Statics

General conditions of equilibrium

The general conditions of equilibrium dictate that a rigid body acted upon by a coplanar or non-coplanar system of forces remains at rest if and only if the vector sum of all forces and the vector sum of all moments about any arbitrary point vanish identically.

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

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Definition and statement

Principle of Virtual Work and General Conditions of Equilibrium

A rigid body acted upon by a system of forces in two or three dimensions is in a state of complete mechanical equilibrium if and only if: (1) The resultant force vector $\vec{R} = \sum \vec{F}_i = \vec{0}$, and (2) The resultant couple or moment vector about any arbitrary point $O$, $\vec{G}_O = \sum (\vec{r}_i \times \vec{F}_i) = \vec{0}$. For coplanar force systems acting in the $xy$-plane, these vector equations reduce to three scalar equations: $\sum F_x = 0$, $\sum F_y = 0$, and $\sum M_O = 0$.

📐 Geometric Interpretation

Geometrically, the vanishing of force components ensures translational equilibrium (no net displacement along any linear axis), while the vanishing of the algebraic sum of moments about any point ensures rotational equilibrium (no net angular acceleration or tendency to rotate about any axis).

⚙️ Algebraic & Structural Meaning

Algebraically, this provides a closed system of independent linear equations in scalar unknowns. For coplanar force systems, exactly three independent scalar equations are available, allowing the determination of up to three unknown constraint reactions or force parameters.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Rigidity of the Body	`Continuous domain of the body`	Strictly Mandatory	If internal deformation occurs, internal forces do zero virtual work only if rigidity is strictly maintained; otherwise, internal strain energy terms must be incorporated.
Completeness of the Force System	`All active and reactive forces`	Strictly Mandatory	Omitting any constraint reaction (such as normal reaction or friction) leads to an underdetermined system and incorrect equilibrium solutions.
Independence of Moment Centers	`Arbitrary point $O$ in the plane/space`	Essential for sufficiency	If the moment equations are taken about collinear points that lie along the line of action of a resultant force, the moment condition alone fails to guarantee full rotational equilibrium unless coupled with independent force summation.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Three Concurrent or Parallel Forces (Lami's Theorem)

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

Rule: If three coplanar forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces. Instantly apply this to string and weight problems without resolving into components.

Application: Directly tested in UP TGT/PGT for tension in strings and inclined plane problems.

Moment Center Selection Trick

$$\sum M_A = 0, \quad \sum M_B = 0, \quad \sum F_x = 0$$

Rule: To find an unknown force, always take the moment about the point of intersection of the lines of action of the other two unknown forces. This eliminates two unknowns instantly in a single equation.

Three-Force Principle

$$Line\, of\, action\, of\, third\, force\, must\, pass\, through\, point\, of\, concurrency$$

Rule: If a rigid body is kept in equilibrium by three coplanar forces, their lines of action must either be all parallel or all concurrent at a single point.

Application: Frequently used in conceptual assertion-reason questions in KVS/NVS PGT.

Inclined Plane Resolution Shortcut

$$W \sin \theta = F, \quad W \cos \theta = R$$

Rule: Instead of resolving horizontally and vertically, always resolve forces parallel and perpendicular to the inclined plane when a body rests on an incline.

Application: Standard recurring pattern in UP PGT Mathematics.

Coplanar Equilibrium Condition Count

$$N_{eq} = 3 \text{ (for 2D)}, \quad N_{eq} = 6 \text{ (for 3D)}$$

Rule: A general coplanar force system permits at most 3 independent scalar equations of equilibrium: $(\sum X = 0, \sum Y = 0, \sum M = 0)$ or $(\sum M_A = 0, \sum M_B = 0, \sum M_C = 0)$ provided $A, B, C$ are not collinear.

Application: Tested in theoretical multiple-choice questions regarding degrees of freedom and statical determinacy.

Rapid Elimination via Symmetry

$$P_1 = P_2 \implies \theta_1 = \theta_2$$

Rule: For symmetric loading or symmetric suspension systems, immediately equate symmetric tension components or reaction forces to eliminate symmetric variables instantly.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Collinear Moment Centers Fallacy

Common Pitfall: Using three moment equations $\sum M_A = 0, \sum M_B = 0, \sum M_C = 0$ when points $A, B, C$ are collinear.

Counterexample: Points $A, B, C$ lying on the line $y = mx + c$

Why it fails: If the moment center points are collinear, the three moment equations are not linearly independent; they fail to ensure translational equilibrium perpendicular to the line.

🛡️ Exam Defense: Ensure that the three points chosen for moment equilibrium do not lie on the same straight line.

⚠️ COMMON ERROR #02

Ignoring Direction of Friction and Normal Reactions

Common Pitfall: Assuming friction always acts upward or normal reaction is simply equal to $W$ on an inclined plane.

Counterexample: Body on an inclined plane with an external pulling force at an angle

Why it fails: Normal reaction depends on all perpendicular components of applied forces, not just weight $W \cos \theta$.

🛡️ Exam Defense: Always draw a complete free-body diagram (FBD) resolving all forces perpendicular and parallel to the surface before applying equilibrium equations.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Lami's Theorem	Three concurrent coplanar forces in equilibrium	$\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$	Force magnitudes proportional to sines of opposite angles	Problem features a point suspended by strings or concurrent forces
General Coplanar Equilibrium	Any system of coplanar forces acting on a rigid body	$\sum X = 0, \sum Y = 0, \sum M = 0$	No net linear force in $x, y$ directions and no net torque	General beams, rods, ladders, or hinged structures
Three-Force Principle	Exactly three coplanar non-parallel forces in equilibrium	Lines of action are concurrent at a single point	Forces form a closed triangle of vectors	A rod or ladder resting against smooth walls and floors
Principle of Virtual Work	A mechanical system in equilibrium with admissible virtual displacements	$/delta W = \sum F_i \cdot \delta r_i = 0$	Total virtual work done by all forces vanishes for any infinitesimal displacement	Complex connected systems, frameworks, or problems avoiding internal reactions

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

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

(A) $\frac{1}{2} \cot \alpha$(B) $\tan \alpha$(C) $\frac{1}{2} \tan \alpha$(D) $\cot \alpha$

📚 Method A: Formal Textbook Derivation

Step 1: Draw FBD with normal reaction at wall $R_1$, normal reaction at floor $R_2$, weight $W$ acting at the midpoint, and limiting friction $F = \mu R_2$ directed away from the wall. Step 2: Apply horizontal equilibrium: $R_1 = \mu R_2$. Step 3: Apply vertical equilibrium: $R_2 = W$. Step 4: Take moments about the lower end: $R_1 (2l \sin \alpha) - W (l \cos \alpha) = 0$. Step 5: Substitute $R_1 = \mu W$ and $R_2 = W$ to get $\mu = \frac{1}{2} \cot \alpha$.

⚡ Method B: 25-Second Exam Shortcut

Direct formula recall for a ladder on a smooth wall and rough floor: $\mu = \frac{1}{2} \cot \alpha$.

💡 Tip: Memorize standard ladder results; they appear verbatim in UP PGT exams.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Three forces $P, Q, R$ acting along the sides of a triangle $ABC$ are in equilibrium. If $P : Q : R = \sqrt{3} : 1 : 2$, what is the angle $A$ of the triangle?

(A) 30^\circ(B) 60^\circ(C) 90^\circ(D) 120^\circ

📚 Method A: Formal Textbook Derivation

Step 1: By Lami's theorem or triangle of forces, forces acting along the sides of a triangle are proportional to the lengths of the sides: $P \propto a$, $Q \propto b$, $R \propto c$. Step 2: Thus, $a : b : c = \sqrt{3} : 1 : 2$. Step 3: Apply the cosine rule for angle $A$: $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{1^2 + 2^2 - (\sqrt{3})^2}{2(1)(2)} = \frac{1 + 4 - 3}{4} = \frac{2}{4} = \frac{1}{2}$. Step 4: Therefore, $A = 60^\circ$? Wait, let's recalculate: sides are $a = \sqrt{3}, b = 1, c = 2$. $\cos A = \frac{1^2 + 2^2 - 3}{4} = \frac{2}{4} = 1/2 \implies A = 60^\circ$. Let's check options: (B) is 60^\circ.

⚡ Method B: 25-Second Exam Shortcut

Recognize that sides $1, \sqrt{3}, 2$ form a $30^\circ-60^\circ-90^\circ$ triangle where hypotenuse is 2, opposite to $90^\circ$. Thus $c=2$ is the hypotenuse opposite to angle $C = 90^\circ$. Then $a=\sqrt{3}$ is opposite to $A = 60^\circ$.

💡 Tip: Always check which side corresponds to which angle in triangle force problems.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If a system of coplanar forces is in equilibrium, then the algebraic sum of the moments of the forces about any point in their plane is:

A Equal to the resultant force

B Zero

C Maximum

D Equal to the sum of forces

Derivation & Analysis:

For any coplanar force system in equilibrium, the moment about any arbitrary point in the plane must vanish identically, i.e., $\sum M = 0$.

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

A Forces are coplanar and parallel

B Forces are concurrent and coplanar in equilibrium

C Forces form a couple

D The body is in rotational motion

Derivation & Analysis:

Lami's theorem specifically applies to three coplanar forces acting concurrently at a single point in equilibrium.

MCQ 3: A weight $W$ is suspended by two strings of lengths $l_1$ and $l_2$ attached to two points at the same horizontal level at distance $d$. The tensions in the strings are inversely proportional to:

A The lengths of the strings

B The sines of the angles the strings make with the vertical

C The horizontal components of the lengths

D The squares of the lengths

Derivation & Analysis:

By Lami's theorem, $\frac{T_1}{\sin \theta_2} = \frac{T_2}{\sin \theta_1} = \frac{W}{\sin(\theta_1 + \theta_2)}$, meaning tensions are inversely proportional to the sine of the angle made by the other string with the vertical.

MCQ 4: The number of independent scalar equations of equilibrium for a non-coplanar (3D) force system acting on a rigid body is:

A 3

B 4

C 5

D 6

Derivation & Analysis:

In 3D space, force equilibrium gives 3 scalar equations $(\sum F_x = 0, \sum F_y = 0, \sum F_z = 0)$ and moment equilibrium gives 3 scalar equations $(\sum M_x = 0, \sum M_y = 0, \sum M_z = 0)$, totaling 6 independent equations.

MCQ 5: If three forces acting in one plane on a rigid body keep it in equilibrium, they must either be all parallel or:

A Perpendicular

B Concurrent

C Equal in magnitude

D Equal to zero

Derivation & Analysis:

This is the classic Three-Force Principle: three coplanar forces in equilibrium are either all parallel or concurrent at a single point.

MCQ 6: A uniform beam of weight $W$ is hinged at one end and supported horizontally by a string at the other end making an angle $\theta$ with the beam. The tension in the string is:

A $W \sin \theta$

B $\frac{W}{2 \sin \theta}$

C $W \cos \theta$

D $2W \sin \theta$

Derivation & Analysis:

Taking moments about the hinge: $T \cdot l \sin \theta - W \cdot \frac{l}{2} = 0 \implies T = \frac{W}{2 \sin \theta}$.

MCQ 7: If the coefficient of friction is $\mu = \tan \lambda$, where $/lambda$ is the angle of friction, then a body placed on an inclined plane of inclination $/alpha$ will begin to slide down if:

A $/alpha < \lambda$

B $/alpha = \lambda$

C $/alpha > \lambda$

D $/alpha = 0$

Derivation & Analysis:

Limiting equilibrium occurs when $/alpha = \lambda$. The body slides down when the angle of inclination exceeds the angle of friction, i.e., $/alpha > \lambda$.

MCQ 8: The principle of virtual work states that the virtual work done by all the forces acting on a mechanical system in equilibrium is:

A Positive

B Negative

C Zero

D Maximum

Derivation & Analysis:

For any system in equilibrium, the total virtual work done by active and constraint forces during any admissible virtual displacement is identically zero.

MCQ 9: Two forces $P$ and $Q$ act at an angle $/alpha$. Their resultant $R$ makes an angle $/theta$ with $P$. If $P = Q$, then $/theta$ is equal to:

A $/alpha$

B $/alpha / 2$

C 2$/alpha$

D $90^\circ - \alpha$

Derivation & Analysis:

When equal forces $P = Q$ act, the resultant bisects the angle between them, so $/theta = \alpha / 2$.

Q10

MCQ 10: Which of the following conditions is necessary and sufficient for a coplanar force system to be in complete equilibrium?

A $\sum F_x = 0$ only

B $\sum M_O = 0$ only

C $\sum F_x = 0, \sum F_y = 0$

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

Derivation & Analysis:

Translational equilibrium requires $\sum F_x = 0$ and $\sum F_y = 0$, and rotational equilibrium requires $\sum M_O = 0$.

Common questions

Because if a rigid body has no net translational force and has no tendency to rotate about one point, it has no tendency to rotate about any other point in space. Varignon's theorem and the independence of moments guarantee that moment equilibrium about any three non-collinear points (or one point combined with two orthogonal force sums) completely locks down all rotational and translational degrees of freedom in 2D.

Students frequently miss accounting for friction direction, miscalculate the line of action of normal reactions on rough surfaces, or fail to take moments about a point that eliminates the maximum number of unknown constraint forces.

By immediately identifying symmetry, applying Lami's theorem for three concurrent forces, choosing moment centers at the intersection of unknown forces to eliminate them instantly, and memorizing standard structural results like ladder friction formulas.

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Lami's theorem

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Lami's Theorem	Three concurrent coplanar forces in equilibrium	\(\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}\)	Force magnitudes proportional to sines of opposite angles	Problem features a point suspended by strings or concurrent forces
General Coplanar Equilibrium	Any system of coplanar forces acting on a rigid body	\(\sum X = 0, \sum Y = 0, \sum M = 0\)	No net linear force in \(x, y\) directions and no net torque	General beams, rods, ladders, or hinged structures
Three-Force Principle	Exactly three coplanar non-parallel forces in equilibrium	Lines of action are concurrent at a single point	Forces form a closed triangle of vectors	A rod or ladder resting against smooth walls and floors
Principle of Virtual Work	A mechanical system in equilibrium with admissible virtual displacements	\(/delta W = \sum F_i \cdot \delta r_i = 0\)	Total virtual work done by all forces vanishes for any infinitesimal displacement	Complex connected systems, frameworks, or problems avoiding internal reactions