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Gravitation - Thrust and Pressure

Grade 9CBSE

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

🔑Concepts

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Thrust is defined as the total force acting perpendicular to a surface. The SI unit of thrust is the Newton (NN).

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Pressure is defined as the thrust per unit area of a surface, expressed as P=FAP = \frac{F}{A}.

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

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For a constant thrust, pressure is inversely proportional to the area. Therefore, a smaller area results in higher pressure (e.g., the sharp edge of a knife).

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Buoyancy (or Buoyant Force) is the upward force exerted by a fluid on any object immersed in it. This force acts in the direction opposite to gravity.

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Archimedes' Principle states that when a body is immersed fully or partially in a fluid, it experiences an upward force equal to the weight of the fluid displaced by it.

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Density is the mass per unit volume of a substance, represented as ρ=mV\rho = \frac{m}{V}. The SI unit is kg/m3kg/m^3.

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Relative Density is the ratio of the density of a substance to the density of water: Relative Density=Density of substanceDensity of water\text{Relative Density} = \frac{\text{Density of substance}}{\text{Density of water}}. It is a dimensionless quantity.

📐Formulae

Pressure(P)=Thrust(F)Area(A)Pressure (P) = \frac{Thrust (F)}{Area (A)}

Density(ρ)=Mass(m)Volume(V)Density (\rho) = \frac{Mass (m)}{Volume (V)}

Relative Density=Density of substanceDensity of water\text{Relative Density} = \frac{\text{Density of substance}}{\text{Density of water}}

Fbuoyant=V×ρfluid×gF_{buoyant} = V \times \rho_{fluid} \times g

💡Examples

Problem 1:

A block of mass 5 kg5\ kg is placed on a table. The dimensions of the block are 20 cm×10 cm×5 cm20\ cm \times 10\ cm \times 5\ cm. Calculate the pressure exerted by the block on the table if it is placed on its side with dimensions 20 cm×10 cm20\ cm \times 10\ cm. (Take g=9.8 m/s2g = 9.8\ m/s^2)

Solution:

Given: m=5 kgm = 5\ kg, g=9.8 m/s2g = 9.8\ m/s^2, Area=20 cm×10 cm=200 cm2=0.02 m2Area = 20\ cm \times 10\ cm = 200\ cm^2 = 0.02\ m^2. Step 1: Calculate Thrust F=m×g=5×9.8=49 NF = m \times g = 5 \times 9.8 = 49\ N. Step 2: Calculate Pressure P=FA=490.02=2450 PaP = \frac{F}{A} = \frac{49}{0.02} = 2450\ Pa.

Explanation:

The thrust is the weight of the block acting downwards. The pressure is calculated by dividing this weight by the surface area in contact with the table.

Problem 2:

The relative density of silver is 10.810.8. If the density of water is 103 kg/m310^3\ kg/m^3, what is the density of silver in SI units?

Solution:

Relative Density=Density of silverDensity of water\text{Relative Density} = \frac{\text{Density of silver}}{\text{Density of water}} 10.8=Density of silver103 kg/m310.8 = \frac{\text{Density of silver}}{10^3\ kg/m^3} Density of silver=10.8×103 kg/m3=10800 kg/m3\text{Density of silver} = 10.8 \times 10^3\ kg/m^3 = 10800\ kg/m^3.

Explanation:

Relative density compares the substance's density to water's density. To find the absolute density, multiply the relative density by the density of water.

Thrust and Pressure - Revision Notes & Key Formulas | CBSE Class 9 Science