AP Physics 1- 8.2 Pressure- Study Notes- New Syllabus
AP Physics 1-8.2 Pressure – Study Notes
AP Physics 1-8.2 Pressure – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Pressure Exerted on a Surface
- Pressure Exerted by a Fluid
Pressure Exerted on a Surface
Pressure is the amount of force exerted per unit area of a surface. It describes how a force is distributed over the surface on which it acts.
Relevant Equation:
\( P = \dfrac{F}{A} \)
- \( P \) = pressure (Pa, Pascals)
- \( F \) = perpendicular (normal) force exerted (N)
- \( A \) = surface area over which the force is distributed (m\(^2\))
Key Points:
- Pressure is a scalar quantity, though the force that causes it has a direction.
- If the same force acts on a smaller area, the pressure increases.
- Only the normal component of the force contributes to pressure (forces parallel to the surface cause shear, not pressure).
Example :
A box of weight \( 200 \, \text{N} \) rests on a table with a base area of \( 0.5 \, \text{m}^2 \). Find the pressure it exerts on the table.
▶️Answer/Explanation
\( P = \dfrac{F}{A} = \dfrac{200}{0.5} = 400 \, \text{Pa} \)
Answer: The pressure on the table is \( 400 \, \text{Pa} \).
Example :
A person of mass \( 60 \, \text{kg} \) is standing on one foot. The area of the foot in contact with the ground is \( 0.03 \, \text{m}^2 \). Calculate the pressure exerted on the ground. (Take \( g = 9.8 \, \text{m/s}^2 \)).
▶️Answer/Explanation
Force due to weight: \( F = m g = 60 \times 9.8 = 588 \, \text{N} \)
Pressure: \( P = \dfrac{F}{A} = \dfrac{588}{0.03} \approx 1.96 \times 10^4 \, \text{Pa} \)
Answer: Pressure on the ground ≈ \( 1.96 \times 10^4 \, \text{Pa} \).
Pressure Exerted by a Fluid
The pressure exerted by a fluid at a given depth is due to the weight of the fluid above that point. Fluid pressure increases with depth and acts equally in all directions (Pascal’s principle).
Relevant Equation:
\( P = P_0 + \rho g h \)
- \( P \) = total pressure at depth \( h \)
- \( P_0 \) = pressure at the surface of the fluid (often atmospheric pressure)
- \( \rho \) = density of the fluid (kg/m\(^3\))
- \( g \) = acceleration due to gravity (9.8 m/s\(^2\))
- \( h \) = depth below the fluid surface (m)
Key Points:
- Fluid pressure increases linearly with depth.
- At the same depth, pressure is the same in all directions (isotropic property of fluids).
- Only the vertical depth matters; shape of the container does not affect pressure at a given depth.
- Units of pressure: Pascal (Pa) = N/m\(^2\).
Example :
Find the pressure at a depth of 5 m in water. Take \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), and atmospheric pressure \( P_0 = 1.01 \times 10^5 \, \text{Pa} \).
▶️Answer/Explanation
Fluid pressure due to water: \( \rho g h = 1000 \times 9.8 \times 5 = 4.9 \times 10^4 \, \text{Pa} \)
Total pressure: \( P = P_0 + \rho g h = 1.01 \times 10^5 + 4.9 \times 10^4 = 1.50 \times 10^5 \, \text{Pa} \)
Answer: Pressure at 5 m depth ≈ \( 1.5 \times 10^5 \, \text{Pa} \).
Example :
A diver is 20 m below the surface of a freshwater lake. Find the gauge pressure (only due to water, ignoring atmospheric pressure).
▶️Answer/Explanation
Gauge pressure: \( P = \rho g h = 1000 \times 9.8 \times 20 = 1.96 \times 10^5 \, \text{Pa} \)
Answer: Gauge pressure at 20 m depth = \( 1.96 \times 10^5 \, \text{Pa} \).