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Physics - Force and Pressure (Turning effect, Fluid pressure, Atmospheric pressure)

Grade 8ICSE

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

🔑Concepts

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The turning effect of a force about a fixed point or axis is called the Moment of Force or Torque. It is determined by the magnitude of the force applied and the perpendicular distance of the line of action of the force from the axis of rotation: τ=F×d\tau = F \times d.

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Pressure is the thrust (normal force) acting per unit area: P=FAP = \frac{F}{A}. The S.I. unit of pressure is the Pascal (PaPa), where 1Pa=1N/m21 Pa = 1 N/m^2.

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Pressure in liquids (Fluid Pressure) depends on the depth (hh), the density of the liquid (ρ\rho), and acceleration due to gravity (gg). It is calculated using the formula P=hρgP = h \rho g.

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Atmospheric Pressure is the pressure exerted by the weight of the air column above the Earth's surface. It decreases with an increase in altitude. Standard atmospheric pressure at sea level is approximately 1.013×105Pa1.013 \times 10^5 Pa or 76cm76 cm of mercury (HgHg).

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Liquid pressure acts in all directions at a given point and increases with depth. It does not depend on the shape or size of the container, only on the vertical height of the liquid column.

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A Barometer is an instrument used to measure atmospheric pressure. Common types include the Simple Barometer and the Fortin's Barometer.

📐Formulae

Moment of Force(τ)=F×d\text{Moment of Force} (\tau) = F \times d

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

Liquid Pressure(P)=hρg\text{Liquid Pressure} (P) = h \rho g

Total Pressure=Patm+hρg\text{Total Pressure} = P_{atm} + h \rho g

💡Examples

Problem 1:

A force of 20N20 N is applied at a perpendicular distance of 0.5m0.5 m from the nut to open it using a wrench. Calculate the moment of force.

Solution:

Given: F=20NF = 20 N, d=0.5md = 0.5 m. Using the formula τ=F×d\tau = F \times d, we get τ=20N×0.5m=10Nm\tau = 20 N \times 0.5 m = 10 N m.

Explanation:

The turning effect depends on the distance from the pivot; applying force further from the nut makes it easier to turn.

Problem 2:

Calculate the pressure exerted by a brick of weight 30N30 N if its base area is 0.02m20.02 m^2.

Solution:

Given: F=30NF = 30 N, A=0.02m2A = 0.02 m^2. Pressure P=FA=300.02=1500PaP = \frac{F}{A} = \frac{30}{0.02} = 1500 Pa.

Explanation:

Pressure is the ratio of the force applied to the area over which it is distributed.

Problem 3:

Find the pressure exerted by water at the bottom of a swimming pool 3m3 m deep. (Take density of water ρ=1000kg/m3\rho = 1000 kg/m^3 and g=9.8m/s2g = 9.8 m/s^2)

Solution:

Given: h=3mh = 3 m, ρ=1000kg/m3\rho = 1000 kg/m^3, g=9.8m/s2g = 9.8 m/s^2. Using P=hρgP = h \rho g, P=3×1000×9.8=29,400PaP = 3 \times 1000 \times 9.8 = 29,400 Pa.

Explanation:

The pressure in a fluid is directly proportional to the depth and the density of the fluid.

Force and Pressure (Turning effect, Fluid pressure, Atmospheric pressure) Revision - Class 8…