8.3 Fluids and Newton's Laws

Fluids obey Newton's laws, with particles experiencing forces and accelerations like solid objects. Their behavior is influenced by internal interactions and external forces, resulting in complex flow patterns and pressure distributions.

Buoyant force, an upward force on immersed objects, arises from fluid pressure increasing with depth. It equals the weight of displaced fluid, as described by Archimedes' principle, and affects objects' behavior in fluids.

Fluid velocity changes

Newton's laws in fluids

Fluid particles follow the same fundamental principles of motion that govern solid objects. Each particle responds to forces with acceleration proportional to the net force and inversely proportional to its mass.

• Fluid particles experience forces and accelerations just like solid objects, but their motion is influenced by the surrounding particles and the fluid's properties
• When a force acts on a fluid particle, it accelerates according to $a = F_{net}/m$, just as with solids
• The collective behavior of countless fluid particles creates the flowing patterns we observe macroscopically

Macroscopic fluid behavior

The bulk movement of fluids emerges from millions of individual particle interactions following Newton's laws, creating observable phenomena like currents, waves, and turbulence.

• Internal particle interactions, such as collisions and attractive/repulsive forces, contribute to the fluid's pressure, viscosity, and flow patterns
• External forces, like gravity, pressure gradients, and surface forces, influence the fluid's motion and shape
• These combined interactions explain why water flows around obstacles, why air creates vortices behind fast-moving objects, and why different fluids flow at different rates

Buoyant force

Upward fluid force

The buoyant force results from pressure differences between the top and bottom of an object submerged in fluid. Because pressure increases with depth ($P = \rho g h$), the bottom of an object experiences greater pressure than the top.

• The buoyant force acts vertically upward on an object immersed in a fluid, opposing the object's weight
• This pressure difference creates a net upward force that we call buoyancy
• You can experience this by pushing a beach ball underwater in a pool - you'll feel the water pushing back

Collective particle forces

The buoyant force we observe macroscopically is actually the sum of countless tiny forces from individual fluid particles colliding with the submerged object.

• Each fluid particle exerts a force on the object's surface due to the fluid pressure, which increases with depth
• The net force from all the fluid particles acting on the object's surface results in the buoyant force
• This microscopic view explains why oddly-shaped objects still experience predictable buoyant forces - it's the sum of all particle interactions across the entire surface

Displaced fluid weight

Archimedes' principle states that the buoyant force equals exactly the weight of the fluid displaced by the object. This explains why some objects float while others sink.

• The magnitude of the buoyant force equals the weight of the fluid displaced by the object, as described by the equation: $F_b = \rho V g$
  • $F_b$ is the buoyant force
  • $\rho$ is the fluid density
  • $V$ is the volume of the displaced fluid
  • $g$ is the acceleration due to gravity
• When an object's density is less than the fluid's density, the buoyant force exceeds the object's weight, causing it to float
• When an object's density is greater than the fluid's density, the buoyant force is less than the object's weight, causing it to sink

Practice Problem 1: Newton's Laws in Fluids

Solution:

The forces acting on the bubble are:

1. Buoyant force ($F_b$) acting upward
2. Weight of the air bubble ($F_g$) acting downward
3. Drag force ($F_d$) acting downward as the bubble rises

Since the bubble moves at constant velocity, the net force must be zero according to Newton's First Law: $F_b - F_g - F_d = 0$

$F_b = F_g + F_d$

The buoyant force equals the weight of the displaced water: $F_b = \rho_{water} V_{bubble} g$

The weight of the air bubble is negligible compared to the buoyant force since $\rho_{air} \ll \rho_{water}$

Therefore, the buoyant force is primarily balanced by the drag force: $F_b \approx F_d$

The bubble reaches terminal velocity because as it accelerates, the drag force increases with speed until it equals the buoyant force, resulting in zero net force and constant velocity.

Practice Problem 2: Buoyant Force

Solution:

Part 1: Block in water The volume of the block is $V = 10 \text{ cm} \times 8 \text{ cm} \times 6 \text{ cm} = 480 \text{ cm}^3 = 4.8 \times 10^{-4} \text{ m}^3$

The mass of the block is $m = \rho V = 650 \text{ kg/m}^3 \times 4.8 \times 10^{-4} \text{ m}^3 = 0.312 \text{ kg}$

For a floating object, the buoyant force equals the weight of the object: $F_b = mg$

The buoyant force also equals the weight of displaced fluid: $F_b = \rho_{water} V_{submerged} g$

Setting these equal: $\rho_{water} V_{submerged} g = mg$

Substituting $m = \rho_{block} V$: $\rho_{water} V_{submerged} g = \rho_{block} V g$

Simplifying: $\frac{V_{submerged}}{V} = \frac{\rho_{block}}{\rho_{water}} = \frac{650 \text{ kg/m}^3}{1000 \text{ kg/m}^3} = 0.65$

So 65% of the block is submerged, meaning 35% remains above water.

Part 2: Block in olive oil Using the same equation: $\frac{V_{submerged}}{V} = \frac{\rho_{block}}{\rho_{oil}} = \frac{650 \text{ kg/m}^3}{920 \text{ kg/m}^3} = 0.707$

So 70.7% of the block is submerged in olive oil, meaning 29.3% remains above the surface.

Practice Problem 3: Macroscopic Fluid Behavior

Solution:

First, convert the flow rate to m³/s: $36 \text{ L/min} = 36 \times 10^{-3} \text{ m}^3/\text{min} = 6 \times 10^{-4} \text{ m}^3/\text{s}$

Using the continuity equation for fluids: $A_1 v_1 = A_2 v_2$ Where:

• $A_1 = 12 \text{ cm}^2 = 12 \times 10^{-4} \text{ m}^2$ (area of wider section)
• $A_2 = 4 \text{ cm}^2 = 4 \times 10^{-4} \text{ m}^2$ (area of narrower section)
• $v_1$ is the speed in the wider section
• $v_2$ is the speed in the narrower section

The flow rate $Q = A_1 v_1 = A_2 v_2 = 6 \times 10^{-4} \text{ m}^3/\text{s}$

For the wider section: $v_1 = \frac{Q}{A_1} = \frac{6 \times 10^{-4} \text{ m}^3/\text{s}}{12 \times 10^{-4} \text{ m}^2} = 0.5 \text{ m/s}$

For the narrower section: $v_2 = \frac{Q}{A_2} = \frac{6 \times 10^{-4} \text{ m}^3/\text{s}}{4 \times 10^{-4} \text{ m}^2} = 1.5 \text{ m/s}$

The speed increases in the narrower section because the same volume of fluid must pass through a smaller area in the same amount of time. This follows from the conservation of mass - the mass flow rate must be constant throughout the pipe. Since fluid cannot be created or destroyed inside the pipe, the product of area and velocity must remain constant. When area decreases, velocity must increase proportionally.