9. Mechanics — Statics

Equilibrium of a body under three forces

The study of a rigid body or particle in static equilibrium under the action of precisely three coplanar forces rests upon fundamental vector principles, Lami's Theorem, and the Triangle Law of Forces.

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

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Definition and statement

Lami's Theorem and Conditions for Three-Force Equilibrium

If a rigid body is in equilibrium under the action of three coplanar, concurrent forces $\vec{F}_1$, $\vec{F}_2$, and $\vec{F}_3$, then: (1) The lines of action of the three forces must be concurrent or parallel. (2) The magnitude of each force is directly proportional to the sine of the angle between the other two forces: $$\frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma}$$ where $\alpha$, $\beta$, and $\gamma$ are the angles opposite to $\vec{F}_1$, $\vec{F}_2$, and $\vec{F}_3$ respectively.

📐 Geometric Interpretation

Geometrically, the vectors representing the three forces can be arranged in magnitude and direction to form a closed triangle (Triangle Law of Forces). The interior angles of this force triangle are supplementary to the angles between the lines of action of the forces.

⚙️ Algebraic & Structural Meaning

Algebraically, the vector sum of the forces must vanish ($\sum \vec{F} = 0$), and the algebraic sum of their moments about any point in their plane must also vanish ($\sum \vec{\tau} = 0$), reducing the static equilibrium equations to scalar trigonometric ratios.

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	`Two-dimensional plane ($$\mathbb{R}^2$$)`	Strictly Mandatory / Necessary	If the forces are non-coplanar and non-concurrent, they constitute a general spatial force system or wrench, and Lami's Theorem fails completely.
Concurrence of Lines of Action	`Point intersection requirement`	Strictly Mandatory	If three forces are coplanar and parallel (non-concurrent), they form a couple or produce net rotation, violating equilibrium unless balanced by a third intersecting force.
Exact count of three forces	$N = 3$	Essential for Lami's ratios	For $N \neq 3$, Lami's direct sine-ratio formula cannot be applied directly without resolving into orthogonal components ($$\sum F_x = 0, \sum F_y = 0$$).

Useful results and shortcuts

Use these identities and results when their conditions apply:

Lami's Ratio Direct Application

$$F_1 : F_2 : F_3 = \sin \alpha : \sin \beta : \sin \gamma$$

Rule: When given angles between three concurrent forces in equilibrium, instantly equate force magnitudes to sines of opposite angles without setting up full $x$-$y$ equilibrium equations.

Application: Directly asked in UP PGT/TGT to find unknown tension in strings or weight suspended by strings.

Lami's Theorem with Right Triangles

$$\frac{T_1}{\sin(90^\circ + \theta)} = \frac{T_2}{\sin(90^\circ + \phi)} = \frac{W}{\sin(180^\circ - (\theta + \phi))}$$

Rule: Simplify sine terms using reduction formulas ($$\sin(90^\circ + \theta) = \cos \theta$$) to instantly obtain trigonometric ratios of string tensions.

Application: Standard recurring pattern in KVS PGT and DSSSB mechanics portions.

Triangle of Forces & Sine Rule Equivalence

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

Rule: Recognize that the force polygon is a triangle whose side lengths are proportional to the magnitudes of the forces, allowing geometric similarity shortcuts.

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

Orthogonal Resolution Alternative ($$\Sigma F_x = 0, \Sigma F_y = 0$$)

$$F_1 \cos \theta_1 + F_2 \cos \theta_2 + F_3 \cos \theta_3 = 0$$

Rule: If angles are cumbersome, align the $x$-axis along one of the unknown forces to eliminate it instantly from the $$\Sigma F_y = 0$$ equation.

Application: Used in complex ladder or string equilibrium problems where Lami's angles are obtuse or difficult to compute.

Lami's Theorem for Equal Forces

$$F_1 = F_2 = F_3 \implies \alpha = \beta = \gamma = 120^\circ$$

Rule: If three forces in equilibrium are equal in magnitude, the angle between any two of them is strictly $120^\circ$.

Application: Quick conceptual elimination in multiple-choice questions.

Rapid Elimination via Triangle Inequality

$$F_1 + F_2 > F_3$$

Rule: Any three forces in equilibrium must form a closed triangle, meaning the sum of any two force magnitudes must strictly exceed the third.

Application: Instantly eliminates 2 absurd options in numerical force magnitude questions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Non-Concurrent Coplanar Forces (Parallel Forces)

Common Pitfall: Applying Lami's theorem blindly to three parallel coplanar forces (e.g., three parallel beam loads).

Counterexample: Three parallel forces $F_1, F_2, F_3$ acting vertically downwards on a beam.

Why it fails: The angle between parallel forces is $0^\circ$ or $180^\circ$, making $\sin \theta = 0$, leading to division by zero in Lami's formula.

🛡️ Exam Defense: Check concurrence: Lami's theorem requires lines of action to intersect at a single point. If parallel, use moment equilibrium ($$\sum M = 0$$).

⚠️ COMMON ERROR #02

Incorrect Angle Identification in Lami's Theorem

Common Pitfall: Taking the angle adjacent to the force instead of the angle between the *other two* forces.

Counterexample: Using the angle between $F_1$ and $F_2$ as the denominator for $F_1$.

Why it fails: Results in inverted or completely incorrect trigonometric ratios, leading to wrong option selection.

🛡️ Exam Defense: Always draw a clear free-body diagram showing vectors acting outwards from the concurrent point, and use angles between adjacent force vectors.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Lami's Theorem	3 coplanar, concurrent forces in equilibrium	$$\frac{F_1}{\sin \alpha} = \frac{F_2}{\sin \beta} = \frac{F_3}{\sin \gamma}$$	Proportionality to sines of opposite angles in force triangle	Mention of three strings, tensions, and concurrent angles
Triangle Law of Forces	3 forces represented in magnitude & direction	Closed vector triangle: $$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0$$	Perimeter closure of vector polygon	Geometrical configuration of forces forming a triangle
Lamé's / Resolution Method	General coplanar concurrent forces ($$N \ge 3$$)	$$\sum F_x = 0, \sum F_y = 0$$	Orthogonal balance of components along axes	Multiple inclined strings or complex force systems
Three Parallel Forces	3 coplanar parallel forces in equilibrium	$$F_1 + F_2 = F_3$$ and $$F_1 d_1 = F_2 d_2$$	Balance of parallel line loads and moment equilibrium	Beam, rod, or weight resting on supports

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

A weight of $$W$$ is suspended by two strings of lengths $$l_1$$ and $$l_2$$ fixed to two points at the same horizontal level at a distance $$l$$ apart. If the strings make angles $\theta$ and $\phi$ with the vertical, find the tension $$T_1$$ in the first string.

(A) $\frac{W \cos \phi}{\sin(\theta + \phi)}$(B) $\frac{W \sin \phi}{\sin(\theta + \phi)}$(C) $\frac{W \sin \theta}{\sin(\theta + \phi)}$(D) $\frac{W \cos \theta}{\sin(\theta + \phi)}$

📚 Method A: Formal Textbook Derivation

Step 1: Draw free-body diagram at the knot where weight $$W$$ is suspended. Step 2: Identify angles between forces: angle between $$T_2$$ and vertical downward weight $$W$$ is $180^\circ - \phi$ (or relative angles). Step 3: Apply Lami's Theorem: $\frac{T_1}{\sin(180^\circ - \phi)} = \frac{W}{\sin(180^\circ - (\theta + \phi))}$ (using appropriate angle relations). Step 4: Simplify to get $T_1 = \frac{W \sin \phi}{\sin(\theta + \phi)}$.

⚡ Method B: 25-Second Exam Shortcut

Use Lami's direct ratio: Tension in string 1 ($T_1$) is proportional to the sine of the angle that string 2 makes with the vertical (i.e., $\sin \phi$), divided by the sine of the included angle between the two strings ($$\sin(\theta + \phi)$$).

💡 Tip: Memorize this standard string suspension result as it appears directly without numerical values in UP PGT.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Three forces acting on a particle are in equilibrium. If their magnitudes are $$10\text{ N}$$, $$10\text{ N}$$, and $10\sqrt{2}\text{ N}$$, the angle between the two smaller forces is:

(A) $30^\circ$(B) $45^\circ$(C) $90^\circ$(D) $120^\circ$

📚 Method A: Formal Textbook Derivation

Step 1: Check triangle condition: $10^2 + 10^2 = 100 + 100 = 200 = (10\sqrt{2})^2$. Step 2: The force vector triangle is a right-angled isosceles triangle. Step 3: The angle between the two equal forces of magnitude $10\text{ N}$ forming the right angle is $90^\circ$.

⚡ Method B: 25-Second Exam Shortcut

Recognize the Pythagorean triplet in magnitudes ($$1^2 + 1^2 = (\sqrt{2})^2$$), which instantly confirms a right-angled triangle of forces, meaning the angle between the orthogonal equal forces is $90^\circ$.

💡 Tip: Always check if force magnitudes satisfy Pythagoras theorem before applying general sine formulas.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Lami's theorem is applicable only when:

A Forces are parallel

B Forces are coplanar and concurrent

C Forces are non-coplanar

D Forces form a couple

Derivation & Analysis:

Lami's theorem requires the forces to be exactly three in number, coplanar, and concurrent.

MCQ 2: If three forces $\vec{F}_1, \vec{F}_2, \vec{F}_3$ in equilibrium satisfy $F_1 = F_2 = F_3$, the angle between any two forces is:

A $60^\circ$

B $90^\circ$

C $120^\circ$

D $180^\circ$

Derivation & Analysis:

By Lami's theorem, equal forces must have equal opposite angles, so $\alpha = \beta = \gamma = \frac{360^\circ}{3} = 120^\circ$.

MCQ 3: A smooth sphere of weight $W$ rests inside a smooth hemispherical cup of radius $r$. The reaction of the cup on the sphere passes through:

A The center of the hemisphere

B The highest point of the hemisphere

C The rim of the cup

D Tangent to the rim

Derivation & Analysis:

For a smooth sphere resting in a smooth cup, the normal reaction at the point of contact always passes through the geometric center of the sphere and cup.

MCQ 4: Three coplanar forces of magnitudes $3\text{ N}$, $4\text{ N}$, and $5\text{ N}$ are in equilibrium. The angle between the $3\text{ N}$ and $4\text{ N}$ forces is:

A $0^\circ$

B $60^\circ$

C $90^\circ$

D $180^\circ$

Derivation & Analysis:

Since $3^2 + 4^2 = 5^2$, the force triangle is a right-angled triangle, making the angle between the two perpendicular legs $90^\circ$.

MCQ 5: If three forces acting at a point are in equilibrium and represented by the sides of a triangle taken in order, then the forces are:

A Parallel

B Proportional to the lengths of the sides

C Equal in magnitude

D Perpendicular to the sides

Derivation & Analysis:

By the Triangle Law of Forces, if three forces acting at a point are represented in magnitude and direction by the sides of a triangle taken in order, they are in equilibrium, and their magnitudes are proportional to the side lengths.

MCQ 6: Two forces $P$ and $Q$ act at an angle $\theta$. If their resultant is equal to $\sqrt{P^2 + Q^2}$, then $\theta$ is:

A $0^\circ$

B $45^\circ$

C $90^\circ$

D $180^\circ$

Derivation & Analysis:

Resultant $R = \sqrt{P^2 + Q^2 + 2PQ \cos \theta}$. Setting $R = \sqrt{P^2 + Q^2}$ gives $2PQ \cos \theta = 0 \implies \cos \theta = 0 \implies \theta = 90^\circ$.

MCQ 7: A body is in equilibrium under three forces. Two of the forces are $P$ and $2P$, and the angle between them is $60^\circ$. The magnitude of the third force is:

A $P\sqrt{3}$

B $P\sqrt{7}$

C $3P$

D $2P$

Derivation & Analysis:

Third force $F_3 = \sqrt{P^2 + (2P)^2 + 2(P)(2P)\cos 60^\circ} = \sqrt{P^2 + 4P^2 + 2P^2(1/2)} = \sqrt{7P^2} = P\sqrt{7}$.

MCQ 8: Assertion (A): Lami's theorem cannot be applied to four coplanar concurrent forces. Reason (R): Lami's theorem relates three forces in equilibrium with the sines of the angles between them.

A Both A and R are true and R is the correct explanation of A

B Both A and R are true but R is not the correct explanation

C A is true, R is false

D A is false, R is true

Derivation & Analysis:

Lami's theorem is strictly formulated for exactly three forces. For four forces, resolution into components is required.

MCQ 9: A string of length $l$ supports a weight $W$ at its midpoint when pulled aside by a horizontal force $F$. The tension in the string is:

A Independent of $l$

B Dependent on the angle of displacement

C Always equal to $W$

D Equal to $F$

Derivation & Analysis:

Tension in strings supporting weights varies inversely with the sine of the angle made with the vertical, depending directly on string geometry.

Q10

MCQ 10: If three concurrent forces are represented by vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ such that $\vec{A} + \vec{B} + \vec{C} = 0$, the angle between $\vec{A}$ and $\vec{B}$ when $|\vec{A}| = 3, |\vec{B}| = 4, |\vec{C}| = 5$ is:

A $0^\circ$

B $90^\circ$

C $180^\circ$

D $60^\circ$

Derivation & Analysis:

Since $3^2 + 4^2 = 5^2$, the vectors form a right-angled triangle, making the angle between $\vec{A}$ and $\vec{B}$ equal to $90^\circ$.

Common questions

No. Lami's theorem strictly requires the lines of action of all three coplanar forces to pass through a single common point (concurrence). If they are parallel, they form a couple or net torque, and moment equilibrium equations must be used instead.

Candidates frequently make sign errors or misidentify the angles between the forces in Lami's theorem by using interior triangle angles directly instead of the angles between the vector lines of action extending outwards from the point.

By immediately memorizing standard geometric configurations (such as strings at equal angles or right-angled force triangles) and applying Lami's sine ratios directly without expanding into full $x$ and $y$ component equations.

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