9. Mechanics — Statics

Coplanar forces

Coplanar forces refer to systems of forces acting in a single two-dimensional plane, governed by the fundamental vector equations of equilibrium in statics. This topic covers resolution, moments, and Lami's theorem.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Varignon's Theorem of Moments (Principle of Moments)

The algebraic sum of the moments of two concurrent coplanar forces about any point in their plane is equal to the moment of their resultant force about the same point.

📐 Geometric Interpretation

Geometrically, the moment of a force about a point represents twice the area of the triangle formed by the point of application, the line of action, and the reference point. Varignon's theorem scales this linear additivity of areas to any coplanar force system.

⚙️ Algebraic & Structural Meaning

Translates vector cross products in 2D space into scalar algebraic equations: $R \cdot d = \sum (F_i \cdot d_i)$, drastically simplifying equilibrium analysis from vector matrices to linear combinations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Coplanarity of Forces	`Plane $\mathbb{R}^2$`	Strictly Mandatory	If forces possess components outside the common plane (spatial force systems), standard 2D equilibrium equations $\sum F_x = 0$, $\sum F_y = 0$, $\sum M = 0$ are insufficient and require 3D spatial torque formulations.
Rigid Body Assumption	`Continuous media / Rigid body`	Strictly Mandatory	Forces cannot be transmitted along their lines of action (Principle of Transmissibility) if the body undergoes internal deformation.
Static Equilibrium Condition	`Time derivative $\frac{dv}{dt} = 0, \frac{d\omega}{dt} = 0$`	Essential for equations	If net force or net moment is non-zero, the system reduces to dynamics ($F = ma$), violating static equilibrium hypotheses.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Lami's Theorem for Three Concurrent Forces

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

Rule: When three coplanar forces are in equilibrium, each force is proportional to the sine of the angle between the other two forces.

Application: Frequently asked in UP PGT/TGT for hanging weights, strings, and inclined plane problems.

Resultant Magnitude Formula

$$R = \sqrt{P^2 + Q^2 + 2PQ \cos \theta}$$

Rule: For two concurrent forces $P$ and $Q$ inclined at angle $\theta$, use this formula directly. If $\theta = 90^\circ$, $R = \sqrt{P^2 + Q^2}$; if $\theta = 120^\circ$ and $P=Q$, then $R=P$.

Application: Direct numerical substitution questions in DSSSB and KVS PGT.

Direction of Resultant (Angle $\alpha$ with $P$)

$$\tan \alpha = \frac{Q \sin \theta}{P + Q \cos \theta}$$

Rule: Instantly find the line of action of the resultant force relative to one of the component vectors.

Application: Standard recurring formula tested in LT Grade and UP TGT mathematics.

Three Forces in Equilibrium Triangle Rule

$$\frac{F_1}{a} = \frac{F_2}{b} = \frac{F_3}{c}$$

Rule: If three forces are represented in magnitude and direction by the sides of a triangle taken in order, they are proportional to the lengths of the triangle's sides ($a, b, c$).

Application: Used to skip trigonometric expansions in geometric statics problems.

Conditions for Equilibrium of Coplanar Non-Concurrent Forces

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

Rule: Instead of taking moments about three separate points, verify force resolution along two orthogonal axes and moments about a single strategic point $O$.

Parallel Like Forces Resultant Position

$$P \cdot x = Q \cdot (d - x) \implies x = \frac{Qd}{P+Q}$$

Rule: The line of action of the resultant divides the distance $d$ between two parallel forces inversely in the ratio of the forces.

Application: Rapid calculation for center of mass and parallel loading problems.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Concurrent and Non-Concurrent Force Equilibrium

Common Pitfall: Applying Lami's theorem to a system of non-concurrent coplanar forces.

Counterexample: Four forces forming a couple or acting along distinct parallel lines on a rigid beam.

Why it fails: Lami's theorem requires all lines of action to intersect at a single point (concurrent forces). Non-concurrent systems require moment equilibrium ($\sum M = 0$).

🛡️ Exam Defense: Check intersection points: if lines of action do not meet at a common point, Lami's theorem is invalid.

⚠️ COMMON ERROR #02

Sign Convention Errors in Moments

Common Pitfall: Mixing up clockwise and counter-clockwise moment signs when summing about a pivot.

Counterexample: Setting $\sum M = P \cdot a + Q \cdot b = 0$ when both act in the same rotational sense.

Why it fails: Moments are vector quantities perpendicular to the plane; opposing rotational senses must carry opposite algebraic signs (e.g., counter-clockwise as positive, clockwise as negative).

🛡️ Exam Defense: Fix a global sign convention (e.g., CCW = $+$) at the very start of every statics problem.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Lami's Theorem	3 Coplanar Concurrent Forces in Equilibrium	$$\frac{A}{\sin \alpha} = \frac{B}{\sin \beta} = \frac{C}{\sin \gamma}$$	Forces proportional to sine of opposite angles	Mention of three strings, rings, or concurrent pulls
Varignon's Theorem	Coplanar Forces with Intersecting Lines of Action	$$R \cdot d = \sum F_i \cdot d_i$$	Additivity of moment areas about a point	Finding resultant location or unknown moment arm
Resultant of Two Forces	Two Concurrent Coplanar Forces	$$R = \sqrt{P^2 + Q^2 + 2PQ \cos \theta}$$	Diagonal of the parallelogram of forces	Angle $\theta$ given between forces $P$ and $Q$
Couple	Two Equal, Unlike, Parallel Coplanar Forces	$$G = F \cdot d$$	Pure rotational effect with zero translation	Forces acting in opposite directions along parallel lines

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Two forces $P$ and $2P$ act at an angle $\theta$ to each other. If the resultant is perpendicular to $P$, find the value of $\theta$.

(A) $30^\circ$(B) $60^\circ$(C) $120^\circ$(D) $150^\circ$

📚 Method A: Formal Textbook Derivation

Step 1: Use the angle formula for the resultant: $\tan \alpha = \frac{Q \sin \theta}{P + Q \cos \theta}$. Step 2: Here $\alpha = 90^\circ$, so $P + Q \cos \theta = 0$. Step 3: Substitute $Q = 2P$: $P + 2P \cos \theta = 0 \implies \cos \theta = -1/2$. Step 4: Therefore, $\theta = 120^\circ$.

⚡ Method B: 25-Second Exam Shortcut

Recall standard result: When resultant is perpendicular to $P$, $\cos \theta = -P/Q$. Here $-P/(2P) = -1/2 \implies \theta = 120^\circ$.

💡 Tip: Memorize the special case where $R \perp P \implies P + Q \cos \theta = 0$ for instant 5-second execution.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Three coplanar forces of magnitudes $2\text{ N}$, $3\text{ N}$, and $4\text{ N}$ act along the sides of an equilateral triangle taken in order. The magnitude of their resultant is:

(A) $0\text{ N}$(B) $2\sqrt{3}\text{ N}$(C) $3\sqrt{3}\text{ N}$(D) $9\text{ N}$

📚 Method A: Formal Textbook Derivation

Step 1: Resolve forces along horizontal ($x$) and vertical ($y$) axes aligned with one side of the triangle. Step 2: Sum forces: $\sum F_x = 2 + 3 \cos 60^\circ - 4 \cos 60^\circ$, etc. Alternatively, note that forces acting along sides of a triangle in order form a couple or polygon. Step 3: Compute exact vector sum yielding $R = 2\sqrt{3}$.

⚡ Method B: 25-Second Exam Shortcut

Polygon law of forces: When forces are proportional to sides of a closed polygon, $R = \sqrt{\sum F_x^2 + \sum F_y^2}$. For equilateral triangle with equal increments, $R = \sqrt{3} \times \text{common difference or specific cyclic sum}$.

💡 Tip: Watch whether forces are 'taken in order' (forming a couple if equal) or meeting at a vertex.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If the resultant of two forces $P$ and $Q$ acting at angle $\theta$ has magnitude $\sqrt{3}Q$ when $P = Q$, then $\theta$ is equal to:

A $30^\circ$

B $60^\circ$

C $90^\circ$

D $120^\circ$

Derivation & Analysis:

$R^2 = P^2 + Q^2 + 2PQ \cos \theta$. Given $P=Q$ and $R = \sqrt{3}Q$: $(\sqrt{3}Q)^2 = Q^2 + Q^2 + 2Q^2 \cos \theta \implies 3Q^2 = 2Q^2(1 + \cos \theta) \implies \frac{3}{2} = 1 + \cos \theta \implies \cos \theta = 1/2 \implies \theta = 60^\circ$.

MCQ 2: Lami's theorem is applicable only when:

A Forces are parallel

B Three forces are coplanar and concurrent

C Four forces are in equilibrium

D Forces form a couple

Derivation & Analysis:

Lami's theorem strictly requires exactly three coplanar forces acting concurrently in static equilibrium.

MCQ 3: The algebraic sum of the moments of two concurrent forces about any point in their plane is equal to:

A Zero

B Moment of their resultant about that point

C Sum of their magnitudes

D Product of their lines of action

Derivation & Analysis:

This is the direct statement of Varignon's Theorem of Moments.

MCQ 4: Two forces acting on a particle have a maximum resultant of $17\text{ N}$ and a minimum resultant of $7\text{ N}$. The magnitude of the forces are:

A $12\text{ N}, 5\text{ N}$

B $10\text{ N}, 7\text{ N}$

C $14\text{ N}, 3\text{ N}$

D $15\text{ N}, 2\text{ N}$

Derivation & Analysis:

Maximum resultant $P + Q = 17$, Minimum resultant $P - Q = 7$. Solving gives $2P = 24 \implies P = 12$, and $Q = 5$.

MCQ 5: A system of coplanar forces in equilibrium satisfies which of the following independent scalar equations?

A Only one

B Only two

C Three ($\sum F_x = 0, \sum F_y = 0, \sum M = 0$)

D Four

Derivation & Analysis:

A general coplanar force system has 3 degrees of freedom in 2D, requiring 3 independent equilibrium equations.

MCQ 6: If three forces in equilibrium are represented by the sides of a triangle $ABC$ taken in order, then each force is proportional to:

A $\cos A, \cos B, \cos C$

B $\sin A, \sin B, \sin C$

C $a, b, c$ (sides of the triangle)

D Median lengths

Derivation & Analysis:

By the triangle law of forces, forces are proportional to the lengths of the respective sides $a, b, c$ they act along.

MCQ 7: The moment of a force $F = 10\text{ N}$ about a point $O$, where the perpendicular distance is $2\text{ m}$, is:

A $5\text{ Nm}$

B $10\text{ Nm}$

C $20\text{ Nm}$

D $40\text{ Nm}$

Derivation & Analysis:

Moment $M = F \cdot d = 10 \times 2 = 20\text{ Nm}$.

MCQ 8: A couple consists of:

A Two equal and like parallel forces

B Two equal and unlike parallel forces

C Two unequal parallel forces

D Three concurrent forces

Derivation & Analysis:

A couple is defined as two equal, parallel forces acting in opposite (unlike) directions with non-coincident lines of action.

MCQ 9: If the angle between two equal forces $P$ is $120^\circ$, the magnitude of their resultant is:

A $P$

B $\sqrt{3}P$

C $2P$

D Zero

Derivation & Analysis:

$R = \sqrt{P^2 + P^2 + 2P^2 \cos 120^\circ} = \sqrt{2P^2 + 2P^2(-1/2)} = \sqrt{2P^2 - P^2} = \sqrt{P^2} = P$.

Q10

MCQ 10: If a body is acted upon by a non-zero couple, it will execute:

A Pure translation

B Pure rotation without translation

C Both translation and rotation

D Rest

Derivation & Analysis:

The vector sum of forces in a couple is zero ($\sum F = 0$), meaning zero translation, but the net moment is non-zero, producing pure rotation.

Common questions

No. Lami's theorem is strictly restricted to exactly three forces. For four or more concurrent forces, resolution into orthogonal $x$ and $y$ components ($\sum F_x = 0, \sum F_y = 0$) must be used.

Sign convention confusion during moment summation and failing to distinguish between concurrent and non-concurrent force systems.

Memorize standard trigonometric ratios for symmetric angles ($60^\circ, 90^\circ, 120^\circ$) and use vector projection shortcuts rather than expanding full tangent formulas.

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