Statics

Lami's theorem

Lami's theorem states that 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.

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 (Sine Rule for Coplanar Concurrent Equilibrium Forces)

If three coplanar forces \mathbf{P}, \mathbf{Q}, \mathbf{R} acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces. Mathematically, \frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}, where \alpha, \beta, \gamma are the angles opposite to forces P, Q, R respectively.

📐 Geometric Interpretation

Geometrically, if three concurrent coplanar forces in equilibrium are represented in magnitude and direction by the sides of a triangle taken in order, or by lines parallel to them, the angles between their lines of action dictate their relative magnitudes via the sine rule applied to the force polygon.

⚙️ Algebraic & Structural Meaning

Algebraically, it reduces a two-dimensional vector equilibrium condition (\sum \mathbf{F} = 0) into a single scalar proportional equality by resolving components along mutually orthogonal axes or applying Lami's ratio directly.

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 (spatial forces), the angle between lines of action in a single plane is undefined, rendering the ratios invalid; spatial equilibrium requires vector cross/dot products or moment equations.
Concurrence at a Single Point	`Point-action (Concurrent system)`	Strictly Mandatory	If lines of action do not intersect at a common point, they form a general force system with a resultant couple, violating simple sine proportional equilibrium.
Exact Equilibrium Condition	`\sum F_x = 0 \text{ and } \sum F_y = 0`	Essential for conclusion	If a net resultant force exists, the polygon of forces remains open, and the ratio \frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma} fails to hold.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Direct Force Ratio Calculation

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

Rule: Directly substitute known angle values (e.g., $30^\circ, 45^\circ, 60^\circ, 90^\circ, 120^\circ$) into the sine denominators to compute unknown tensions in strings instantly.

Application: Frequently asked in UP PGT/TGT when a weight hangs from two strings at given angles to the vertical.

Orthogonal / Right-Angle Triangle Reduction

$$P^2 + Q^2 = R^2 \iff \text{Angle between } P \text{ and } Q \text{ is } 90^\circ$$

Rule: When two forces are perpendicular, the third force balances their vector sum via Pythagoras theorem, and Lami's theorem gives $\frac{P}{\sin(90 + \theta)} = \frac{Q}{\cos \theta}$.

Application: Directly tested in KVS PGT and LT Grade mechanics sections.

Symmetric Loading Shortcut

$$P = Q \implies \alpha = \beta$$

Rule: If two of the three forces are equal in magnitude ($P = Q$), the angles opposite to them ($\alpha$ and $\beta$) are equal, meaning the third force bisects the exterior angle.

Application: Standard string-tension symmetry question in UP TGT.

Supplementary Angle Identity for Sine

$$\sin(180^\circ - \theta) = \sin \theta$$

Rule: When internal angles between forces sum to $360^\circ$, use the exterior angles or supplementary interior angles directly without recalculating.

Application: Saves 30 seconds in multi-step equilibrium calculations.

Lami's Theorem via Component Resolution

$$P \sin \theta_1 = Q \sin \theta_2$$

Rule: Equate components perpendicular to any one force to instantly bypass simultaneous equations.

Rapid Elimination via Extreme Angles

$$F \propto \frac{1}{\sin \theta}$$

Rule: The largest force is always opposite to the smallest angle, and vice-versa. Eliminate options violating this inequality instantly.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Using Angles Between Forces vs Angles with Strings

Common Pitfall: Candidates use the angle of the string with the vertical or horizontal directly as the angle in Lami's denominator instead of the angle between the two opposing forces.

Counterexample: \text{Using } \theta \text{ instead of } 180^\circ - \theta \text{ or } 90^\circ + \theta

Why it fails: Lami's theorem requires the angle between the lines of action of the *other two* forces, not the inclination to a coordinate axis.

🛡️ Exam Defense: Always draw a free-body diagram and calculate angles between lines of action meeting at the concurrent point so they sum to $360^\circ$.

⚠️ COMMON ERROR #02

Applying to Non-Concurrent Force Systems

Common Pitfall: Applying Lami's theorem to parallel forces or forces whose lines of action do not intersect at a single point.

Counterexample: \text{Three parallel forces } P, Q, R \text{ in a beam}

Why it fails: Parallel forces have an angle of $0^\circ$ or $180^\circ between them, making \sin 0^\circ = 0$ (division by zero undefined).

🛡️ Exam Defense: Verify concurrence: check that all force vectors intersect at a single spatial/plane point before writing Lami's equations.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Lami's Theorem	3 coplanar, concurrent forces in equilibrium	\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}	Ratio of force to sine of opposite angle in force triangle	Three concurrent strings/tensions in equilibrium
Triangle Law of Forces	3 forces in equilibrium	\frac{A}{a} = \frac{B}{b} = \frac{C}{c}	Forces represented by sides of a closed triangle in order	Geometric length ratios given
Polygon of Forces	n coplanar concurrent forces in equilibrium	\sum \mathbf{F} = 0	Closed polygon formed by force vectors	Multiple forces acting at a single point
Varignon's Theorem	Coplanar concurrent/non-concurrent forces	Moment of resultant = Sum of moments	Sum of torques about a point equals torque of resultant	Levers, beams, and turning effects

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 $l_1$ and $l_2$ fastened to two points in a horizontal line at a distance $l$ apart. If the strings are equally inclined to the vertical, find the tension in the strings when $W = 50\text{ N}$ and the angle between the strings is $60^\circ$.

(A) $50\text{ N}$(B) $25\sqrt{3}\text{ N}$(C) $50/\sqrt{3}\text{ N}$(D) $100\text{ N}$

📚 Method A: Formal Textbook Derivation

Step 1: Identify concurrent forces: Tension $T_1$, Tension $T_2$, and Weight $W = 50$. Step 2: By symmetry, $T_1 = T_2 = T$. Step 3: Angle between strings is $60^\circ$, so angles between weight and each tension are $180^\circ - 30^\circ = 150^\circ$ or similar geometry. Using Lami's theorem: $\frac{T}{\sin(90 + 30)} = \frac{W}{\sin 60}$. Step 4: Compute $T = \frac{50 \sin 120}{\sin 60} = 50$ or $\frac{50}{2 \cos 30} = \frac{50}{\sqrt{3}}$.

⚡ Method B: 25-Second Exam Shortcut

For symmetric strings at angle $2\theta$, each tension $T = \frac{W}{2 \cos \theta}$. Here $2\theta = 60^\circ \implies \theta = 30^\circ$. Thus $T = \frac{50}{2 \cos 30^\circ} = \frac{50}{2(\sqrt{3}/2)} = \frac{50}{\sqrt{3}}$.

💡 Tip: Memorize the symmetric string tension formula: $T = \frac{W}{2\cos\theta}$ where $\theta$ is the angle with the vertical.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Three forces $P, Q, R$ acting at a point are in equilibrium. If the angle between $P$ and $Q$ is $120^\circ$ and the angle between $Q$ and $R$ is $150^\circ$, find the ratio $P : Q : R$.

(A) $1 : \sqrt{2} : 2$(B) $1 : \sqrt{3} : 2$(C) $2 : \sqrt{3} : 1$(D) $\sqrt{3} : 1 : 2$

📚 Method A: Formal Textbook Derivation

Step 1: Sum of angles around a point is $360^\circ$. Angle between $P$ and $Q$ is $120^\circ$, between $Q$ and $R$ is $150^\circ$, so angle between $R$ and $P$ is $360^\circ - (120^\circ + 150^\circ) = 90^\circ$. Step 2: Apply Lami's theorem: $\frac{P}{\sin 150^\circ} = \frac{Q}{\sin 90^\circ} = \frac{R}{\sin 120^\circ}$. Step 3: Evaluate sines: $\sin 150^\circ = 1/2$, $\sin 90^\circ = 1$, $\sin 120^\circ = \sqrt{3}/2$. Step 4: Ratio $P:Q:R = \frac{1}{2} : 1 : \frac{\sqrt{3}}{2} = 1 : 2 : \sqrt{3}$? Wait, let's recheck: $\frac{P}{\sin 150} = \frac{Q}{\sin 90} = \frac{R}{\sin 120} \implies P : Q : R = \frac{1}{2} : 1 : \frac{\sqrt{3}}{2} = 1 : 2 : \sqrt{3}$. Let's verify options: option B is $1 : \sqrt{3} : 2$. Wait, angle opposite to $P$ is angle between $Q$ and $R$ ($150^\circ$). Angle opposite to $Q$ is angle between $R$ and $P$ ($90^\circ$). Angle opposite to $R$ is angle between $P$ and $Q$ ($120^\circ$). Thus ratios are $\sin 150^\circ : \sin 90^\circ : \sin 120^\circ = 1/2 : 1 : \sqrt{3}/2 = 1 : 2 : \sqrt{3}$.

⚡ Method B: 25-Second Exam Shortcut

Write down the three angles opposite to forces: $150^\circ, 90^\circ, 120^\circ$. Take their sines: $1/2, 1, \sqrt{3}/2$. Multiply by 2 to get integers: $1, 2, \sqrt{3}$.

💡 Tip: Always ensure the three angles sum to $360^\circ$ before applying Lami's theorem ratios.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Lami's theorem is applicable only when the number of forces acting at a point is:

A Two

B Three

C Four

D Any number

Derivation & Analysis:

Lami's theorem specifically applies to exactly three coplanar concurrent forces in equilibrium.

MCQ 2: If three forces $P, Q, R$ 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:

If magnitudes are equal, $\sin \alpha = \sin \beta = \sin \gamma$, which for three forces summing to $360^\circ$ implies each angle between them is $120^\circ$.

MCQ 3: A weight $W$ is suspended by a string and pulled aside by a horizontal force $F$ until the string makes an angle $30^\circ$ with the vertical. The tension $T$ in the string is:

A $W \cos 30^\circ$

B $W / \cos 30^\circ$

C $W \sin 30^\circ$

D $W / \sin 30^\circ$

Derivation & Analysis:

By Lami's theorem, $\frac{T}{\sin 90^\circ} = \frac{W}{\sin(180^\circ - 30^\circ)} \implies T = \frac{W}{\sin 150^\circ} = \frac{W}{\cos 30^\circ}$ or $W/(\sqrt{3}/2)$.

MCQ 4: Which of the following conditions is NOT required for Lami's theorem?

A Forces must be coplanar

B Forces must be concurrent

C Forces must be equal in magnitude

D System must be in equilibrium

Derivation & Analysis:

Lami's theorem applies to forces of any magnitude, provided they are concurrent, coplanar, and in equilibrium.

MCQ 5: If three concurrent forces are represented by the sides of a triangle in magnitude, direction, and order, then:

A They are in equilibrium

B Their resultant is maximum

C They are perpendicular

D They are parallel

Derivation & Analysis:

By the triangle law of forces, forces represented in magnitude and direction by the sides of a triangle taken in order are in equilibrium.

MCQ 6: In Lami's theorem equation $\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$, the angles $\alpha, \beta, \gamma$ are:

A Angles between the forces and the horizontal

B Angles between the respective pairs of the other two forces

C Angles of the force triangle

D Interior angles of a polygon

Derivation & Analysis:

By definition, each denominator is the sine of the angle between the other two forces.

MCQ 7: Two forces $P$ and $2P$ act at an angle of $120^\circ$. The magnitude of their resultant is:

A $P$

B $P\sqrt{3}$

C $2P$

D $3P$

Derivation & Analysis:

Resultant $R = \sqrt{P^2 + (2P)^2 + 2(P)(2P)\cos 120^\circ} = \sqrt{P^2 + 4P^2 - 2P^2} = \sqrt{3P^2} = P\sqrt{3}$? Wait: $\cos 120 = -1/2$, so $4P^2 - 2P^2 = 2P^2$, total $3P^2$, so $R = P\sqrt{3}$. Correct option is B ($P\sqrt{3}$).

MCQ 8: If the angle between two equal forces in equilibrium is $120^\circ$, each force is equal to:

A The resultant

B Half the resultant

C Twice the resultant

D Zero

Derivation & Analysis:

Two equal forces at $120^\circ$ have a resultant equal in magnitude to each of the component forces.

MCQ 9: Assertion (A): Lami's theorem cannot be applied to four concurrent coplanar forces in equilibrium. Reason (R): Lami's theorem is strictly formulated for exactly three forces.

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 of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Lami's theorem is mathematically restricted to three forces; for four or more forces, polygon of force or resolution methods are required.

Q10

MCQ 10: If three forces $P, Q, R$ are in equilibrium such that $P = 3\text{ N}, Q = 4\text{ N}$, and the angle between $P$ and $Q$ is $90^\circ$, then $R$ is:

A $3\text{ N}$

B $4\text{ N}$

C $5\text{ N}$

D $7\text{ N}$

Derivation & Analysis:

Since $P$ and $Q$ are at $90^\circ$, their resultant is $\sqrt{3^2 + 4^2} = 5\text{ N}$. For equilibrium, $R$ must equal this resultant in magnitude, i.e., $5\text{ N}$.

Common questions

No, Lami's theorem is strictly valid for exactly three concurrent coplanar forces in equilibrium. For four or more forces, one must use resolution of forces into rectangular components (\sum X = 0, \sum Y = 0) or the polygon of forces.

Candidates frequently mistake the angle a string makes with the vertical or horizontal as the denominator angle in Lami's formula, instead of calculating the angle between the two opposing lines of action.

Memorize standard symmetric tension relations (like $T = \frac{W}{2\cos\theta}$) and utilize the proportionality rule where the largest force opposes the smallest angle without performing full trigonometric expansions.

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