Lecture Notes: Forces, Equilibrium, and Beams (Newton's Laws, Transmissibility, Vector Decomposition, Moments, and Structural Supports)

Newton's Laws

• First Law (Law of Inertia)

  • An object at rest stays at rest and an object in motion stays in uniform straight-line motion unless acted upon by a net external force.
  • Often called the law of inertia because it describes an object's tendency to maintain its state of motion or rest unless a force changes it.
  • Unbalanced force is required to change velocity; equal-and-opposite forces cancel and do not change motion.
  • On Earth, motion changes due to friction and air resistance; in deep space away from gravity, an object would continue moving with constant velocity until another force acts.
  • Summary: an object moves with constant velocity or stays still when there is no net force applied.

• Second Law (Law of Acceleration)

  • Formula: F = m a
  • Force is proportional to the acceleration required to change the velocity of an object with constant mass.
  • Units: Newtons (N), where 1~ ext{N} = 1~ ext{kg} \, ext{m}/ ext{s}^2
  • Default SI units: mass in kilograms (kg), acceleration in ext{m}/ ext{s}^2, force in N.
  • Example from transcript (with note on a minor calculation discrepancy):
  • Object mass: m = 20~ ext{kg}, acceleration due to gravity: a = 10~ ext{m}/ ext{s}^2
  • Correct force: F = m a = 20 \times 10 = 200\ ext{N}
  • Transcript shows an intermediate step leading to 2000\ ext{N} due to an extra multiplication; correct result for a vertical gravitational force is F = 200\ ext{N}
  • Note on units: Force is a derived unit, equal to the force needed to accelerate 1 kg at 1 m/s^2, i.e., 1~ ext{N} = 1~ ext{kg} \cdot ext{m}/ ext{s}^2.

• Third Law (Action-Reaction)

  • Statement: for every action there is an equal and opposite reaction.
  • More precise: if object A exerts a force on object B, then B exerts an equal-magnitude, opposite-direction force on A.
  • Forces come in pairs: e.g., pushing on a wall, gravitational action between Earth and Sun.
  • Normal force example: when an object rests on the ground, gravity pulls downward with weight, while the ground exerts an upward normal force of equal magnitude to balance (in static equilibrium).
  • Walking example: feet push down on the floor; floor pushes back up, enabling forward motion.
  • Key point: the action and reaction forces are equal in magnitude but opposite in direction.

Law of Transmissibility (Transmissibility Principle)

• Core idea: a force may be considered to act at any point along its line of action, as long as its magnitude and direction are unchanged.
• Implication: moving the point of application along the line of action does not change the external reaction forces on a rigid body; the resulting motion (acceleration or rotation) remains the same.
• Definitions:
  • LOA: Line of Action
  • The force acts the same whether applied at Point A, B, or C if along the same line of action.
• Practical consequence: forces with the same magnitude and direction on the same line of action are equivalent in terms of the resulting motion.
• Example wording from transcript: The force acts the same weather it is at points A, B, or C; the action remains the same.

Vector Forces and Components

• When a force F is diagonal, it can be decomposed into two perpendicular components along the x and y axes: Fx, Fy, such that: oldsymbol{F} = Fx \, oldsymbol{i} + Fy \, oldsymbol{j} and Fx^2 + Fy^2 = F^2.
• Common relationship (for a force at angle \theta\0 with the +x axis):
  • F_x = F \cos \theta
  • F_y = F \sin \theta
• Alternative perspective used in the notes (angle referenced to axes):
  • If the angle to horizontal is \theta, then
  • F_x = F \cos \theta
  • F_y = F \sin \theta
• Worked examples (from transcript):
  • Example: Force F = 100\ \,\text{kN} at 30° above the +x axis.
  • F_x = 100 \cos 30^{\circ} = 86.60\ \text{kN}
  • F_y = 100 \sin 30^{\circ} = 50.00\ \text{kN}
  • Another example: Force F = 100\ \text{kN} with angle such that vertical component angle to vertical is 60° (i.e., 30° to horizontal).
  • Fy = 100 \cos(60^{\circ}) = 50\ \text{kN} or equivalently Fy = 100 \sin(30^{\circ}) = 50\ \text{kN}
• Sign conventions:
  • Components in the +x and +y directions are taken as positive.
  • If the diagonal force points left-to-right or right-to-left, the x-component direction follows the diagonal.
  • In the course notes, the downward y-component is often taken as the positive direction for Fy, but you should adopt a consistent coordinate system.
• How to draw and read components:
  • Drop perpendiculars from the force head to the x and y axes to form the components.
  • Represent each component as a vector along its axis; magnitude is the component value, with sign indicating direction along the axis.
• Quick reminder: for a force F at angle θ relative to the +x axis,
  • F_x = F \cos \theta
  • F_y = F \sin \theta

Moment (Torque) and Turning Effect

• Definition: The moment (torque) of a force about a rotation point is the rotational effect produced by the force. It can be defined as
  • M = ext{perpendicular force} \times ext{perpendicular distance}
  • In general: M = \boldsymbol{r} \times \boldsymbol{F} where r is the position vector from the rotation point to the line of action of the force.
• Units: Newton-meters (N·m).
• Examples from transcript:
  • Moment A: if a 20 N force acts at 1 m perpendicular distance: M_A = 20~ ext{N} \times 1~\text{m} = 20~\text{N} \cdot \text{m}
  • Moment B: 10 N at 2 m: M_B = 10~\text{N} \times 2~\text{m} = 20~\text{N} \cdot \text{m}
• Important note: The sign (clockwise vs counterclockwise) for moments depends on the chosen convention. In many statics problems, clockwise moments are taken as positive, counterclockwise as negative (or vice versa) depending on the text; be consistent.
• Connection to equilibrium: Moments about a point must balance (net moment = 0) for static equilibrium.

Equilibrium in Two Dimensions

• Three static-equilibrium conditions in 2D:
  • \sum F_x = 0 (horizontal forces balance; right is positive)
  • \sum F_y = 0 (vertical forces balance; up is positive by convention in the notes)
  • \sum M_{\text{point}} = 0 (sum of moments about a chosen point equals zero; clockwise can be taken as positive depending on convention)
• Example structure from transcript (illustrative values):
  • Given horizontal forces: left-to-right positive, some numbers such as 20 kN, 10 kN, 25 kN, 5 kN; a static balance condition is shown as
  • -25 + 20 + 10 - 5 = 0
  • Given vertical forces: upward positive; a different set of numbers can yield a similar balance, e.g., -25 + 20 + 10 - 5 = 0 (notes show same sum for vertical count by convention)
  • Moment balance example: rotations around a point must balance so that the sum of clockwise moments equals the sum of counterclockwise moments.
• Seesaw illustration (balance):
  • Weight of A times its lever arm equals weight of B times its lever arm at equilibrium: WA \cdot LA = WB \cdot LB
• Practical interpretation: Equilibrium means balance of all forces so the object does not accelerate or rotate.
• Additional resource: a video link is provided in the notes for visual intuition on equilibrium.

Equilibrium: Real-World Contexts and Examples

• Intuition for equilibrium:
  • Balance examples: balanced scales – equal weights on both sides yield no net force.
  • Floating ball in water – gravity downward is balanced by buoyant force upward.
  • Tug-of-war – equal opposing forces yield no movement.
• Key takeaway: Equilibrium means net force is zero and there is no net moment about any point, leading to either rest or uniform motion with no rotation.

Structural Supports in Beams

• Roller supports
  • Allow thermal expansion/contraction and some horizontal movement, but prevent vertical translation.
  • Reaction is a single vertical force, perpendicular to the surface (upward or downward).
  • Do not resist horizontal translation or rotation; used at one end of long spans (e.g., bridges).
  • Visual concept: a structure on rollers remains in place unless a lateral load causes rolling.
• Pinned (or anchored) supports
  • Restrain translation in all directions (both x and y components) but do not resist rotation (no moment reaction).
  • Reaction is a single force with horizontal and vertical components, which together form the resultant at the pin.
  • Two supports are generally required to resist the moment in a frame; pinned supports do not supply a moment reaction themselves.
• Fixed (rigid) supports
  • Maintain the angular relationship between connected members; provide both force and moment resistance.
  • Resist translation in x and y and all rotation (moment) about the connection point.
  • Commonly used in cantilevers and in moment-resisting frames (e.g., steel frames, concrete frames).
• Takeaway about supports:
  • Different supports provide different combinations of translational and rotational restraint.
  • The choice of supports determines the distribution of horizontal and vertical reactions and moments in a structure.

Simply Supported Beam System

• Definition: A beam with two supports, one pinned (or anchored) and one roller; the beam cannot translate vertically at either end but can rotate freely.
• Purpose of roller vs pinned ends:
  • Roller end allows axial expansion/contraction (horizontal movement), while pinned end resists horizontal reaction.
• Case 1 (A pinned, B roller) vs Case 2 (A roller, B pinned):
  • In Case 1: A provides both vertical and horizontal reactions; B provides only vertical reaction.
  • In Case 2: A provides only vertical reaction; B provides vertical and horizontal reactions.
• Free Body Diagram (FBD) steps:
  • Step 1: Identify support types and reactions (which ends carry horizontal vs vertical reactions).
  • Step 2: Convert all loads to equivalent point loads if needed.
  • Step 3: Apply equilibrium equations: \sum Fx = 0, \quad \sum Fy = 0, \quad \sum M_A = 0 (moments about A are used to solve for unknowns).
• Important note about FBD conventions:
  • The line of action of a force through a rotation point makes its moment equal to zero.
  • The horizontal reaction at one end and vertical reactions at ends are solved using the three equilibrium equations.

Beam Analysis: Worked Examples (Case Studies)

• Setup conventions used in the transcript:

  • A beam with point loads and supports at A and B.
  • Distances provided as segments along the beam; loads are converted to equivalent point loads if not already vertical.

• General procedure (three-step approach):

  • Step 1: Identify support reactions (which supports provide horizontal vs vertical reactions).
  • Step 2: Draw Free Body Diagram (FBD) with loads and reactions.
  • Step 3: Solve using equilibrium equations:
  • \sum F_x = 0
  • \sum F_y = 0
  • \sum M_A = 0 (moments about A; clockwise typically treated as positive depending on convention).

• Example 1 (Case 1): 12 kN vertical load, A anchored, B roller

  • Given: A anchored at left, B roller at right; load: 12 kN vertical at some location; distances: 5 m to load from A and 10 m from load to B (as per transcript’s diagram).
  • Step 1: Reactions: A provides Fay (vertical) and Fax (horizontal); B provides Fby (vertical only).
  • Step 2: Convert to vertical loads: vertical load is 12 kN (no horizontal component from vertical load).
  • Step 3: Equations:
  • \sum F_x = 0: Fax = 0
  • \sum F_y = 0: Fay + Fby = 12
  • \sum M_A = 0: 12\text{ kN} \times 5\text{ m} - Fby \times 10\text{ m} = 0
  • Result: Fby = 6\text{ kN}, \quad Fay = 6\text{ kN}, \quad Fax = 0\text{ kN}.

• Example 2 (Case 2): 20 kN load at some distance; A roller, B anchored

  • Given: A roller (vertical only), B anchored (vertical + horizontal).
  • Step 1: Reactions: Fay (vertical at A), Fby (vertical at B), Fbx (horizontal at B). No horizontal at A (since A is roller).
  • Step 2: Convert to equivalent vertical load: only vertical component if load is not vertical; transcript uses a vertical load of 20 kN (implies load is vertical for this step).
  • Step 3: Equations:
  • \sum F_x = 0: -Fbx = 0 \Rightarrow Fbx = 0
  • \sum M_A = 0: 20\text{ kN} \times 4\text{ m} - Fby \times 10\text{ m} = 0 \Rightarrow Fby = 8\text{ kN}
  • \sum F_y = 0: Fay + Fby = 20\text{ kN} \Rightarrow Fay = 12\text{ kN}

• Example 3 (Case 2): 20 kN diagonal load at 30° to horizontal

  • Load components:
  • The vertical component: F_y = F \cos (90^{\circ} - \theta) = 20 \cos(60^{\circ}) = 10\text{ kN}
  • The horizontal component: F_x = F \cos (\theta) = 20 \cos (30^{\circ}) = 17.32\text{ kN}
  • Step 1:
  • Horizontal reaction: F_{bx} = 17.32\text{ kN} (direction depending on chosen axis; can be + or - depending on assumed direction)
  • Step 2: Moment about A:
  • Fy(4\text{ m}) - F{by}(10\text{ m}) = 0 (note: the horizontal component does not contribute to moment about A if its line of action passes through A)
  • Result: F_{by} = 4\text{ kN}
  • Step 3: Vertical forces: F{ay} + F{by} = 20\text{ kN} \Rightarrow F_{ay} = 6\text{ kN}

• Example 4 (Case 1): 20 kN diagonal + 30 kN diagonal

  • The vertical load is 20 kN at 30° to the horizontal; decompose:
  • F_y = 20 \cos(60^{\circ}) = 10\text{ kN}
  • F_x = 20 \cos(30^{\circ}) = 17.32\text{ kN}
  • Step 1: Horizontal equilibrium: F_{ax} = 0? (The transcript shows two valid sign conventions depending on assumed direction, with
  • either F{ax} = -17.32\text{ kN} or F{ax} = +17.32\text{ kN} depending on chosen positive direction.)
  • Step 2: Moment about A: using the given distances (2 m, 6 m, 10 m), solve for Fby:
  • 10\text{ kN} \times 2\text{ m} + 30\text{ kN} \times 6\text{ m} - F_{by} \times 10\text{ m} = 0
  • Result: F_{by} = 20\text{ kN}
  • Step 3: Vertical forces: F{ay} + F{by} = 40\text{ kN} with given vertical contributions; substituting gives F_{ay} = 20\text{ kN}
  • Notes: Horizontal reaction can be ±17.32 kN depending on sign convention; there are two mathematically valid solutions corresponding to the chosen sign for the horizontal reaction.

• Summary of procedure for beam problems:

  • Identify end conditions (Case 1 vs Case 2) to determine which end carries horizontal reaction.
  • Resolve non-vertical loads into components; replace diagonal loads with their horizontal and vertical components in the FBD.
  • Write and solve the three equilibrium equations: \sum Fx = 0, \quad \sum Fy = 0, \quad \sum M_A = 0 (moments about a chosen point, typically A).
  • Check consistency: the vertical reactions should sum to the total vertical load, horizontal reactions should balance as per the diagonal force components, and moments should balance.

• Practical notes on interpretation and signage:

  • The choice of positive directions for x and y (and the direction chosen for the moment sign) affects the algebraic sign of the solutions; a negative result simply indicates the assumed direction was opposite to the actual direction.
  • In problems with diagonals, it is often helpful to first resolve the diagonal into its x and y components, then apply the standard equilibrium equations.

• Real-world relevance and connections:

  • Beams with pinned, roller, and fixed supports are fundamental in civil and structural engineering (bridges, buildings, frames).
  • Transmissibility and vector decomposition are foundational for understanding how loads transfer through structures and how to analyze them using components.
  • The equilibrium framework underpins design choices to ensure safety and stability in structures and mechanisms.

• Quick reference formulas:

  • Newton's Second Law: F = m a
  • 1 Newton: 1~\text{N} = 1~\text{kg} \, \text{m}/\text{s}^2
  • Force components (diagonal force at angle θ):
  • F_x = F \cos \theta
  • F_y = F \sin \theta
  • Moment (torque): M = F \cdot d_{ ext{perp}} or \boldsymbol{M} = \boldsymbol{r} \times \boldsymbol{F}
  • Equilibrium in 2D:
  • \sum Fx = 0, \quad \sum Fy = 0, \quad \sum M = 0
  • Relationship for equilibrium examples: e.g., weights in seesaw: WA LA = WB LB