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Motion, Forces and Energy - Effects of forces

Grade 12IGCSEPhysics

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

🔑Concepts

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A resultant force acting on an object causes it to accelerate. According to Newton's Second Law, the acceleration aa is directly proportional to the resultant force FF and inversely proportional to the mass mm: F=maF = ma.

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Hooke's Law states that the extension Δx\Delta x of a spring is directly proportional to the applied load FF, provided the limit of proportionality is not exceeded: F=kΔxF = k \Delta x, where kk is the spring constant.

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The moment of a force is its turning effect about a pivot. It is calculated as the product of the force FF and the perpendicular distance dd from the pivot to the line of action of the force: Moment=F×d\text{Moment} = F \times d.

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For an object to be in static equilibrium, two conditions must be satisfied: the resultant force in any direction must be zero (∑F=0∑ F = 0) and the sum of clockwise moments must equal the sum of anticlockwise moments about any point (∑M=0∑ M = 0).

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Centripetal force is the resultant force required to keep an object moving in a circle. It acts towards the center of the circle, perpendicular to the object's velocity, and is given by Fc=mv2rF_c = \frac{mv^2}{r}.

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Friction is a force between two surfaces that impedes motion and results in heating. Air resistance (drag) is a form of friction that increases with the speed of an object moving through a fluid.

📐Formulae

F=maF = ma

F=kΔxF = k \Delta x

Moment=F×d\text{Moment} = F \times d

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

Fc=mv2rF_c = \frac{mv^2}{r}

∑Clockwise Moments=∑Anticlockwise Moments\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}

💡Examples

Problem 1:

A uniform beam of length 2.0 m2.0\text{ m} is pivoted at its center. A weight of 50 N50\text{ N} is placed 0.8 m0.8\text{ m} to the left of the pivot. Calculate the force FF required at the right end (1.0 m1.0\text{ m} from the pivot) to maintain equilibrium.

Solution:

Using the Principle of Moments: Clockwise Moments=Anticlockwise Moments\text{Clockwise Moments} = \text{Anticlockwise Moments}. F×1.0 m=50 N×0.8 mF \times 1.0\text{ m} = 50\text{ N} \times 0.8\text{ m}. F=40 Nm1.0 m=40 NF = \frac{40\text{ Nm}}{1.0\text{ m}} = 40\text{ N}.

Explanation:

Since the beam is uniform and pivoted at its center, the weight of the beam acts through the pivot and exerts no moment. We equate the clockwise moment (from force FF) to the anticlockwise moment (from the 50 N50\text{ N} weight).

Problem 2:

A spring with a spring constant k=200 N/mk = 200\text{ N/m} is compressed by 0.05 m0.05\text{ m}. Calculate the force exerted by the spring.

Solution:

F=kΔx=200 N/m×0.05 m=10 NF = k \Delta x = 200\text{ N/m} \times 0.05\text{ m} = 10\text{ N}.

Explanation:

According to Hooke's Law, the force is the product of the spring constant and the displacement (compression or extension).

Problem 3:

An object of mass 5 kg5\text{ kg} is pulled by a force of 25 N25\text{ N} to the right, while a frictional force of 10 N10\text{ N} acts to the left. Determine the acceleration aa of the object.

Solution:

Resultant Force Fres=25 N−10 N=15 NF_{res} = 25\text{ N} - 10\text{ N} = 15\text{ N}. Using F=maF = ma, 15=5×a15 = 5 \times a. a=155=3 m/s2a = \frac{15}{5} = 3\text{ m/s}^2.

Explanation:

The acceleration is determined by the net (resultant) force acting on the mass. Friction opposes the pulling force, so it is subtracted.

Effects of forces - Revision Notes & Key Formulas | IGCSE Grade 12 Physics