IB MYP 4-5 Physics- Pascal’s principle- Study Notes - New Syllabus
IB MYP 4-5 Physics-Pascal’s principle- Study Notes
Key Concepts
- Pascal’s principle
Hydraulic Systems (Pascal’s Principle)
Hydraulic Systems (Pascal’s Principle)
Pascal’s Principle:
Pascal’s principle states that when pressure is applied to an enclosed incompressible fluid, the pressure change is transmitted equally and undiminished to all parts of the fluid and the walls of the container.
\(\text{Pressure applied at one point} = \text{Pressure at every other point in the fluid}\)
Mathematical Expression:
\( P = \dfrac{F}{A} \)
Where: – \( P \) = Pressure (Pa) – \( F \) = Force applied (N) – \( A \) = Area of surface (m²)
Key Points of Pascal’s Principle:
- The fluid must be incompressible (liquid, not gas).
- The fluid must be completely enclosed.
- Pressure is transmitted equally in all directions.
Hydraulic Systems:
Hydraulic systems use Pascal’s principle to multiply force. They consist of two pistons connected by a pipe filled with fluid.
If \( F_1 \) is applied to a small piston with area \( A_1 \), and \( F_2 \) is produced on a large piston with area \( A_2 \):
\( \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2} \)
Therefore: \( F_2 = F_1 \times \dfrac{A_2}{A_1} \)
Force Multiplication:
- If \( A_2 > A_1 \), the output force \( F_2 \) is greater than the input force \( F_1 \).
- The trade-off is that the large piston moves a shorter distance compared to the small piston.
Applications of Pascal’s Principle:
- Hydraulic car brakes
- Hydraulic car lifts
- Hydraulic jacks
- Aircraft hydraulic control systems
Advantages of Hydraulic Systems:
- High force multiplication with small input force.
- Smooth and precise control of motion.
- Can transmit power over long distances without much loss.
Disadvantages of Hydraulic Systems:
- Leakage of fluid can cause failure.
- Requires maintenance and clean fluid.
- Cannot be used with compressible fluids (like air) for the same effect.
Example:
A hydraulic lift has a small piston of area \( 0.01 \, m^2 \) and a large piston of area \( 0.5 \, m^2 \). If a force of \( 200 \, N \) is applied to the small piston, find the force exerted by the large piston.
▶️ Answer/Explanation
Given: \( A_1 = 0.01 \, m^2 \), \( A_2 = 0.5 \, m^2 \), \( F_1 = 200 \, N \)
Using \( \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2} \):
\( \dfrac{200}{0.01} = \dfrac{F_2}{0.5} \)
\( 20000 = \dfrac{F_2}{0.5} \)
\( F_2 = 20000 \times 0.5 = \boxed{10000 \, N} \)
Example:
A hydraulic press has a small piston of radius \( 0.05 \, m \) and a large piston of radius \( 0.2 \, m \). If the small piston is pushed with a force of \( 300 \, N \), determine the force on the large piston.
▶️ Answer/Explanation
First, find areas: \( A_1 = \pi r_1^2 = \pi (0.05)^2 = 0.00785 \, m^2 \) \( A_2 = \pi r_2^2 = \pi (0.2)^2 = 0.1256 \, m^2 \)
Using \( \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2} \):
\( \dfrac{300}{0.00785} = \dfrac{F_2}{0.1256} \)
\( 38216.56 = \dfrac{F_2}{0.1256} \)
\( F_2 = 38216.56 \times 0.1256 = \boxed{4800 \, N} \)