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Motion, Forces and Energy - Moments

Grade 12IGCSEPhysics

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 called the pivot or fulcrum. It is a vector quantity, though at IGCSE level, it is primarily treated as clockwise or anticlockwise.

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

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The Principle of Moments states that for an object in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about that same point: ∑Mclockwise=∑Manticlockwise\sum M_{clockwise} = \sum M_{anticlockwise}.

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An object is in static equilibrium only if two conditions are met: the resultant force acting on the object is zero (sumF=0\\sum F = 0) and the resultant moment acting on the object is zero (sumM=0\\sum M = 0).

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The Center of Gravity (CGCG) is the point through which the entire weight (W=mgW = mg) of an object appears to act. For a uniform object, the CGCG is located at its geometric center.

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Stability of an object is determined by the position of its CGCG. An object will topple if the vertical line acting downwards from its CGCG falls outside its base area.

📐Formulae

M=F×dM = F \times d

∑Mclockwise=∑Manticlockwise\sum M_{clockwise} = \sum M_{anticlockwise}

W=m×gW = m \times g

Resultant Moment=∣∑MCW−∑MACW∣\text{Resultant Moment} = |\sum M_{CW} - \sum M_{ACW}|

💡Examples

Problem 1:

A uniform plank of length 4.0 m4.0\text{ m} and weight 100 N100\text{ N} is supported by a pivot at its center. A child weighing 300 N300\text{ N} sits at a distance of 1.5 m1.5\text{ m} from the pivot. Calculate the force FF required at the opposite end of the plank to keep it horizontal.

Solution:

  1. Identify the pivot: The center of the plank (2.0 m2.0\text{ m} from either end).
  2. Identify moments: The child creates an anticlockwise moment: MACW=300 N×1.5 m=450 NmM_{ACW} = 300\text{ N} \times 1.5\text{ m} = 450\text{ Nm}.
  3. The force FF at the end (2.0 m2.0\text{ m} from pivot) creates a clockwise moment: MCW=F×2.0 mM_{CW} = F \times 2.0\text{ m}.
  4. Apply Principle of Moments: 450=F×2.0⇒F=4502.0=225 N450 = F \times 2.0 \Rightarrow F = \frac{450}{2.0} = 225\text{ N}.

Explanation:

Since the plank is uniform and pivoted at the center, the weight of the plank acts through the pivot and creates zero moment. We only balance the moments created by the child and the applied force FF.

Problem 2:

A non-uniform rod ABAB of length 2.0 m2.0\text{ m} is balanced horizontally on a pivot placed 0.8 m0.8\text{ m} from end AA when a mass of 5.0 kg5.0\text{ kg} is hung from end AA. If the mass of the rod is 2.0 kg2.0\text{ kg}, find the distance of the center of gravity from end AA.

Solution:

  1. Convert mass to weight using g=9.8 m/s2g = 9.8\text{ m/s}^2 or 10 m/s210\text{ m/s}^2. Let g=10 m/s2g = 10\text{ m/s}^2. Weight at A=50 NA = 50\text{ N}. Weight of rod = 20 N20\text{ N}.
  2. Distance from AA to pivot =0.8 m= 0.8\text{ m}. Anticlockwise moment =50 N×0.8 m=40 Nm= 50\text{ N} \times 0.8\text{ m} = 40\text{ Nm}.
  3. Let the CGCG be xx meters from the pivot. Clockwise moment =20 N×x= 20\text{ N} \times x.
  4. Principle of Moments: 40=20x⇒x=2.0 m40 = 20x \Rightarrow x = 2.0\text{ m} from the pivot.
  5. Total distance from A=0.8 m+2.0 m=2.8 mA = 0.8\text{ m} + 2.0\text{ m} = 2.8\text{ m}.

Explanation:

In this problem, the rod is non-uniform, so the center of gravity is not at the midpoint. We find the relative position of the CGCG to the pivot first, then add it to the pivot's position from the end.

Moments - Revision Notes & Key Formulas | IGCSE Grade 12 Physics