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Physics: Motion and Mechanics - Moments, Levers, Torque, and Centre of Mass

Grade 8IB

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

🔑Concepts

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The turning effect of a force is known as a Moment or Torque. It is calculated by multiplying the force applied by the perpendicular distance from the pivot (fulcrum).

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The Principle of Moments states that for an object to be in equilibrium (balanced), the sum of the clockwise moments about a pivot must be equal to the sum of the anticlockwise moments about the same pivot: ∑Mclockwise=∑Manticlockwise\sum M_{clockwise} = \sum M_{anticlockwise}.

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Levers are simple machines consisting of a rigid beam and a pivot. They are categorized into three classes: Class 1 (Pivot in the middle, e.g., seesaw), Class 2 (Load in the middle, e.g., wheelbarrow), and Class 3 (Effort in the middle, e.g., tweezers).

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The Centre of Mass (or Centre of Gravity) is the point through which the entire weight of an object appears to act. For a symmetrical object of uniform density, the centre of mass is at its geometric center.

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An object's Stability depends on the position of its centre of mass. An object is stable if its centre of mass is low and its vertical line of action falls within its base area. If the line of action falls outside the base, the object will topple.

📐Formulae

M=F×dM = F \times d

Moment=Force (N)×Perpendicular distance from pivot (m)\text{Moment} = \text{Force (N)} \times \text{Perpendicular distance from pivot (m)}

F1×d1=F2×d2F_1 \times d_1 = F_2 \times d_2

Unit of Moment=N⋅m\text{Unit of Moment} = N \cdot m

💡Examples

Problem 1:

A mechanic uses a wrench to loosen a nut. He applies a force of 50N50 N at a distance of 0.25m0.25 m from the center of the nut. Calculate the moment of the force.

Solution:

M=F×dM = F \times d M=50×0.25M = 50 \times 0.25 M=12.5N⋅mM = 12.5 N \cdot m

Explanation:

The moment is found by multiplying the applied force by the perpendicular distance from the pivot point (the nut).

Problem 2:

A uniform seesaw is balanced at its center. A child with a weight of 400N400 N sits 1.5m1.5 m from the pivot. How far from the pivot must a second child weighing 300N300 N sit on the opposite side to balance the seesaw?

Solution:

Using the Principle of Moments: Anticlockwise Moment=Clockwise Moment\text{Anticlockwise Moment} = \text{Clockwise Moment} F1×d1=F2×d2F_1 \times d_1 = F_2 \times d_2 400×1.5=300×d2400 \times 1.5 = 300 \times d_2 600=300×d2600 = 300 \times d_2 d2=600300d_2 = \frac{600}{300} d2=2md_2 = 2 m

Explanation:

To achieve equilibrium, the clockwise and anticlockwise moments must be equal. We solve for the unknown distance d2d_2 by dividing the total anticlockwise moment by the weight of the second child.

Problem 3:

A heavy crate is placed on a plank. If the total weight of the system is 800N800 N acting through the centre of mass, and the support forces at two ends are FAF_A and FBF_B, what is the sum of these forces?

Solution:

FA+FB=800N\begin{array}{r} F_A + F_B = 800 N \end{array}

Explanation:

For an object in vertical equilibrium, the sum of the upward forces must equal the sum of the downward forces (the total weight acting through the centre of mass).