AP Physics 2 Unit 8 – Fluids

1. Finding the Density of a Small Object

Use a triple-beam balance to find the mass of the object.
Fill a graduated cylinder with water.
Submerge the object in the cylinder to find the amount of water displaced.
Calculate the density using the formula:
$\text{Density} = \frac{\text{Mass of the object}}{\text{Volume of water displaced}}$

2. Gauge Pressure and Absolute Pressure at the Bottom of a Container

a. Deriving Gauge Pressure

Given:
- Height of fluid above the bottom: $h$
- Area of the base of the container: $A$
The force exerted by the fluid due to gravity is given by:
$F = m imes g = \rho_f imes V imes g$
Volume $V$ is given by:
$V = A imes h$
Thus, the gauge pressure $P_G$ at the bottom can be derived as follows:
$P_G = \frac{F}{A} = \frac{\rho_f \times A \times h \times g}{A} = \rho_f g h$

b. Absolute Pressure

Absolute pressure $P$ at the bottom combines the gauge pressure and atmospheric pressure:
$P = P_G + P_0$ ,
where $P_0$ is the atmospheric pressure.
Thus, the absolute pressure is:
$P = \rho_f g h + P_0$

3. Buoyant Force on a Submerged Lead Cube

A solid lead cube is attached to a string and completely submerged in water.
The cube is at rest, indicating that the forces are balanced.
According to Archimedes’ principle, the buoyant force $F_b$ acting on the block equals the weight of the fluid displaced.
The tension in the string is non-zero, indicating:
- The buoyant force is greater than the weight of the block:
  F_b > W_{block}
When the cube is at rest:
$T + F_b = W_{block}$ (where $T$ is the tension)
Therefore, it can be concluded that the buoyant force exceeds the weight of the block.

4. Microscopic Causes of Forces

The normal force exerted by the table on the resting block and the force exerted by water on the submerged block can be understood microscopically.
For the block on the table:
- The molecules in the table's surface repel the electrons in the block due to electromagnetic forces, generating a normal force.
For the submerged block:
- Water molecules similarly repel the electrons of the block, exerting a buoyant force upwards.
In both scenarios, the forces arise from the electromagnetic repulsions between electrons.

5. Height Difference in a Flowing Fluid

A fluid flows with speed $v_1$ through a horizontal section of pipe with area $A_1$ and pressure $P_1$ .
The flow continues to a second section with pressure $P_2$ , area $A_2$ , and unknown speed $v_2$ .
To determine the height difference $h$ between the two sections:
- Apply Bernoulli’s equation:
  $P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$
Rearranging allows for finding $h$ , considering $y_2$ is set to zero.

6. Pressure and Force in Three Containers

a. Fluid Pressures

When comparing three cylinders filled with the same liquid to equal heights:
- Pressure at the bottom governed by:
  $P = \rho_f g h$
- Since heights are identical, fluid pressures at the bottom of each container are equal:
- Rank: $P_1 = P_2 = P_3$

b. Forces Exerted by the Liquid

The force exerted on the bottom depends on pressure and area:
$F = P imes A$
Since pressures are identical:
- Greater area yields greater force:
- Rank by area:
- F_1 > F_2 > F_3

7. Flow Rate and Speed in Pipes

a. Continuity Equation for Different Cross-Section Areas

For a pipe, if $A_2 = \frac{1}{2} A_1$ , then:
Using the equation of continuity:
$A_1 v_1 = A_2 v_2$
leads to:
$v_2 = 2 v_1$

b. Diameter Relations

If the diameter of the first pipe is twice that of the second (thus, area:i.e., $A_2 = \frac{1}{4} A_1$ ):
Using the continuity equation:
$A_1 v_1 = A_2 v_2$ leads to:
$v_2 = 4 v_1$

8. Pressure Changes When a Person Exits the Pool

a. Height of Water

When a person exits the pool, the volume of water displaced decreases, thus lowering the height of the water:
- Hence, pressure (calculated as $P = \rho_f g h$ ) decreases due to reduction in height.

b. Buoyant Force Effects

While submerged, the person exerts a downward force equal to the buoyant force experienced:
- According to Newton's Third Law, the downward force increases the pressure at the bottom of the pool.
Upon exit, the buoyant force disappears:
The pressure at the bottom decreases accordingly since the equal downward force is no longer present.

Hints and Answers Summary

To find density: $\frac{\text{Mass of object}}{\text{Volume of displaced water}}$ .
a. $P_G = \rho_f g h$ ; b. $P = \rho_f g h + P_0$
Buoyant force > weight of block
Similar microscopic causes in normal forces.
Use Bernoulli’s Equation.
a. Same pressure; b. Force relates to area.
a. $v_2 = 2 v_1$ ; b. $v_2 = 4 v_1$
a. Pressure decreases due to height; b. Pressure decreases after the person exits.

Practice Problems

Finding Density:
- A cube with a mass of 300 grams is submerged in water. If the volume of water displaced is 100 cm³, calculate the density of the cube.
  - Answer: 3 g/cm³
Gauge Pressure Calculation:
- If the height of a liquid column is 5 meters and the density of the fluid is 1000 kg/m³, calculate the gauge pressure at the bottom of the container. (Assume g = 9.8 m/s²)
  - Answer: P ext{_G} =
    ho ext{_f} g h = 1000 imes 9.8 imes 5 = 49000 ext{Pa}
Buoyant Force:
- A wooden block with a weight of 20 N is floating in water. Calculate the buoyant force acting on it.
  - Answer: 20 N (buoyant force equals the weight of the water displaced, which is equal to the weight of the block in equilibrium).
Bernoulli’s Equation:
- A fluid with a pressure of 2000 Pa, flowing with a velocity of 3 m/s, enters a pipe of a larger cross-sectional area. If the pressure in the larger section is 1500 Pa, calculate the speed of the fluid in this section assuming the height difference is negligible.
  - Answer: Using Bernoulli’s equation:
    P ext{_1} + rac{1}{2}
    ho v ext{_1}^2 = P ext{_2} + rac{1}{2}
    ho v ext{_2}^2
    Rearranging gives:
    $v ext{_2} = \frac{A ext{_1}}{A ext{_2}} v ext{_1}$ (using continuity equation). Substitute values to find velocity in section 2.
Effect on Pressure when Exiting the Pool:
- A person with a weight of 600 N exits a pool. How does this affect the pressure at the bottom of the pool? Assume the pressure contributes additional downward force while submerged.
  - Answer: The pressure decreases after the person exits since the buoyant force is lost; the previous additional force at the bottom (equal to their weight) is also lost, thus reducing the overall pressure.