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9. Mechanics — Statics

Equilibrium under a system of coplanar forces

Equilibrium under a system of coplanar forces deals with the necessary and sufficient conditions for a rigid body or particle in a single plane to remain at rest or in unaccelerated motion. This topic covers resolving forces into orthogonal components and applying Lamami's theorem or Varignon's theorem.

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Definition and statement

Conditions for Equilibrium of Coplanar Forces
A system of coplanar forces acting on a rigid body is in equilibrium if and only if the algebraic sum of the components of all the forces along any two mutually perpendicular directions is zero, and the algebraic sum of the moments of all the forces about any point in the plane is zero. Mathematically, $$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_O = 0$$

📐 Geometric Interpretation

Geometrically, the force polygon closes (meaning vector addition of forces yields a zero vector) and the total turning effect (moment) about any arbitrary pivot point $O$ vanishes, preventing any rotational acceleration.

⚙️ Algebraic & Structural Meaning

Provides three independent scalar equations of equilibrium for a planar system, enabling the exact determination of up to three unknown forces, reactions, or geometric parameters.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Coplanar Nature of Forces Plane $\mathbb{R}^2$ Strictly Mandatory / Necessary If forces possess out-of-plane components, 2D equilibrium equations fail, necessitating spatial equilibrium equations ($\sum F_z = 0$, moments about $y$ and $z$ axes).
Rigidity of the Body Continuum / Rigid Body Strictly Mandatory Deformable bodies undergo internal strain adjustments, rendering pure static equilibrium equations insufficient without material constitutive laws.
Concurrent or Non-concurrent classification Point of intersection / Line of action Essential for method selection Applying Lamis theorem ($\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$) to non-concurrent forces leads to severe algebraic errors.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Lamis Theorem for Three Concurrent Forces

$$\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$$

Rule: If three coplanar forces keep a particle in equilibrium, each force is proportional to the sine of the angle between the other two forces.

Application: Directly asked in UP TGT/PGT to find tension in strings or reactions at hinges.
02

Three Non-Concurrent Coplanar Forces Equilibrium

$$P \parallel Q \text{ or all three forces must be concurrent}$$

Rule: If three non-parallel coplanar forces are in equilibrium, their lines of action must meet at a single point (concurrent).

Application: Use this result in the stated type of problem.
03

Varignons Theorem of Moments

$$R \cdot p = P_1 \cdot p_1 + P_2 \cdot p_2 + \dots$$

Rule: The algebraic sum of the moments of two concurrent forces about any point is equal to the moment of their resultant about the same point.

Application: Used to bypass calculating coordinates of points of application in complex truss/beam problems.
04

Resolution along Inclined Plane

$$\sum F_{\parallel} = W \sin \theta - F = 0, \quad \sum F_{\perp} = N - W \cos \theta = 0$$

Rule: Instead of horizontal/vertical, always resolve along and perpendicular to an inclined plane when friction or normal reaction is requested.

Application: Standard UP PGT mechanics problem template involving blocks on rough inclined planes.
05

Four-Force Equilibrium Determinant Trick

$$\begin{vmatrix} X_1 & Y_1 & M_1 \\ X_2 & Y_2 & M_2 \\ X_3 & Y_3 & M_3 \end{vmatrix} = 0$$

Rule: For three forces whose line equations and components are known, consistency requires the determinant of coefficients to vanish.

Application: High-level PGT questions involving analytical statics.
06

Symmetry & Equal Tension Elimination

$$T_1 = T_2 \text{ when geometry is symmetric}$$

Rule: If a symmetrical body is suspended by two identical strings making equal angles with the vertical, the tensions are strictly equal ($T_1 = T_2$).

Application: Instantly eliminates two options in suspension bridge and weight-hanging problems.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Applying Lamis Theorem to Non-Concurrent Systems

Common Pitfall: Candidates blindly apply Lamis theorem ($\frac{P}{\sin \alpha} = \dots$) to forces acting along different lines on a rigid beam.

Counterexample: A uniform ladder leaning against a smooth wall and rough floor.

Why it fails: The forces are non-concurrent (normal reaction from wall, weight at center, normal and friction at floor do not intersect at a single point). Lamis theorem fails completely.

🛡️ Exam Defense: Check concurrency before writing Lamis theorem. If lines of action form a triangle or do not meet, use $\sum F_x = 0, \sum F_y = 0, \sum M = 0$.
⚠️ COMMON ERROR #02

Ignoring Direction of Friction in Limiting Equilibrium

Common Pitfall: Assuming friction always acts upwards or opposite to gravity without checking the impending direction of motion.

Counterexample: A body on an inclined plane pushed upwards vs pulled downwards.

Why it fails: Friction always opposes impending motion. Setting $F = \mu R$ with the wrong sign reverses the equilibrium equation.

🛡️ Exam Defense: Determine where the body would slide if friction were absent; set friction in the exact opposite direction.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Lamis TheoremThree concurrent coplanar forces in equilibrium$P/\sin\alpha = Q/\sin\beta = R/\sin\gamma$Sides of triangle proportional to sines of opposite anglesKeyword: 'Three strings', 'suspended particle', 'angles given'
Varignons TheoremCoplanar forces meeting at or forming a system with a moment center$R \cdot d = \sum (P_i \cdot d_i)$Total moment equals sum of component momentsKeyword: 'Moment about a point', 'resultant line of action'
General Coplanar EquilibriumGeneral rigid body in 2D plane$\sum F_x = 0, \sum F_y = 0, \sum M_O = 0$Zero net translation and zero net rotationKeyword: 'Ladder', 'beam', 'rod resting on floor/wall'
Three Non-Concurrent ForcesEquilibrium under exactly 3 non-parallel forcesLines of action are concurrent or parallelAll three force lines intersect at one common pointKeyword: 'Rod pivoted at one end', 'three forces in plane'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

A weight $W$ is supported by two strings of lengths $3\text{ m}$ and $4\text{ m}$ fastened to two points in the same horizontal line at a distance $5\text{ m}$ apart. Find the tensions $T_1$ and $T_2$ in the strings.

(A) $T_1 = \frac{3}{5}W, T_2 = \frac{4}{5}W$(B) $T_1 = \frac{4}{5}W, T_2 = \frac{3}{5}W$(C) $T_1 = \frac{3}{4}W, T_2 = W$(D) $T_1 = W, T_2 = W$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Check side lengths $3, 4, 5$, which form a right-angled triangle ($3^2 + 4^2 = 5^2$). Step 2: Let angles with the horizontal be $\theta_1$ and $\theta_2$. Then $\cos\theta_1 = \frac{3}{5}, \sin\theta_1 = \frac{4}{5}$ and $\cos\theta_2 = \frac{4}{5}, \sin\theta_2 = \frac{3}{5}$. Step 3: Resolve vertically: $T_1 \sin\theta_1 + T_2 \sin\theta_2 = W$. Resolve horizontally: $T_1 \cos\theta_1 = T_2 \cos\theta_2$. Step 4: Solve simultaneous equations to get $T_1 = \frac{3}{5}W$ and $T_2 = \frac{4}{5}W$.
⚡ Method B: 25-Second Exam Shortcut
Recognize the $3-4-5$ Pythagorean triangle right angle at the weight. By Lamis theorem or triangle of forces, tensions are proportional to the opposite sides: $T_1$ is opposite to side $4$ or proportional to $3/5$ scale. Match with options immediately.
💡 Tip: Whenever side lengths form a Pythagorean triplet ($3-4-5, 5-12-13$), immediately use triangle geometry instead of lengthy trigonometric expansions.
CANONICAL PROBLEM 02 Dual-Method Breakdown

A uniform ladder of weight $W$ and length $2l$ rests with its upper end against a smooth vertical wall and lower end on a rough horizontal floor coefficient of friction $\mu$. If the ladder is at an angle $\alpha$ to the horizontal in limiting equilibrium, then:

(A) $\tan \alpha = \frac{1}{2\mu}$(B) $\tan \alpha = 2\mu$(C) $\tan \alpha = \mu$(D) $\tan \alpha = \frac{1}{\mu}$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Draw free body diagram. Normal reaction from wall $N_1$ at top, weight $W$ downwards at midpoint, normal reaction $N_2$ and limiting friction $F = \mu N_2$ at the floor. Step 2: $\sum F_x = 0 \implies N_1 - \mu N_2 = 0$. Step 3: $\sum F_y = 0 \implies N_2 - W = 0 \implies N_2 = W$. Step 4: Take moments about the bottom point: $N_1 \cdot (2l \sin \alpha) - W \cdot (l \cos \alpha) = 0$. Step 5: Substitute $N_1 = \mu W$ and solve for $\tan \alpha$: $2 \mu \sin \alpha = \cos \alpha \implies \tan \alpha = \frac{1}{2\mu}$.
⚡ Method B: 25-Second Exam Shortcut
Standard derived result in UP PGT statics: For smooth wall and rough floor, $\tan \alpha = \frac{1}{2\mu}$. Memorize this exact formula to answer in 5 seconds.
💡 Tip: Memorize standard ladder configurations: (1) Smooth wall, rough floor $\to \tan \alpha = \frac{1}{2\mu}$; (2) Both rough $\to$ use general moment equation.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If three forces acting in one plane upon a rigid body keep it in equilibrium, then they must either meet in a point or be:
A Perpendicular
B Parallel
C Equal in magnitude
D Zero
Derivation & Analysis:

By statics principles, three coplanar forces in equilibrium are either concurrent or completely parallel.

Q2
MCQ 2: Lamis theorem is applicable only when the number of forces in equilibrium is:
A Two
B Three
C Four
D Any number
Derivation & Analysis:

Lamis theorem is specifically formulated for three concurrent coplanar forces.

Q3
MCQ 3: The algebraic sum of the moments of two concurrent forces about any point in their plane is equal to the moment of their resultant about the same point. This is known as:
A Lamis Theorem
B Varignons Theorem
C Triangle Law of Forces
D Polygon Law of Forces
Derivation & Analysis:

Varignons theorem states that the moment of a resultant force equals the sum of moments of its components.

Q4
MCQ 4: A body of weight $W$ is placed on a rough inclined plane of inclination $\alpha$ to the horizon. If the coefficient of friction is $\mu = \tan \lambda$, the condition for sliding down is:
A $\alpha < \lambda$
B $-\alpha = \lambda$
C $\alpha \ge \lambda$
D $\alpha = 0$
Derivation & Analysis:

Limiting equilibrium occurs when angle of inclination $\alpha$ equals the angle of friction $\lambda$. For motion downwards, $\alpha \ge \lambda$.

Q5
MCQ 5: Two forces $P$ and $Q$ act at an angle $\theta$. If their resultant is $R = \sqrt{P^2 + Q^2}$, the angle $\theta$ is:
A $0^\circ$
B $90^\circ$
C $180^\circ$
D $45^\circ$
Derivation & Analysis:

When $\theta = 90^\circ$, $R = \sqrt{P^2 + Q^2 + 2PQ \cos 90^\circ} = \sqrt{P^2 + Q^2}.

Q6
MCQ 6: The unit of moment of a force in SI units is:
A Newton ($N$)
B Newton-second ($N \cdot s$)
C Newton-meter ($N \cdot m$)
D Joule per second ($J/s$)
Derivation & Analysis:

Moment is force multiplied by perpendicular distance, measured in Newton-meters ($N \cdot m$).

Q7
MCQ 7: If a system of coplanar forces is in equilibrium, then the sum of resolved parts along two perpendicular directions must be:
A Maximum
B Zero
C Equal
D Unity
Derivation & Analysis:

For translational equilibrium, $\sum F_x = 0$ and $\sum F_y = 0$.

Q8
MCQ 8: The angle of friction $\lambda$ and coefficient of friction $\mu$ are related by:
A $\mu = \sin \lambda$
B $\mu = \cos \lambda$
C $\mu = \tan \lambda$
D $\mu = \cot \lambda$
Derivation & Analysis:

By definition, the coefficient of friction equals the tangent of the angle of friction, $\mu = \tan \lambda$.

Q9
MCQ 9: If three forces $P, Q, R$ acting at a point are in equilibrium and $P = Q = R$, the angle between any two of them is:
A $60^\circ$
B $90^\circ$
C $120^\circ$
D $180^\circ$
Derivation & Analysis:

By Lamis theorem, $\frac{P}{\sin \alpha} = \frac{P}{\sin \beta} = \frac{P}{\sin \gamma} \implies \sin \alpha = \sin \beta = \sin \gamma \implies \alpha = \beta = \gamma = 120^\circ$.

Q10
MCQ 10: The necessary and sufficient conditions for the equilibrium of a rigid body in a plane are:
A Only $\sum F_x = 0$
B Only $\sum M = 0$
C $\sum F_x = 0, \sum F_y = 0, \sum M = 0$
D $\sum M_1 = 0, \sum M_2 = 0$
Derivation & Analysis:

Three independent scalar equations are required for 2D rigid body equilibrium: two for translation and one for rotation.

Common questions

No. Lamis theorem requires all three forces to pass through a single common point of concurrency. If they form a triangle of forces (non-concurrent), vector addition or moment equations must be used.

Sign errors in taking moments and incorrect identification of the direction of friction in rough surface problems are the leading causes of lost marks.

Memorize standard formula results (like $\tan \alpha = \frac{1}{2\mu}$ for a ladder on a smooth wall and rough floor) and take moments about the point where two unknown reactions intersect to eliminate them instantly.

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