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Motion, Forces and Energy - Turning effects (Moments)

Grade 11IGCSEPhysics

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

🔑Concepts

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The moment of a force is a measure of its turning effect about a specific point, known as the pivot or fulcrum.

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A moment depends on two factors: the magnitude of the force (FF) and the perpendicular distance (dd) from the line of action of the force to the pivot.

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The SI unit for a moment is the Newton-metre (N⋅mN \cdot m).

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Principle of Moments: For an object to be in rotational equilibrium, the sum of the clockwise moments about any point must equal the sum of the anticlockwise moments about that same point: ∑Mclockwise=∑Manticlockwise\sum M_{clockwise} = \sum M_{anticlockwise}.

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Center of Mass: The point through which the entire weight of an object is considered to act. For a uniform object, the center of mass is located at its geometric center.

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An object will tip over if its center of mass acts outside the boundary of its base, creating a resultant moment that causes rotation.

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Stability can be increased by broadening the base of the object or lowering its center of mass.

📐Formulae

M=F×dM = F \times d

∑(F×d)clockwise=∑(F×d)anticlockwise\sum (F \times d)_{clockwise} = \sum (F \times d)_{anticlockwise}

W=m×gW = m \times g

💡Examples

Problem 1:

A uniform beam of length 4.0 m4.0\text{ m} is pivoted at its center. A force of 60 N60\text{ N} is applied downwards at a distance of 1.5 m1.5\text{ m} to the left of the pivot. What force FF must be applied downwards at a distance of 2.0 m2.0\text{ m} to the right of the pivot to maintain equilibrium?

Solution:

Using the Principle of Moments: Anticlockwise Moment=Clockwise Moment\text{Anticlockwise Moment} = \text{Clockwise Moment}. 60 N×1.5 m=F×2.0 m60\text{ N} \times 1.5\text{ m} = F \times 2.0\text{ m}. 90 N⋅m=F×2.0 m90\text{ N}\cdot\text{m} = F \times 2.0\text{ m}. F=902.0=45 NF = \frac{90}{2.0} = 45\text{ N}.

Explanation:

Since the beam is uniform and pivoted at the center, the weight of the beam acts exactly at the pivot, meaning its moment is 00 and it does not affect the balance. We equate the turning effect on the left to the turning effect on the right.

Problem 2:

A see-saw is balanced by a girl of mass 40 kg40\text{ kg} sitting 2.0 m2.0\text{ m} from the pivot and a boy sitting 1.6 m1.6\text{ m} from the pivot on the opposite side. Calculate the weight of the boy (assume g=9.8 m/s2g = 9.8\text{ m/s}^2).

Solution:

First, calculate the girl's weight: Wg=m×g=40 kg×9.8 m/s2=392 NW_g = m \times g = 40\text{ kg} \times 9.8\text{ m/s}^2 = 392\text{ N}. Apply Principle of Moments: Wgirl×dgirl=Wboy×dboyW_{girl} \times d_{girl} = W_{boy} \times d_{boy}. 392 N×2.0 m=Wboy×1.6 m392\text{ N} \times 2.0\text{ m} = W_{boy} \times 1.6\text{ m}. 784 N⋅m=Wboy×1.6 m784\text{ N}\cdot\text{m} = W_{boy} \times 1.6\text{ m}. Wboy=7841.6=490 NW_{boy} = \frac{784}{1.6} = 490\text{ N}.

Explanation:

The downward force provided by each person is their weight (mgmg). To balance the see-saw, the clockwise and anticlockwise moments created by these weights must be equal.

Turning effects (Moments) - Revision Notes & Key Formulas | IGCSE Grade 11 Physics