraction of the object that will be submerged when it floats!

Try it yourself

Practice Problem 1

A spherical weather balloon with a radius of 2.0 m is filled with helium. The balloon itself has a mass of 5 kg. It is released from the ground in Chicago, where the density of air is approximately 1.225 kg/m³. What is the initial net force on the balloon? (Volume of a sphere = (4/3)πr³; Density of helium ≈ 0.179 kg/m³).

Practice Problem 2

A rectangular barge with dimensions 20 m long, 8 m wide, and 3 m high has a total mass of 120,000 kg. It's floating in a saltwater harbor in Dallas (salt water density ≈ 1025 kg/m³). How deep will the bottom of the barge sit below the water's surface? This depth is called the "draft."

TL;DR

In simple terms, buoyancy is the upward push a fluid (like water or air) exerts on an object, making it feel lighter or even float. This push comes from pressure differences in the fluid.

Syntax pattern

F_b = ρ_fluid * V_displaced * g

Key terms

Buoyant force Archimedes' principle Ap physics 1 Fluid dynamics Pressure and depth Sink or float Displaced fluid Fluid density Newton's laws in fluids ===END===

You can now…

8.3.A: Describe the conditions under which a fluid's velocity changes.
8.3.B: Describe the buoyant force exerted on an object interacting with a fluid.

Essential knowledge (exam-tested)

8.3.A.1: Newton's laws can be used to describe the motion of particles within a fluid.
8.3.A.2: The macroscopic behavior of a fluid is a result of the internal interactions between the fluid's constituent particles and external forces exerted on the fluid.
8.3.B.1: The buoyant force is a net upward force exerted on an object by a fluid.
8.3.B.2: The buoyant force exerted on an object by a fluid is a result of the collective forces exerted on the object by the particles making up the fluid.
8.3.B.3: The magnitude of the buoyant force exerted on an object by a fluid is equivalent to the weight of the fluid displaced by the object. Relevant equation: F_b = ρVg

Concept map

flowchart TD
    A[Object enters a fluid] --> B{Displaces a volume of fluid};
    B --> C{Fluid exerts pressure on object};
    C --> D[Pressure increases with depth];
    D --> E[Pressure on bottom > Pressure on top];
    E --> F[Creates a net upward force];
    F --> G[Buoyant Force, F_b];
    G --> H{Compare F_b to Weight F_g};
    H --> I[If F_b = F_g, it floats];
    H --> J[If F_b < F_g, it sinks];

Read what Saavi narrates

Have you ever tried to push a beach ball under the water at a pool party? It’s surprisingly hard! The moment you let go, it rockets back to the surface. Or think about how you feel lighter when you’re swimming. That upward push from the water is a real force we call the buoyant force.

This topic is all about understanding that force. It's not magic, it's just a consequence of fluid pressure. We'll see how the random motion of trillions of tiny water molecules creates a net upward force on any object in the water.

Let's walk through a classic problem. Imagine a solid aluminum cube, about a foot and a half on each side, and it's fully submerged in a tank of fresh water. We want to find the buoyant force acting on it.

The key equation is... Buoyant Force equals the density of the fluid, times the volume of the displaced fluid, times g.

First, we need the volume. The cube has sides of 0.5 meters, so its volume is 0.5 cubed, which is 0.125 cubic meters. Since it's fully submerged, that's also our displaced volume.

The density of fresh water is a standard value you should know, it's 1000 kilograms per cubic meter. And g is 9.8.

So, we just multiply them together. 1000... times 0.125... times 9.8. That gives us a buoyant force of 1225 Newtons. That's the upward push the water exerts on the cube.

Now, here's a very common mistake I see every year. Students will use the density of the object, the aluminum, in the buoyant force formula. But that's wrong. The buoyant force is exerted by the fluid. It's the water's reaction to being pushed aside. So you must always use the density of the fluid.

Remembering that one detail will save you a lot of points. Keep practicing with free-body diagrams, and you'll master this in no time. You've got this.

Back to AP Physics 1: Algebra-Based Take a mock exam

Settings & Accessibility

Fluids and Newton's Laws

Why this matters

Diagram

Concept map

Core explanation

From Tiny Collisions to a Giant Push

The Origin of the Buoyant Force

Archimedes' Principle: The "Eureka!" Moment

Putting It All Together: Sink or Float?

Worked examples

Calculating Buoyant Force

Why Do Things Float?

Try it yourself

Practice Problem 1

Practice Problem 2

Practice — 8 questions