Forces and Moments

Overview

This topic explores how objects interact through forces and how these forces produce both linear and rotational effects. Understanding forces and moments is essential for analyzing structures, machines, and motion in physics and engineering contexts.

Contents

• Types of forces
• Hooke’s Law and Elastic Forces
• Moment and torque
• Couples and the Principle of Moments
• Equilibrium Conditions

Learning Outcomes

By the end of this section, students should be able to:

• Describe forces on masses, charges, and currents in various fields
• Understand forces such as normal, frictional, buoyant, and viscous
• Apply Hooke’s Law: F = k × x
• Define and apply moment of a force and torque
• Understand the concept of a couple
• Apply the principle of moments and define the centre of gravity
• Analyze equilibrium using free-body diagrams and vector methods

1. Types of Forces

Contact forces (require touching):

• Normal force — acts perpendicular to the surface; balances compression (e.g., table on book).
• Friction — opposes relative motion along the surface; kinetic: f_k = μ_k × N, static: f_s ≤ μ_s × N.
• Tension — pull transmitted by an ideal (massless, inextensible) string; same magnitude along the string.
• Spring force — linear elastic restoring force: F = k × x (toward equilibrium; k in N·m^-1).

Non-contact (field) forces (act at a distance):

• Gravitational force — attraction between masses: F = G × (m₁ × m₂) / r² (near Earth’s surface: weight W = m × g).
• Electrostatic force — between charges: F = k × (|q₁ × q₂|) / r² (repel/attract depending on signs).
• Magnetic force — on a moving charge: F = q × v × B × sin(θ) (or on a wire F = I × L × B × sin(θ)).
• Buoyant force — upward fluid force: F_b = ρ × g × V_disp (Archimedes’ principle).
• Viscous/drag force — resistance in fluids; low speed: F ∝ v (e.g., 6 × π × η × r × v); high speed: F ∝ v².

Tip: When drawing a free-body diagram (FBD), show each force as an arrow from the object’s centre; label magnitudes/angles and use up to two axes so you can write ∑ F_x = 0, ∑ F_y = 0 (and ∑ M = 0 for rotation) clearly.

2. Moment and Torque

What is a moment? The moment of a force (turning effect) about a pivot measures how strongly the force tends to rotate an object.

• Scalar (perpendicular) definition: For a force applied perpendicularly at a distance r from the pivot, M = F × r with SI unit N·m.
• General (angled) case: When the force makes an angle θ to the lever arm:
  M = F × r × sin(θ)     or     M = F_⊥ × r
  where F_⊥ = F × sin(θ) is the component of the force perpendicular to the lever arm.
• Vector definition (torque): τ = r × F; |τ| = r × F × sin(θ). Direction given by the right-hand rule (thumb = τ, fingers curl from r to F).
• Units: Newton–metre (N·m). Moments/torques are not “joules” even though 1 N·m = 1 J dimensionally; energy is a scalar work quantity, torque is an axial pseudovector quantity.
• Sign convention: In 2D statics, take counter-clockwise positive (CCW +), clockwise negative (CW −). Be consistent across the page.

Rotational effect (moment) M = F × r for perpendicular force; in general M = F × r × sin(θ).

Worked Example (perpendicular force):
A 10 N force is applied perpendicularly at 0.5 m from a hinge.
M = F × r = 10 × 0.5 = 5.0 N·m (CCW if force tends to rotate counter-clockwise).

Worked Example (angled force):
A 20 N force acts at 30° to a spanner of length 0.40 m.
M = F × r × sin(θ) = 20 × 0.40 × sin(30°) = 20 × 0.40 × 0.5 = 4.0 N·m.
Only the perpendicular component F_⊥ = 20 × sin(30°) = 10 N produces rotation; the along-spanner component does not contribute to moment.

SEO note for learners: Engineers often search “moment vs torque,” “lever arm,” and “right-hand rule.” Remember: moment is the turning effect; torque is the vector form τ = r × F; use M = F × r × sin(θ) for angled forces, and N·m as the SI unit.

3. Couples and the Principle of Moments

Couple:

• Two equal and opposite forces acting on different lines of action.
• Produces pure rotation (no net force, only turning effect).

Principle of Moments:

• For an object in rotational equilibrium: Sum of clockwise moments = Sum of anticlockwise moments.
• The diagram below illustrates the moment due to two equal forces F acting at distance D apart in opposite directions, forming a pure couple.

Pure Couple

Example:
A uniform beam is balanced on a pivot. A 20 N weight is 1.5 m from the pivot on one side. What force must balance it 1 m from the pivot on the other side?
Clockwise Moment = 20 × 1.5 = 30 N·m
Let force F × 1 = 30 ⇒ F = 30 N

4. Equilibrium Conditions

Translational Equilibrium (no net force):

• Object remains at rest or moves with constant velocity (Newton’s First Law).
• In 2D: ∑ F_x = 0, ∑ F_y = 0. In 3D, also ∑ F_z = 0.

Rotational Equilibrium (no net moment/torque):

• No angular acceleration; object does not start/stop spinning.
• About any chosen pivot: ∑ M = 0.

Free-Body Diagrams (FBDs):

• Show the body as a simple shape (point/box/beam) and draw all external forces acting on it.
• Label magnitudes and directions; choose and mark a sign convention for moments (e.g., counter-clockwise +).

Block on a horizontal surface: typical forces F (applied), f (friction), m × g (weight), N (normal).

Friction models (useful in FBDs):

• Static friction (no slip): |f| ≤ μ_s × N (at impending motion: f = μ_s × N).
• Kinetic friction (sliding): f_k = μ_k × N opposing motion.

Inclined Plane Decomposition (angle θ to horizontal):

• Weight: m × g acts vertically downward.
• Components along/normal to plane: m × g × sin(θ) (down the plane), m × g × cos(θ) (into the plane).
• Normal reaction (no lift-off): N = m × g × cos(θ).
• If at rest (no slip): ∑ F_parallel = 0 ⇒ f = m × g × sin(θ) with |f| ≤ μ_s × N.

Inclined plane: resolve m × g into m × g × sin(θ) (parallel) and m × g × cos(θ) (normal).

Beam Equilibrium (supports & moments):

A beam is in equilibrium when both translation and rotation balance: ∑ F_x = 0, ∑ F_y = 0, ∑ M = 0.
For a simply supported beam with reactions R_A, R_B, and a point load P at distance a from support A:

Beam equilibrium: use ∑ F = 0 and ∑ M = 0 to solve for reactions.

Worked Example (Ladder idea, FBD thinking):
A ladder leans against a rough wall and rests on a rough floor. Draw the FBD: weight m × g at the ladder’s centre, normal reactions at wall and floor, and friction forces at both contacts opposing possible slip. Apply ∑ F_x = 0, ∑ F_y = 0, ∑ M = 0 about a convenient pivot to find unknown reactions. Check that friction forces obey |f| ≤ μ × N.

Some Interesting Examples

Example 1: Equilibrium of Forces — Vector Method

Problem. Two concurrent forces act at point O: A = 80 N directed horizontally to the right, and B = 60 N directed vertically upward. Find the single reaction force R (magnitude and direction) which, when applied at O, will keep the point in equilibrium (∑ F = 0).

Section 1 — Graphical (tip-to-tail) vector addition

1. From O, draw vector A to the right (length proportional to 80 N).
2. From the tip of A, draw vector B upward (length proportional to 60 N).
3. The closing vector from the tip of B back to O is the reaction R, satisfying A + B + R = 0 ⇒ R = −(A + B).

Vector addition (tip-to-tail): the reaction is equal in magnitude and opposite in direction to the resultant.

Section 2 — Resolved-component (analytic) method

1. Resolve forces into Cartesian components (+x right, +y upward): A = (80, 0) N, B = (0, 60) N.
2. Resultant to oppose: A + B = (80, 60) N ⇒ R = −(A + B) = (-80, -60) N.
3. Magnitude of reaction: |R| = √((-80)² + (-60)²) = √(6400 + 3600) = √(10000) = 100 N.
4. Direction of reaction: θ_R = arctan(R_y / R_x) = arctan(-60 / -80). Since R_x < 0 and R_y < 0 (third quadrant), add 180°: θ_R = 180° + 36.87° = 216.87° (or 36.9° below the -x direction).

Component method: sum components, then compute magnitude and direction.

Example 2: Centre of Gravity (Barycentre) of the Sun–Earth–Moon System

Problem. Three bodies lie approximately on a line: Sun (M_⊙ = 1.989 × 10³⁰ kg at x = 0), Earth (M_E = 5.972 × 10²⁴ kg at x = 1.496 × 10¹¹ m), and Moon (M_m = 7.347 × 10²² kg at x = 1.49984 × 10¹¹ m).

1D Barycentre Formula: x_CG = (∑ M_i × x_i) / (∑ M_i)

1. Total mass: M_tot = 1.989006045 × 10³⁰ kg.
2. Weighted sum of positions: ∑ M_i × x_i ≈ 9.044305245 × 10³⁵ kg·m.
3. Centre of gravity: x_CG = 9.044305245 × 10³⁵ / 1.989006045 × 10³⁰ ≈ 4.55 × 10⁵ m (455 km from the Sun’s centre, deep inside the Sun).

Barycentre (1D approximation; not to scale)

Example 3: Maximum Incline Before Sliding

Problem. A box of mass m = 10 kg rests on a rough plane with static friction coefficient μ = 0.20. What is the largest angle θ_max for which the box does not slide?

1. Resolve forces: W_parallel = m × g × sin(θ), W_normal = m × g × cos(θ), f_max = μ × m × g × cos(θ).
2. Limiting equilibrium balance: m × g × sin(θ) = μ × m × g × cos(θ).
3. Solve for angle: tan(θ_max) = μ = 0.20 ⇒ θ_max = arctan(0.20) ≈ 11.3°.

Free-body diagram on an incline at limiting equilibrium.

Key Concepts Recap

Interactive Review Questions

Additional Resources

To deepen your understanding, you are strongly encouraged to explore the materials provided on the following pages:

• Classical Mechanics — Principles governing rigid body systems.
• Dynamics and Kinematics — Motion resulting from non-zero net forces and torques.
• Newton’s First Law of Motion — Translational and rotational equilibrium concepts.
• Newton’s Second Law of Motion — Force components and angular acceleration.
• Newton’s Third Law of Motion — Interaction force pairs in static systems.
• Fundamental Forces of Nature — Gravitational and electromagnetic fields.
• Kinematics — Trajectory and displacement analysis.
• Statics — Advanced structural load and reaction analysis.
• Solid Mechanics — Internal shear forces and bending moments in beams.

End-of-Page Comprehensive Practice Module