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Forces and Motion - Moments, Levers, and Centre of Mass

Grade 9IB

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

🔑Concepts

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Moment of a Force: The turning effect of a force about a pivot point. It depends on the magnitude of the force applied and the perpendicular distance from the pivot to the line of action of the force. Its SI unit is Newton-metres (N m\text{N m}).

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Principle of Moments: For an object in equilibrium, the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about that same point. This is expressed as ∑Mclockwise=∑Manticlockwise\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}}.

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Levers: Simple machines that use a pivot (fulcrum) to multiply force or distance. They are categorized into three classes: Class 1 (Pivot in middle), Class 2 (Load in middle), and Class 3 (Effort in middle).

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Centre of Mass: The unique point of an object where the entire mass of the body may be considered to be concentrated. For uniform symmetrical objects, the centre of mass is at the geometric centre.

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Stability and Toppling: Stability is a measure of how likely an object is to tip over. An object is stable if its centre of mass is low and it has a wide base. An object will topple if the vertical line through its centre of mass falls outside its base area.

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Equilibrium: An object is in total equilibrium if there is no resultant force acting on it (translational equilibrium) and no resultant moment acting on it (rotational equilibrium).

📐Formulae

M=F×dM = F \times d

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

W=m×gW = m \times g

Mechanical Advantage=LoadEffort\text{Mechanical Advantage} = \frac{\text{Load}}{\text{Effort}}

💡Examples

Problem 1:

A see-saw is 4.0 m4.0\text{ m} long and balanced at its center. A child with a weight of 300 N300\text{ N} sits 1.5 m1.5\text{ m} from the pivot. Where should a second child weighing 450 N450\text{ N} sit to balance the see-saw?

Solution:

Anticlockwise Moment=Clockwise MomentF1×d1=F2×d2300×1.5=450×d2450=450×d2d2=450450d2=1.0 m\begin{array}{r} \text{Anticlockwise Moment} = \text{Clockwise Moment} \\ F_1 \times d_1 = F_2 \times d_2 \\ 300 \times 1.5 = 450 \times d_2 \\ 450 = 450 \times d_2 \\ d_2 = \frac{450}{450} \\ d_2 = 1.0\text{ m} \end{array}

Explanation:

To achieve equilibrium, the anticlockwise moment produced by the first child must be balanced by the clockwise moment produced by the second child. By rearranging the formula for the moment, we find the distance d2d_2 is 1.0 m1.0\text{ m} from the pivot.

Problem 2:

A uniform beam of mass 5 kg5\text{ kg} and length 2 m2\text{ m} is supported at one end. Calculate the moment produced by the weight of the beam about that support (Use g=10 N/kgg = 10\text{ N/kg}).

Solution:

W=m×g=5×10=50 Nd=length2=22=1 mM=F×d=50×1=50 N m\begin{array}{r} W = m \times g = 5 \times 10 = 50\text{ N} \\ d = \frac{\text{length}}{2} = \frac{2}{2} = 1\text{ m} \\ M = F \times d = 50 \times 1 = 50\text{ N m} \end{array}

Explanation:

For a uniform beam, the weight acts through its centre of mass, which is at its geometric midpoint. Therefore, the distance from the end support to the centre of mass is 1 m1\text{ m}. The moment is the weight multiplied by this distance.