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Physics: Motion and Mechanics - Pressure in Solids and Fluids

Grade 8IB

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

🔑Concepts

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Pressure is defined as the force applied perpendicular to the surface of an object per unit area over which that force is distributed. The SI unit of pressure is the Pascal (PaPa), where 1Pa=1N/m21 Pa = 1 N/m^2.

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In solids, pressure depends on the magnitude of the force and the area of contact. For a constant force, the pressure is inversely proportional to the area: P∝1AP \propto \frac{1}{A}. This explains why sharp knives cut better than blunt ones.

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Pressure in fluids (liquids and gases) increases with depth because of the weight of the fluid above. It acts equally in all directions at any given depth.

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The density of the fluid (ρ\rho) and the gravitational field strength (gg) also determine the pressure in a liquid. Higher density results in higher pressure at the same depth.

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Pascal's Law states that when pressure is applied to an enclosed fluid, it is transmitted equally in all directions throughout the fluid. This principle is the basis for hydraulic systems.

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

📐Formulae

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

P=ρghP = \rho g h

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

Ptotal=Patm+ρghP_{total} = P_{atm} + \rho g h

💡Examples

Problem 1:

A person weighing 600N600 N stands on one foot. The area of the sole of their shoe is 0.015m20.015 m^2. Calculate the pressure exerted on the ground.

Solution:

P=6000.015=40,000PaP = \frac{600}{0.015} = 40,000 Pa

Explanation:

Using the formula P=FAP = \frac{F}{A}, we divide the force (weight) of 600N600 N by the contact area of 0.015m20.015 m^2 to find the pressure in Pascals.

Problem 2:

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

Solution:

P=1000×9.8×3=29,400PaP = 1000 \times 9.8 \times 3 = 29,400 Pa

Explanation:

The pressure in a fluid is calculated using P=ρghP = \rho g h. Substituting the values: 1000kg/m3×9.8m/s2×3m1000 kg/m^3 \times 9.8 m/s^2 \times 3 m gives 29,400Pa29,400 Pa.

Problem 3:

A hydraulic lift has a small piston with an area of 0.05m20.05 m^2 and a large piston with an area of 0.8m20.8 m^2. If a force of 200N200 N is applied to the small piston, what is the force exerted by the large piston?

Solution:

2000.05=F20.8  ⟹  F2=200×0.80.05=3200N\frac{200}{0.05} = \frac{F_2}{0.8} \implies F_2 = \frac{200 \times 0.8}{0.05} = 3200 N

Explanation:

According to Pascal's Law, the pressure is constant: P1=P2P_1 = P_2. Therefore, F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}. Solving for F2F_2 gives the output force.

Problem 4:

An object experiences an atmospheric pressure of 101,325Pa101,325 Pa and a fluid pressure of 45,000Pa45,000 Pa. Calculate the total pressure using vertical addition.

Solution:

101325+45000146325\begin{array}{r} 101325 \\ + 45000 \\ \hline 146325 \end{array} Total Pressure = 146,325Pa146,325 Pa

Explanation:

The total pressure is the sum of the atmospheric pressure and the gauge pressure (fluid pressure).