Pressure
The equation for pressure is:
p = \frac{F}{A}
where:
- p is the pressure in pascals (Pa) or newtons per square metre (N/m^2).
- F is the force acting perpendicular to the surface in newtons (N). For a solid resting on a horizontal surface, this force is typically its weight.
- A is the area of contact in square metres (m^2) or square centimetres (cm^2).
Why this definition matters:
Pressure is not just force; it is how that force is distributed. A large force spread over a huge area creates low pressure, while a small force concentrated on a tiny point creates high pressure.
Unit: The SI unit of pressure is the pascal (Pa).
1 \text{ Pa} = 1 \text{ N/m}^2
When using the equation p = F/A, you must ensure units are consistent. If force is in Newtons and area is in cm^2, the resulting pressure will be in N/cm^2. To convert to pascals, multiply by 10,000 (since 1 m^2 = 10,000 cm^2).
Question: Calculate the pressure exerted by the block on the table.
Solution:
- Identify the force: The force is the weight, so F = 12 \text{ N}.
- Identify the area: The contact area is A = 0.014 m^2.
- Apply the formula:
p = \frac{F}{A} = \frac{12}{0.014} - Calculate:
p \approx 857 \text{ Pa}
Note: Always check if the question asks for pressure in Pa or N/cm^2. If the area was given in cm^2, you would need to convert it to m^2 first to get Pascals.
The Correct Understanding: Pressure depends only on the contact area (the face touching the surface). You must identify which face is resting on the ground and calculate its area (length \times width) only. Volume is irrelevant for pressure calculations unless you are calculating density or mass first.
The Reasoning: Examiners look for the link between Area and Pressure. You must explicitly state that increasing the area reduces the pressure for the same force.
Correct Phrasing:
'Wide tyres increase the contact area with the ground. Since the weight (force) is constant, a larger area results in lower pressure, preventing the tractor from sinking into soft ground.'
Incorrect Phrasing: 'Wide tyres reduce the weight.' (Weight is constant; only pressure changes.)
Pressure beneath the surface of a liquid changes based on two factors:
- Depth: Pressure increases as depth increases. This is because there is more liquid above the point, so the weight of the liquid column pushing down is greater.
- Density: Pressure increases as the density of the liquid increases. Denser liquids have more mass per unit volume, so a column of dense liquid weighs more than a column of light liquid at the same depth.
Key Relationship:
- Deeper = Higher Pressure
- Denser = Higher Pressure
This relates to the molecular model: pressure is caused by collisions of molecules. At greater depths, there are more layers of molecules above, increasing the force per unit area.
\Delta p = \rho g \Delta h
where:
- \Delta p is the change in pressure (or gauge pressure) in pascals (Pa).
- \rho (rho) is the density of the liquid in kilograms per cubic metre (kg/m^3).
- g is the gravitational field strength (approx. 9.8 N/kg or 10 N/kg depending on the question).
- \Delta h is the change in depth (vertical distance from the surface) in metres (m).
Important Note: This formula calculates the pressure due to the liquid only. It does not include atmospheric pressure acting on the surface unless 'total pressure' is specified.
Solution:
- Identify variables:
- \rho = 1000 kg/m^3
- g = 9.8 N/kg (use value given in question)
- h = 10 m
- Apply formula:
p = \rho g h
p = 1000 \times 9.8 \times 10 - Calculate:
p = 98,000 \text{ Pa}
Extension: If the question asks for total pressure, add atmospheric pressure (\approx 101,000 Pa) to this result.
The Correct Understanding: The formula p = \rho g h gives the gauge pressure (pressure from the liquid alone). If a question asks for the 'total pressure' at a point submerged in an open container, you must add the atmospheric pressure acting on the surface:
P_{total} = P_{atm} + \rho g h
Always read the question carefully to see if it asks for 'pressure due to the liquid' or 'total pressure'.
The Reasoning: Examiners accept explanations that link macroscopic weight to microscopic force. You must mention the weight of the liquid column.
Correct Phrasing:
'Pressure increases with depth because the weight of the liquid above the point increases. This greater weight exerts a larger downward force per unit area.'
Alternative Phrasing:
'There are more layers of liquid molecules above the deeper point, resulting in more frequent and forceful collisions per unit area.'
Avoid: Saying 'gravity pulls harder on the water.' Gravity is constant; it's the amount of water above that changes.
\Delta p = 1000 \times 9.8 \times 0.4
\Delta p = 3920 Pa