AP Physics 1- 8.3 Fluids and Newton’s Laws- Study Notes- New Syllabus
AP Physics 1-8.3 Fluids and Newton’s Laws – Study Notes
AP Physics 1-8.3 Fluids and Newton’s Laws – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Conditions under which a Fluid’s Velocity Changes
- Buoyant Force Exerted on an Object in a Fluid
Conditions under which a Fluid’s Velocity Changes
The velocity of a fluid can change when the cross-sectional area of the flow path or the pressure conditions of the fluid system are altered. This behavior is explained by the principle of continuity and Bernoulli’s principle.
Continuity Equation:
\( A_1 v_1 = A_2 v_2 \)
- \( A \) = cross-sectional area of the pipe or channel
- \( v \) = velocity of the fluid
- If the area decreases (\( A_2 < A_1 \)), the velocity increases (\( v_2 > v_1 \)).
- If the area increases, the velocity decreases.
Bernoulli’s Principle:
Velocity can also change due to pressure differences within a flowing fluid. According to Bernoulli’s equation:
\( P + \dfrac{1}{2} \rho v^2 + \rho g h = \text{constant} \)
- An increase in velocity corresponds to a decrease in pressure (for constant height).
- A decrease in velocity corresponds to an increase in pressure.
Example :
Water flows through a pipe that narrows from a diameter of 10 cm to 5 cm. If the velocity in the wider section is \( 2 \, \text{m/s} \), find the velocity in the narrower section.
▶️Answer/Explanation
Using continuity equation:
\( A_1 v_1 = A_2 v_2 \)
\( \pi (0.05)^2 (2) = \pi (0.025)^2 v_2 \)
\( 0.005 = 0.00196 v_2 \)
\( v_2 \approx 2.55 \, \text{m/s} \)
Example :
Airplane wings are designed so air flows faster over the top surface than underneath. Explain why this produces lift.
▶️Answer/Explanation
By Bernoulli’s principle, faster airflow on the top reduces pressure compared to the bottom. This pressure difference creates an upward lift force, allowing the airplane to fly.
Buoyant Force Exerted on an Object in a Fluid
When an object is placed in a fluid, it experiences an upward force called the buoyant force. This force is due to the pressure difference exerted by the fluid on the top and bottom surfaces of the object.
Archimedes’ Principle:
The buoyant force on an object is equal to the weight of the fluid displaced by the object.
\( F_b = \rho_f V_d g \)
- \( F_b \) = buoyant force
- \( \rho_f \) = density of the fluid
- \( V_d \) = volume of fluid displaced
- \( g \) = acceleration due to gravity
| Condition | Outcome |
|---|---|
| \( F_b > W \) (Buoyant force greater than weight) | Object floats upward |
| \( F_b < W \) (Buoyant force less than weight) | Object sinks |
| \( F_b = W \) (Buoyant force equals weight) | Object remains neutrally buoyant |
Example :
A wooden block of volume \( 0.02 \, \text{m}^3 \) floats in water. Find the buoyant force acting on it. (Take \( \rho_{water} = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \)).
▶️Answer/Explanation
\( F_b = \rho_f V_d g = 1000 \times 0.02 \times 9.8 \)
\( F_b = 196 \, \text{N} \)
Answer: Buoyant force = 196 N
Example :
A steel ball of volume \( 5 \times 10^{-4} \, \text{m}^3 \) is fully submerged in water. Find the buoyant force on it.
▶️Answer/Explanation
\( F_b = \rho_f V_d g = 1000 \times 5 \times 10^{-4} \times 9.8 \)
\( F_b = 4.9 \, \text{N} \)
Answer: Buoyant force = 4.9 N