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Newton's Laws of Motion - Turning Forces (Moment of Force/Torque)-advanced

Grade 9CBSE

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

🔑Concepts

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Moment of Force (Torque) is the measure of the turning effect of a force about a pivot or axis of rotation. It is defined as the product of the magnitude of the force (FF) and the perpendicular distance (dd) of the line of action of the force from the axis of rotation.

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The S.I. unit of moment of force is Newton-meter (N⋅mN \cdot m). In the C.G.S. system, it is dyne⋅cmdyne \cdot cm. 1N⋅m=107dyne⋅cm1 N \cdot m = 10^7 dyne \cdot cm.

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Direction and Sign Convention: By convention, an anti-clockwise moment is taken as positive (++), while a clockwise moment is taken as negative (−-).

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Principle of Moments: For a body to be in rotational equilibrium, the sum of clockwise moments about any point must be equal to the sum of anti-clockwise moments about that same point. Mathematically: ∑Clockwise Moments=∑Anti-clockwise Moments\sum \text{Clockwise Moments} = \sum \text{Anti-clockwise Moments}.

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A Couple consists of two equal and opposite parallel forces acting along different lines of action. The moment of a couple (or torque of the couple) is the product of either force and the perpendicular distance between the two forces (called the arm of the couple).

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Factors affecting the moment of force: (i) The magnitude of the force applied (FF), and (ii) The perpendicular distance of the line of action of the force from the axis of rotation (dd). To increase the turning effect, one can either increase the force or increase the distance from the pivot.

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Center of Gravity (C.G.): This is the point through which the entire weight of the body acts. For a uniform body, the weight can be considered as a single force acting downwards at its geometric center.

📐Formulae

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

Moment of Couple=F×Arm of Couple\text{Moment of Couple} = F \times \text{Arm of Couple}

∑τanti-clockwise=∑τclockwise (At Equilibrium)\sum \tau_{\text{anti-clockwise}} = \sum \tau_{\text{clockwise}} \text{ (At Equilibrium)}

1 kgf⋅m=9.8N⋅m1 \text{ kgf} \cdot m = 9.8 N \cdot m

1 gf⋅cm=980dyne⋅cm1 \text{ gf} \cdot cm = 980 dyne \cdot cm

💡Examples

Problem 1:

A uniform meter scale is balanced at its 50 cm50 \text{ cm} mark. A weight of 40 gf40 \text{ gf} is placed at the 10 cm10 \text{ cm} mark. Where should a weight of 80 gf80 \text{ gf} be placed to balance the scale?

Solution:

Let the 80 gf80 \text{ gf} weight be placed at a distance xx from the pivot (50 cm50 \text{ cm} mark) on the right side.

Distance of 40 gf40 \text{ gf} from pivot = 50 cm−10 cm=40 cm50 \text{ cm} - 10 \text{ cm} = 40 \text{ cm} (Anti-clockwise moment)

Anti-clockwise moment = F1×d1=40 gf×40 cm=1600 gf⋅cmF_1 \times d_1 = 40 \text{ gf} \times 40 \text{ cm} = 1600 \text{ gf} \cdot cm

Clockwise moment = F2×x=80 gf×xF_2 \times x = 80 \text{ gf} \times x

According to the Principle of Moments: 1600=80×x1600 = 80 \times x x=160080=20 cmx = \frac{1600}{80} = 20 \text{ cm}

The weight must be placed 20 cm20 \text{ cm} to the right of the pivot. Position on scale = 50 cm+20 cm=70 cm50 \text{ cm} + 20 \text{ cm} = 70 \text{ cm}.

Explanation:

To balance the scale, the anti-clockwise turning effect produced by the 40 gf40 \text{ gf} weight must be countered by an equal clockwise turning effect from the 80 gf80 \text{ gf} weight.

Problem 2:

Calculate the moment of a couple where two equal forces of 5N5 N act in opposite directions at the ends of a rod 0.5 m0.5 \text{ m} long.

Solution:

Given: Force F=5NF = 5 N Arm of the couple d=0.5 md = 0.5 \text{ m}

Formula for Moment of Couple: Moment=F×d\text{Moment} = F \times d Moment=5×0.5\text{Moment} = 5 \times 0.5 Moment=2.5N⋅m\text{Moment} = 2.5 N \cdot m

The moment of the couple is 2.5N⋅m2.5 N \cdot m.

Explanation:

A couple produces rotation without translation. The total moment is the sum of the moments of individual forces, which simplifies to the product of one force and the distance between them.