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Force and Pressure - Pressure: Definition and Calculation

Grade 8CBSE

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

🔑Concepts

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Pressure is defined as the force acting perpendicularly on a unit area of a surface. The perpendicular force is also referred to as thrust.

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The SI unit of pressure is the Pascal (PaPa), which is equivalent to one Newton per square metre (1N/m21 N/m^2).

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Pressure is directly proportional to the applied force (FF) and inversely proportional to the area (AA) over which the force acts.

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For the same amount of force, a smaller area of contact results in higher pressure. This explains why tools like knives and needles have sharp edges/points (small Asmall\ A).

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Liquids and gases (fluids) also exert pressure on the walls of their containers. Liquid pressure increases with depth.

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Atmospheric pressure is the pressure exerted by the weight of air in the atmosphere. At sea level, it is approximately 101,325 Pa101,325\ Pa.

📐Formulae

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

1 Pa=1 N/m21\ Pa = 1\ N/m^2

Force(F)=P×AForce (F) = P \times A

Area(A)=FPArea (A) = \frac{F}{P}

💡Examples

Problem 1:

A force of 200 N200\ N is applied over an area of 5 m25\ m^2. Calculate the pressure exerted on the surface.

Solution:

Given: Force F=200 NF = 200\ N, Area A=5 m2A = 5\ m^2. Using the formula P=FAP = \frac{F}{A}, we get P=200 N5 m2=40 N/m2P = \frac{200\ N}{5\ m^2} = 40\ N/m^2 or 40 Pa40\ Pa.

Explanation:

The pressure is calculated by dividing the total perpendicular force by the total area over which it is distributed.

Problem 2:

A solid block weighs 60 N60\ N. It is placed on a table such that its area of contact is 0.03 m20.03\ m^2. Find the pressure exerted by the block on the table.

Solution:

Given: Weight (Force) F=60 NF = 60\ N, Area A=0.03 m2A = 0.03\ m^2. Pressure P=FA=600.03=60×1003=2000 PaP = \frac{F}{A} = \frac{60}{0.03} = \frac{60 \times 100}{3} = 2000\ Pa.

Explanation:

The weight of the object acts as the force. Converting the decimal area into a fraction helps in simplifying the calculation to find the pressure in Pascals.

Problem 3:

Why do school bags have broad straps instead of thin strings?

Solution:

Broad straps have a larger area (AA). According to the formula P=FAP = \frac{F}{A}, pressure is inversely proportional to area. A larger area reduces the pressure exerted on the student's shoulders for the same weight (force) of the bag.

Explanation:

By increasing the area of contact, the distribution of force becomes wider, resulting in less physical pain or pressure on the shoulders.