FLUID PHYSICS

Lab: Buoyancy Challenge

Lab: Buoyancy Challenge Data Chart

Upward Force Produced on an Object by a Surrounding Fluid Buoyancy Upward Force Produced on an Object by a Surrounding Fluid

Fb = Weight of Displaced Water Buoyancy Fb = Weight of Displaced Water

Fb = Real Weight – Apparent Weight Buoyancy Fb = Real Weight – Apparent Weight

Decrease an object’s density, it floats! Buoyancy Factors Decrease an object’s density, it floats!

Increase an object’s density, it sinks! Buoyancy Factors Increase an object’s density, it sinks!

Buoyancy Factors Fluid’s Density

F(b) = Force of Buoyancy F (b) = Object’s Weight – Apparent Weight Buoyancy Formulas F(g) = Object’s Weight F(b) = Force of Buoyancy ρ(o) = Object’s Density ρ(f) = Fluid’s Density F (b) = Object’s Weight – Apparent Weight F(g) ρ(o) F(b) ρ(f)

Sample Problem King Hiero sent his gold crown out to be fixed. The king heard that his crown had been altered and no longer was pure gold. The king measured the crown’s weight to be 7.84 N. When he immersed it in water, his scale read 6.86 N. Was his crown tampered with?

Calculate the force of buoyancy. Sample Problem Calculate the force of buoyancy. Step 1 F(b) = F(g) – Apparent Weight Step 2 F(b) = 7.84N – 6.86N Step 3 F(b) = 0.98N

Calculate the crown’s density. Sample Problem Calculate the crown’s density. Step 1 ρ(o) = F(g) · ρ(f) / F(b) Step 2 ρ(o) = (7.84N) (1000 kg/m^3) / .98N Step 3 ρ(o) = 8000 kg/m^3

Lab: Buoyant Force in Liquids Interactive

Lab: Buoyant Force in Liquids Data Chart 1 Trial # ρ(o) (g/cm^3) F(b) (N) F(g) F(m) 1 2 3 4

Lab: Buoyant Force in Liquids Data Chart 2 Trial # ρ(l) (g/cm^3) F(b) (N) F(g) F(m) 1 2 3

Lab: Buoyant Force in Liquids Calculation Table Trial # F(g) / F(b) ρ(o) /ρ(l) 1 2 3

Lab: Archimedes’ Challenge

Lab: Archimedes’ Challenge Data Chart Metal Sample F(g) (N) F(apparent) F(b) 1 2 3 4

Lab: Archimedes’ Challenge Calculation Table Metal Sample ρ(o) (kg/m^3) Identity 1 2 3 4

Air Pressure Demonstration

Pressure Definition Force / Unit Area

Pressure Force / Unit Area

Pressure Formula P = F/a Units of Measurement N/m^2 (Newtons per meters squared) Pa (Pascals) Atmospheres kg/ms^2 (kilograms per meter seconds squared)

Sample Problem Determine the area that a force of 2.1 N would act on to produce a pressure of 300,000 N/m^2. Step 1 P = F/a Step 2 300,000 N/m^2 = 2.1N / a Step 3 a = .000007m^2

Lab: Delude’s Hummer

Lab: Delude’s Hummer Data Chart 1

Lab: Delude’s Hummer Data Chart 2 Sample Tire Length (in) Width Left Front Right Front Left Rear Right Rear

Lab: Delude’s Hummer Calculation Table Sample Tire Area (square inches) Force (lbs) Left Front Right Front Left Rear Right Rear

Pressure applied to a fluid in a closed container is equally transmitted to every point of the fluid and the walls of the container.

Hydraulic System

You can lift heavy loads with minimal force. Hydraulic System You can lift heavy loads with minimal force.

The fluid pressure here… equals the fluid pressure here. P1 = P2 F1/A1 = F2/A2

Sample Problem What force is needed to raise a 14,500 N car if the radius is the large piston is 17.0 cm and the radius of the small piston is 4.0 cm? Step 1 F1/A1 = F2/A2 Step 2 F1 /50.24 = 14,500N/907.46 Step 3 F1 = 800N

Static Fluid Pressure = Fluid Density • g • Height P = ρ • g • h

Sample Problem 1 The Empire State Building is 366 meters high. How much pressure will you need to pump fresh water to King Kong on the top of the building?

Sample Problem 1 The Empire State Building is 366 meters high. How much pressure will you need to pump fresh water to King Kong on the top of the building? Step 1 P = ρgh Step 2 P = (1.00 x 10^3)(9.81)(366) P = 3.59 x 10^6 Pa Step 3

Sample Problem 2 The Hoover Dam is 221 meters tall. What is the pressure at the base of the dam?

Sample Problem 2 The Hoover Dam is 221 meters tall. What is the pressure at the base of the dam? Step 1 P(total) = P(atmosphere) + ρgh Step 2 P(total) = 1.01 x 10^5 +(1.00 x 10^3)(9.81)(221) Step 3 P(total) = 2.27 x 10^6 Pa

What goes in a pipe must come out of the pipe! Continuity Equation What goes in a pipe must come out of the pipe!

Continuity Equation Flow Rate = Fluid Velocity x Area Flow Rate is constant throughout the pipe. Wide Area • Velocity = Narrow Area • Velocity

Step 1: Determine each pipe’s area Sample Problem Water flows through a horizontal pipe at a velocity of 1 m/s. If the pipe narrows to ¼ its original diameter, what will be the water’s velocity? Step 1: Determine each pipe’s area Pipe 2: a = πr^2 Pipe 1: a = πr^2 Pipe 2: a = π(1)^2 Pipe 1: a = π(4)^2 Pipe 1: a = 16π Pipe 2: a = 1π

Step 2: Set up Continuity Equation Sample Problem Water flows through a horizontal pipe at a velocity of 1 m/s. If the pipe narrows to ¼ its original diameter, what will be the water’s velocity? Step 2: Set up Continuity Equation A1v1 = A2v2

Step 3: Substitute Values & Solve Sample Problem Water flows through a horizontal pipe at a velocity of 1 m/s. If the pipe narrows to ¼ its original diameter, what will be the water’s velocity? Step 3: Substitute Values & Solve (16π)(1m/s) = (1π)v2 v2 = 16m/s