7.1 Force Systems, Coplanar Equilibrium & Lami's Theorem

6.1 Force Systems, Coplanar Equilibrium & Lami's Theorem

Engineering Mechanics forms the bedrock of mechanical engineering design, structural integrity, and heavy machinery operations across Coal India Limited's open-cast and underground installations. Whether evaluating hoist cable tensions in deep vertical shafts, conveyor gantry reactions, or dragline boom equilibrium, mastering force resolution and equilibrium criteria is paramount for the CIL Management Trainee CBT.

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1. Classification of Force Systems

A force system represents an assemblage of two or more forces acting simultaneously upon a rigid body. Force systems are categorized based on the spatial orientation of their lines of action:

                          ┌─────────────────────────────┐
                          │     Force Systems in 3D     │
                          └──────────────┬──────────────┘
                                         │
                  ┌──────────────────────┴──────────────────────┐
                  │                                             │
       ┌──────────▼──────────┐                       ┌──────────▼──────────┐
       │   Coplanar Systems  │                       │ Non-Coplanar (Space)│
       └──────────┬──────────┘                       └──────────┬──────────┘
                  │                                             │
     ┌────────────┼────────────┐                   ┌────────────┼────────────┐
     │            │            │                   │            │            │
┌────▼─────┐ ┌────▼─────┐ ┌────▼─────────┐   ┌─────▼────┐ ┌─────▼────┐ ┌─────▼─────────┐
│Concurrent│ │ Parallel │ │Non-Concurrent│   │Concurrent│ │ Parallel │ │Non-Concurrent│
└──────────┘ └──────────┘ └──────────────┘   └──────────┘ └──────────┘ └──────────────┘

Taxonomy of Force Configurations

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2. Resultant of Coplanar Force Systems

The resultant ($\vec{R}$) is a single equivalent force that produces the identical translational and rotational effect on a rigid body as the original system of forces.

Analytical Resolution Method

For a system of $n$ coplanar forces $\vec{F}_1, \vec{F}_2, \dots, \vec{F}_n$ inclined at angles $\theta_1, \theta_2, \dots, \theta_n$ to the positive $x$-axis:

$$R_x = \sum_{i=1}^{n} F_{ix} = \sum_{i=1}^{n} F_i \cos\theta_i$$ $$R_y = \sum_{i=1}^{n} F_{iy} = \sum_{i=1}^{n} F_i \sin\theta_i$$ $$\lvert \vec{R} \rvert = R = \sqrt{R_x^2 + R_y^2}$$ $$\theta_R = \tan^{-1}\\left(\frac{R_y}{R_x}\\right)$$

Parallelogram Law of Forces

For two concurrent forces $\vec{P}$ and $\vec{Q}$ with an included angle $\theta$:

$$R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$$ $$\tan\alpha = \frac{Q \sin\theta}{P + Q \cos\theta}$$

where $\alpha$ is the inclination of the resultant $\vec{R}$ with respect to the line of action of force $\vec{P}$.

Special Cases:

• $\theta = 0^{\circ}$ (Collinear, same direction): $R_{\text{max}} = P + Q$
• $\theta = 180^{\circ}$ (Collinear, opposite direction): $R_{\text{min}} = \lvert P - Q \rvert$
• $\theta = 90^{\circ}$ (Orthogonal forces): $R = \sqrt{P^2 + Q^2}$, $\tan\alpha = \frac{Q}{P}$

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3. Varignon's Theorem of Moments

Varignon's Theorem (Principle of Moments) states: The algebraic sum of the moments of any number of coplanar forces about any point in their plane is equal to the moment of their resultant force about the same point.

Mathematical Vector Proof

Consider $n$ concurrent forces $\vec{F}_1, \vec{F}_2, \dots, \vec{F}_n$ acting at a point $A$ defined by position vector $\vec{r}$ relative to origin $O$:

$$\vec{R} = \sum_{i=1}^{n} \vec{F}_i$$

Taking the vector cross product with $\vec{r}$ about point $O$:

$$\vec{M}_O = \vec{r} \times \vec{R} = \vec{r} \times \\left(\sum_{i=1}^{n} \vec{F}_i\\right) = \sum_{i=1}^{n} \\left(\vec{r} \times \vec{F}_i\\right) = \sum_{i=1}^{n} \vec{M}_{Oi}$$

In scalar 2D Cartesian terms, for forces with components $(F_{ix}, F_{iy})$ acting at coordinates $(x_i, y_i)$:

$$M_O = \sum_{i=1}^{n} (x_i F_{iy} - y_i F_{ix}) = R \cdot d$$

where $d$ is the perpendicular distance from moment center $O$ to the line of action of the resultant $\vec{R}$.

Couples and Pure Moments

A couple consists of two equal, parallel, and oppositely directed non-collinear forces ($+F$ and $-F$) separated by a perpendicular distance $d$.

• Resultant force: $\vec{R} = \vec{F} + (-\vec{F}) = \vec{0}$ (Zero translational tendency).
• Moment of a couple: $M_C = F \cdot d$. The moment of a couple is a free vector; its magnitude, sign, and rotational effect are identical about any reference point in the plane.

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4. Conditions of Static Equilibrium

A rigid body is in static equilibrium when it experiences neither linear acceleration nor angular acceleration under the applied load system.

2D Coplanar General Force System

The necessary and sufficient conditions are expressed in three independent scalar equations:

$$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_O = 0$$

Alternative Sets of Equilibrium Equations

To simplify calculations, alternative sets of 3 equations may be used:

1. Two Moment Equations and One Force Equation: $$\sum M_A = 0, \quad \sum M_B = 0, \quad \sum F_x = 0$$ Constraint: Line $AB$ connecting moment centers $A$ and $B$ must not be perpendicular to the $x$-axis.
2. Three Moment Equations: $$\sum M_A = 0, \quad \sum M_B = 0, \quad \sum M_C = 0$$ Constraint: Points $A, B,$ and $C$ must not be collinear (must form a non-degenerate triangle).

3D Spatial Force System Equilibrium

In three-dimensional Cartesian space, a rigid body possesses 6 degrees of freedom. Static equilibrium requires satisfying 6 independent scalar equations:

$$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0$$ $$\sum M_x = 0, \quad \sum M_y = 0, \quad \sum M_z = 0$$

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5. Lami's Theorem for Three Concurrent Coplanar Forces

Lami's Theorem is an analytical corollary of the Sine Rule applied to the closed triangle of forces in static equilibrium.

                           F2
                            ^
                             \
                              \   β
                               \-----
                               /  O  \
                       α      /   |   \    γ
                             /    |    \
                            /     |     \
                           v      v      v
                          F1      F3

Statement

If three coplanar, concurrent, non-collinear forces acting at a point keep a rigid body in static equilibrium, then the magnitude of each force is directly proportional to the sine of the angle included between the other two forces.

$$\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}$$

where:

• $\alpha$ is the angle between $\vec{F}_2$ and $\vec{F}_3$
• $\beta$ is the angle between $\vec{F}_1$ and $\vec{F}_3$
• $\gamma$ is the angle between $\vec{F}_1$ and $\vec{F}_2$
• $\alpha + \beta + \gamma = 360^{\circ}$

Mathematical Derivation via Triangle of Forces

Since $\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$, these vectors form a closed triangle. The interior angles of the triangle corresponding to exterior angles $\alpha, \beta, \gamma$ are $(180^{\circ} - \alpha), (180^{\circ} - \beta),$ and $(180^{\circ} - \gamma)$. Applying the Law of Sines:

$$\frac{F_1}{\sin(180^{\circ} - \alpha)} = \frac{F_2}{\sin(180^{\circ} - \beta)} = \frac{F_3}{\sin(180^{\circ} - \gamma)}$$

Since $\sin(180^{\circ} - \theta) = \sin\theta$, this reduces directly to Lami's equation.

Mandatory Validity Criteria for Lami's Theorem

1. Exactly three forces must act on the body or particle.
2. The three forces must be strictly coplanar and concurrent (intersecting at one point).
3. The forces must be non-collinear.
4. All three force vectors must be drawn either pointing away from the concurrence node (diverging) or pointing toward the node (converging). Never mix inward and outward directions when reading angles.

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6. Free-Body Diagram (FBD) Methodology & Support Reactions

A Free-Body Diagram (FBD) is a graphical representation of an isolated physical body or sub-system, stripped of all surrounding physical constraints, showing all applied external loads, body forces (gravity), and reaction forces exerted by removed supports.

Standard Support Reactions in 2D Planar Statics

| Support Type | Graphical Symbol | Unknown Reactions | Description & Orientation | |---|---|:---:|---|| | Frictionless Roller / Rocker | Small wheels on ground | $1$ | Single normal force ($R_n$) perpendicular to contact surface | | Smooth Contact Surface | Flat interface | $1$ | Normal compressive force ($N$) directed into the body | | Frictionless Pin / Hinge | Triangle with pivot pin | $2$ | Two orthogonal components ($R_x, R_y$) or reaction $R$ at unknown angle $\theta$ | | Fixed / Built-in / Encastre | Embedded into wall | $3$ | Horizontal force ($R_x$), vertical force ($R_y$), and reactive moment ($M_R$) | | Weightless Cable / Link | Flexible cord or slender bar | $1$ | Single axial tensile force along the centerline directed away from body | | Collar on Smooth Rod | Sleeve sliding on rod | $1$ | Single normal force perpendicular to the guide rod axis |

  Roller Support           Pinned Hinge Support            Fixed / Built-In Support
       ▲ Rn                     ▲ Ry                               ▲ Ry
       │                        │                                  │   ┌───
  ─────┴─────              ─────┼─────                        ─────┼───│ MR (↺)
  ◯ ◯ ◯ ◯ ◯ ◯                   │                                  │   └───
  ═══════════             ──────┴─────── ► Rx                 ─────┴───── ► Rx

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7. Cable-Pulley Systems & Tension Transmission

Cables and ropes transmit loads purely via axial tension. In classical mechanics analysis:

• Ideal Cable: Completely flexible (zero bending stiffness), inextensible (length $L = \text{const}$), and massless ($m = 0$).
• Ideal Pulley / Sheave: Massless and mounted on frictionless bearings.

Fundamental Principles

1. Uniform Tension: Across an ideal frictionless pulley, cable tension magnitude remains invariant on either side: $T_1 = T_2 = T$.
2. Pin Reaction on Pulley: For a cable turning through an angle $\beta$ over a pulley, the resultant bearing force $\vec{R}_p$ acting on the pulley axle is: $$R_p = \sqrt{T^2 + T^2 + 2T^2 \cos\beta} = 2T \cos\\left(\frac{\beta}{2}\\right)$$ For a $90^{\circ}$ bend: $R_p = T\sqrt{2}$. For a $180^{\circ}$ wrap: $R_p = 2T$.

Mechanical Advantage in Pulley Systems

• Mechanical Advantage ($MA$): Ratio of output load lifted ($W$) to input effort applied ($P$): $$MA = \frac{W}{P}$$
• Velocity Ratio ($VR$): Ratio of displacement of effort ($d_P$) to displacement of load ($d_W$): $$VR = \frac{d_P}{d_W}$$
• Efficiency ($\eta$): $\eta = \frac{MA}{VR} \times 100\%$. For ideal frictionless systems, $\eta = 100\% \implies MA = VR$.

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8. Step-by-Step Worked Engineering Calculations

Worked Example 6.1.1: Sphere Resting in a V-Groove (Lami's Theorem)

Problem: A smooth, homogeneous cylinder of weight $W = 1200\text{ N}$ and radius $r = 250\text{ mm}$ rests in a symmetric V-shaped trough whose inclined surfaces make angles of $30^{\circ}$ and $60^{\circ}$ with the horizontal. Calculate the normal contact reactions $R_A$ and $R_B$ exerted by surfaces $A$ ($30^{\circ}$ plane) and $B$ ($60^{\circ}$ plane) on the cylinder.

                          Normal Reaction RB
                             ^ (30° to horiz)
                              \
                               \   O (Cylinder Center)
                               /   │
                 (60° to horiz)    │
               Normal Reaction RA  │ Weight W = 1200 N
                                   v

Step-by-step Solution:

1. Identify Concurrency and Angles:

   • Normal reaction $R_A$ is perpendicular to the $30^{\circ}$ surface $\implies R_A$ makes an angle of $90^{\circ} - 30^{\circ} = 60^{\circ}$ with the horizontal (i.e., $30^{\circ}$ with the vertical).
   • Normal reaction $R_B$ is perpendicular to the $60^{\circ}$ surface $\implies R_B$ makes an angle of $90^{\circ} - 60^{\circ} = 30^{\circ}$ with the horizontal (i.e., $60^{\circ}$ with the vertical).
   • Downward gravity force $W = 1200\text{ N}$ acts vertically downward along negative $y$.

2. Calculate Included Angles between Force Vectors:

   • Angle between $R_A$ and $R_B$: Since $R_A$ is at $+60^{\circ}$ to horizontal and $R_B$ is at $+150^{\circ}$ (i.e. $30^{\circ}$ to negative horizontal), the included angle between $R_A$ and $R_B$ is $\gamma = 180^{\circ} - (60^{\circ} + 30^{\circ}) = 90^{\circ}$.
   • Angle between $R_B$ and $W$: $\alpha = 90^{\circ} + 60^{\circ} = 150^{\circ}$.
   • Angle between $R_A$ and $W$: $\beta = 90^{\circ} + 30^{\circ} = 120^{\circ}$.
   • Check sum: $\alpha + \beta + \gamma = 150^{\circ} + 120^{\circ} + 90^{\circ} = 360^{\circ}$.

3. Apply Lami's Theorem:

   $$\frac{R_A}{\sin\alpha} = \frac{R_B}{\sin\beta} = \frac{W}{\sin\gamma}$$ $$\frac{R_A}{\sin 150^{\circ}} = \frac{R_B}{\sin 120^{\circ}} = \frac{1200}{\sin 90^{\circ}}$$

4. Evaluate Numerically:

   $$R_A = 1200 \times \sin 150^{\circ} = 1200 \times 0.5 = 600\text{ N}$$ $$R_B = 1200 \times \sin 120^{\circ} = 1200 \times \frac{\sqrt{3}}{2} = 600\sqrt{3} \approx 1039.23\text{ N}$$

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Worked Example 6.1.2: Minimum Force to Pull a Heavy Roller Over a Curb

Problem: A heavy cylindrical roller of weight $W = 4000\text{ N}$ and radius $R = 500\text{ mm}$ rests against a rectangular step (curb) of height $h = 100\text{ mm}$. Determine the minimum horizontal pull force $P$ applied at the center axle of the roller required to just lift the roller over the curb.

                  P (Pull)
           ──────────────► O (Radius R)
                          /│
                         / │ (R - h)
                        /  │
                    R  /   │
                      /    │
                     ▼     ┴
                    A (Step Corner)

Step-by-step Solution:

1. Impending Motion Condition: At the instant the roller begins to lift over the step corner $A$, contact with the flat floor is lost $\implies N_{\text{floor}} = 0$. The only contact reaction acts at corner $A$.

2. Geometric Dimensions about Pivot Point $A$:

   • Vertical distance from axle $O$ to corner $A$: $y = R - h = 500 - 100 = 400\text{ mm}$.
   • Horizontal distance from axle $O$ to corner $A$ (lever arm of gravity $W$): $$x = \sqrt{R^2 - (R - h)^2} = \sqrt{500^2 - 400^2} = \sqrt{250000 - 160000} = \sqrt{90000} = 300\text{ mm}$$

3. Apply Moment Equilibrium about Pivot $A$ ($\Sigma M_A = 0$):

   • Clockwise overturning moment by pull force $P$: $M_P = P \times (R - h) = P \times 400\text{ mm}$.
   • Counter-clockwise stabilizing moment by self-weight $W$: $M_W = W \times x = 4000 \times 300\text{ mm} = 1{,}200{,}000\text{ N}\cdot\text{mm}$.

4. Calculate Minimum Pull $P$:

   $$P \times 400 = 4000 \times 300 \implies P = \frac{1{,}200{,}000}{400} = 3000\text{ N}$$

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9. CIL MT Exam Traps & Rapid Solver Strategies

Exam Trap Alert: Direction of Vectors in Lami's Theorem A frequent mistake in PSU exams is applying Lami's theorem when one vector is pointing towards the node while two are pointing away. Always extend vectors across the node to ensure all three are diverging before reading angles!

Alternative Set Rule: When replacing $\Sigma F_y = 0$ with moment equations $\Sigma M_A = 0$ and $\Sigma M_B = 0$, verify that the line connecting $A$ and $B$ is not perpendicular to the $x$-axis; otherwise, the equations are linearly dependent and cannot yield a unique solution.