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Motion, Forces and Energy - Pressure

Grade 11A LevelPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Pressure is defined as the force exerted per unit area. It is a scalar quantity.

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The SI unit for pressure is the Pascal (PaPa), where 1 Pa=1 N/m21\ Pa = 1\ N/m^2.

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In solids, for a constant force, the pressure is inversely proportional to the surface area. This explains why sharp knives cut better than blunt ones (P∝1AP \propto \frac{1}{A}).

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In liquids, pressure increases with depth because of the weight of the liquid above. The pressure at a point in a fluid acts equally in all directions.

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Pressure in a liquid depends on the density of the liquid (ρ\rho), the gravitational field strength (gg), and the depth (hh).

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Atmospheric pressure is caused by the weight of the air above the Earth's surface. At sea level, it is approximately 1.01×105 Pa1.01 \times 10^5\ Pa.

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A barometer is used to measure atmospheric pressure, often using a column of mercury. Standard atmospheric pressure is 760 mmHg760\ mmHg.

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A manometer is used to measure the pressure difference between two gases or a gas and the atmosphere by observing the difference in liquid levels (hh).

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Pascal's Principle states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the vessel.

📐Formulae

P=FAP = \frac{F}{A}

ΔP=ρgΔh\Delta P = \rho g \Delta h

Ptotal=Patmos+ρghP_{total} = P_{atmos} + \rho gh

F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}

💡Examples

Problem 1:

A rectangular block of wood has a weight of 600 N600\ N and dimensions 2 m×0.5 m×0.1 m2\ m \times 0.5\ m \times 0.1\ m. Calculate the maximum pressure it can exert on the ground.

Solution:

Pmax=FAminP_{max} = \frac{F}{A_{min}} Amin=0.5 m×0.1 m=0.05 m2A_{min} = 0.5\ m \times 0.1\ m = 0.05\ m^2 P=600 N0.05 m2=12,000 PaP = \frac{600\ N}{0.05\ m^2} = 12,000\ Pa

Explanation:

To exert maximum pressure, the block must be placed on its smallest surface area. We identify the two smallest dimensions to calculate the minimum area.

Problem 2:

Calculate the pressure exerted by water at the bottom of a swimming pool that is 3 m3\ m deep. (Density of water ρ=1000 kg/m3\rho = 1000\ kg/m^3, g=9.8 m/s2g = 9.8\ m/s^2)

Solution:

P=ρghP = \rho g h P=1000 kg/m3×9.8 m/s2×3 mP = 1000\ kg/m^3 \times 9.8\ m/s^2 \times 3\ m P=29,400 PaP = 29,400\ Pa

Explanation:

The pressure in a fluid depends on the density, gravity, and the height of the fluid column. This value represents the pressure due to the water alone (gauge pressure).

Problem 3:

A hydraulic jack has an input piston with an area of 0.02 m20.02\ m^2 and an output piston with an area of 0.5 m20.5\ m^2. If a force of 150 N150\ N is applied to the input piston, what is the weight of the load that can be lifted?

Solution:

P1=P2⇒F1A1=F2A2P_1 = P_2 \Rightarrow \frac{F_1}{A_1} = \frac{F_2}{A_2} 150 N0.02 m2=F20.5 m2\frac{150\ N}{0.02\ m^2} = \frac{F_2}{0.5\ m^2} F2=150×0.50.02=3750 NF_2 = \frac{150 \times 0.5}{0.02} = 3750\ N

Explanation:

According to Pascal's Principle, the pressure is transmitted equally throughout the hydraulic fluid. Because the output area is larger, the output force is magnified.