Comprehensive Notes on Moment of Force, Rotational Dynamics, and Equilibrium

Moment of Force Definitions and Fundamentals

• Definition of Moment of Force: The product of a force and the perpendicular distance from the point of application of the force to the axis of rotation or pivot.
• Torque Symbol: Represented by the torque symbol $\tau$ (or $T$).
• General Formula:   $\text{Moment} = F \times d$   where $F$ is the applied force and $d$ is the perpendicular distance from the pivot line.
• Weight Component: The weight of an object acts vertically downwards through its center of gravity.
• Clockwise Moment (CWM):   $\text{CWM} = W \times d_{AP}$   where $W$ is the weight acting downwards and $d_{AP}$ is the perpendicular distance from the pivot point $A$ to the line of action of the weight.
• Anticlockwise Moment (ACWM):   $\text{ACWM} = T \times d_1$   where $T$ is the tension force and $d_1$ is the perpendicular distance from the pivot to the line of action of the tension force.

Resolving Forces at Angles

• Forces Applied at an Angle $\theta$: When a force is applied at an angle $\theta$ relative to the line connecting the application point to the pivot, the force must be resolved into rectangular components.
• Alternate Angles ($Z$-Angles Rule): Geometric alternate interior angles ($Z$-angles) are used to identify the angle $\theta$ relative to the orthogonal axes.
• Force Components:
  • Horizontal/Parallel Component: $F \cos(\theta)$
  • Vertical/Perpendicular Component: $F \sin(\theta)$
• Moment Contribution of Components:
  • The parallel component $F \cos(\theta)$ passes directly through or parallel without perpendicular displacement to the pivot, yielding a perpendicular distance of zero. Its moment contribution is $0$.
  • The perpendicular component $F \sin(\theta)$ creates the rotational moment about the pivot:     $\text{Moment} = F \sin(\theta) \times d$
• Three Structural Representations of Resolution:
  • Resolving the force directly at the point of application into $F \cos(\theta)$ and $F \sin(\theta)$.
  • Resolving the distance vector into perpendicular components relative to the force.
  • Extending the line of action of the force and drawing a perpendicular line from the pivot.

Rules for Determining Perpendicular Distance

• Locating Perpendicular Distance: Draw a straight line parallel to the given force vector such that it passes directly through the pivot point.
• Point of Application: The point of application for the resolved components of a force remains identical to the original point where the actual force is applied.
• Selecting Perpendicular Lines: Always match a horizontal force with its vertical perpendicular distance to the pivot, and a vertical force with its horizontal perpendicular distance to the pivot.

Step-by-Step Worked Example Problems

Example 1: Tension in an Inclined Plank System

• System Parameters:
  • Pivot located at one end of a plank.
  • Downward weight $W = 50\,\text{N}$ located at distance $d_1 = 2\,\text{m}$ from the pivot.
  • Tension $T$ applied at an angle $\theta = 30^\circ$ at a total distance $d_{\text{total}} = 2\,\text{m} + 2\,\text{m} = 4\,\text{m}$ from the pivot.
  • System is in static equilibrium.
• Calculation Step-by-Step:
  • Clockwise Moment:     $\text{CWM} = 50\,\text{N} \times 2\,\text{m} = 100\,\text{N}\cdot\text{m}$
  • Anticlockwise Moment:     $\text{ACWM} = T \sin(30^\circ) \times 4\,\text{m}$
  • Value of $\sin(30^\circ) = 0.5$
  • Equating CWM and ACWM for equilibrium:     $100\,\text{N}\cdot\text{m} = T \times 0.5 \times 4\,\text{m}$$100\,\text{N}\cdot\text{m} = T \times 2\,\text{m}$$T = \frac{100\,\text{N}\cdot\text{m}}{2\,\text{m}} = 50\,\text{N}$

Example 2: Horizontal and Vertical Force Moments on a Structure

• System Parameters:
  • Pivot located at the top of a vertical structure.
  • Total vertical length $= 12\,\text{m}$.
  • Horizontal displacement line $= 4\,\text{m}$.
• Calculating Clockwise Moment due to Force $F_1$:
  • Horizontal force $F_1 = 20\,\text{N}$.
  • Perpendicular vertical distance from pivot $d = 12\,\text{m} - 2\,\text{m} = 10\,\text{m}$.
  • Clockwise Moment:     $\text{Moment}_{F1} = 20\,\text{N} \times 10\,\text{m} = 200\,\text{N}\cdot\text{m}$
• Calculating Anticlockwise Moment due to Force $F_2$:
  • Horizontal force $F_2 = 5\,\text{N}$.
  • Perpendicular vertical distance to line of action $= 4\,\text{m}$.
  • Anticlockwise Moment:     $\text{Moment}_{F2} = 7 \times 4\,\text{m} = 28\,\text{N}\cdot\text{m}$
• Calculating Anticlockwise Moment due to Force $F_3$:
  • Force $F_3 = 10\,\text{N}$.
  • Distance $d = 5\,\text{m}$.
  • Moment:     $\text{Moment}_{F3} = 10\,\text{N} \times 5\,\text{m} = 50\,\text{N}\cdot\text{m}$
• Resultant Moment Calculation:
  • Subtract total clockwise moments from total anticlockwise moments (or vice versa) to find net torque about the pivot.

Example 3: Frictional Force on a Ladder Resting Against a Smooth Wall

• System Parameters:
  • Ladder resting against a smooth (frictionless) vertical wall and rough horizontal ground.
  • Height against wall $h = 6\,\text{m}$.
  • Horizontal base distance $= 16\,\text{m}$ (divided into two equal halves of $8\,\text{m}$ and $8\,\text{m}$).
  • Weight of ladder $W = 200\,\text{N}$ acting vertically downwards at its midpoint ($8\,\text{m}$ horizontally from wall/ground points).
  • Wall reaction force $= 120\,\text{N}$.
  • Ground reaction force $R_1 = 80\,\text{N}$.
  • Friction force $F$ acts horizontally at the rough ground base to prevent slipping.
  • Consider the frictionless contact point at the wall as the pivot point.
• Calculation Step-by-Step:
  • Clockwise Moment (due to weight $W$):     $\text{CWM}_W = 200\,\text{N} \times 8\,\text{m} = 1600\,\text{N}\cdot\text{m}$
  • Clockwise Moment (due to friction $F$ acting at vertical height $6\,\text{m}$ from wall contact point):     $\text{CWM}_F = F \times 6\,\text{m}$
  • Anticlockwise Moment (due to ground reaction acting at full length $16\,\text{m}$):     $\text{ACWM} = 120\,\text{N} \times 16\,\text{m} = 1920\,\text{N}\cdot\text{m}$
  • Setting up Equilibrium Equation:     $\text{CWM}_W + \text{CWM}_F = \text{ACWM}$$1600 + (F \times 6) = 1920$$F \times 6 = 1920 - 1600$$F \times 6 = 320$$F = \frac{320}{6} = 53.33\,\text{N}$

Example 4: Equilibrium of a Suspended Uniform Plank

• System Parameters:
  • Uniform plank of total length $12\,\text{m}$ suspended in equilibrium.
  • Downward force $F_1 = 100\,\text{N}$ at distance $30\,\text{cm}$.
  • Unknown weight $W$ acting downwards at distance $10\,\text{cm}$.
  • Force $400\,\text{N}$ at distance $10\,\text{cm}$.
  • Force $3\,\text{N}$ at distance $20\,\text{cm}$.
• Pivot Selection Protocol: Explicitly choose and specify the pivot point (e.g., point $D$) before setting up moment equations.
• Direction Differentiation:
  • Clockwise motion follows standard rotational direction of clock hands.
  • Anticlockwise motion follows opposite rotational direction.
  • Avoid common misconceptions by visually tracing force vector paths relative to the selected pivot point.
• Vertical Force Balance:
  • Total Upward Forces = Total Downward Forces
  • $T + F_D = F_{\text{down1}} + F_{\text{down2}} + F_{\text{down3}}$
  • Sum of downward forces: $100\,\text{N} + 106\,\text{N} + 400\,\text{N} = 606\,\text{N}$ (or $603\,\text{N}$).

Example 5: Resultant Moment on a Hinged Horizontal Bar (PQ)

• System Parameters:
  • Horizontal bar $PQ$ of length $L = 50\,\text{cm}$ hinged at end $P$
  • Downward vertical force $F_{\text{down}} = 5\,\text{N}$ acting at end $Q$ ($50\,\text{cm}$ from $P$).
  • Inclined upward force $F_{\text{inc}} = 16\,\text{N}$ acting at angle $\theta = 30^\circ$ at end $Q$ ($50\,\text{cm}$ from $P$).
• Unit Conversion:
  • Distance $d = 50\,\text{cm} = 50 \times 10^{-2}\,\text{m} = 0.5\,\text{m}$
• Calculation Step-by-Step:
  • Clockwise Moment (due to $5\,\text{N}$ downward force):     $\text{CWM} = 5\,\text{N} \times (50 \times 10^{-2}\,\text{m}) = 5\,\text{N} \times 0.5\,\text{m} = 2.5\,\text{N}\cdot\text{m}$
  • Anticlockwise Moment (due to vertical component of $16\,\text{N}$ force):     $\text{Vertical Component} = 16 \sin(30^\circ) = 16 \times 0.5 = 8\,\text{N}$$\text{ACWM} = 8\,\text{N} \times 0.5\,\text{m} = 4.0\,\text{N}\cdot\text{m}$
  • Net Resultant Moment about $P$:     $\text{Resultant Moment} = \text{ACWM} - \text{CWM}$$\text{Resultant Moment} = 4.0\,\text{N}\cdot\text{m} - 2.5\,\text{N}\cdot\text{m} = 1.5\,\text{N}\cdot\text{m}\quad (\text{Anticlockwise})$

Principles and Conditions of Equilibrium

Classification of Equilibrium

• General Definition: An object is in equilibrium if and only if its linear and angular acceleration are zero ($a = 0$, $\alpha = 0$).
• Static Equilibrium: An object is at rest ($v = 0$) and has zero acceleration ($a = 0$).
• Dynamic Equilibrium: An object moves with constant uniform velocity ($v = \text{constant}$, $v \neq 0$) and has zero acceleration ($a = 0$).

Necessary Conditions for Equilibrium

• First Condition of Equilibrium (Translational Equilibrium):

  • The vector sum of all external forces acting on the object must equal zero.
  • $\sum F = 0$
  • Resolved into Cartesian components:     $\sum F_x = 0$$\sum F_y = 0$

• Second Condition of Equilibrium (Rotational Equilibrium):

  • The algebraic sum of all moments (torques) acting about any arbitrary pivot point must equal zero.
  • $\sum \tau = 0$
  • Alternatively stated:     $\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}$

Law of Triangle of Forces

• Statement: If three concurrent forces acting at a single point on a body are represented in magnitude and direction by the three sides of a triangle taken in one order (forming a closed vector loop), then the object is in equilibrium.
• Converse Principle: If an object is in equilibrium under the action of three concurrent forces, these three forces can be geometrically represented in magnitude and direction by the three sides of a closed triangle taken in sequential order.