Statics

Rules of triangle

The Rules of Triangle in Statics form the foundational geometric and vector principles—such as Lami's Theorem, the Triangle Law of Forces, and Sine/Cosine rules—governing concurrent and non-concurrent coplanar forces in equilibrium.

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

📖 Topic progress

○ Mark as complete

Definition and statement

Triangle Law of Forces and Lami's Theorem

If three concurrent forces acting at a point are in equilibrium, the magnitude of each force is directly proportional to the sine of the angle between the other two forces: $$\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$$

📐 Geometric Interpretation

Geometrically, if three forces in equilibrium are represented in magnitude and direction by the sides of a triangle taken in order, their vector sum is zero, closing the polygon and forming a closed triangle with interior angles related to force directions.

⚙️ Algebraic & Structural Meaning

Algebraically, it translates vector equilibrium equations $\sum \vec{F} = 0$ into solvable trigonometric ratios using sine and cosine rules, eliminating unknown angles or magnitudes efficiently.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Coplanar Forces	`Geometric Plane (2D space)`	Strictly Mandatory	If forces are non-coplanar, the triangle law fails because three forces cannot form a single closed plane triangle; spatial equilibrium requires tetrahedral or general vector methods.
Concurrency at a Point	`Single Point of Application`	Strictly Mandatory	If lines of action do not intersect at a single point, moments are produced, resulting in rotation rather than pure translational equilibrium.
Equilibrium State ($\sum F = 0$)	`Balanced System`	Essential for conclusion	If the resultant is non-zero, the forces form an open vector chain and cannot be represented by the closed sides of a triangle.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Lami's Theorem Direct Proportion

$$P : Q : R = \sin \alpha : \sin \beta : \sin \gamma$$

Rule: Directly equate ratios of forces to sines of opposite angles without setting up bulky simultaneous equations.

Application: Frequently asked in UP PGT for finding unknown force magnitudes given mutually inclined strings.

Right-Angled Force Triangle

$$P^2 + Q^2 = R^2 \implies \theta = 90^\circ$$

Rule: When two forces are perpendicular, the resultant magnitude is given instantly by Pythagoras theorem without cosine rule expansion.

Application: Standard recurring pattern in UP TGT mechanics sections.

Equal Forces Equilibrium

$$P = Q = R \implies \alpha = \beta = \gamma = 120^\circ$$

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

Application: Direct theoretical question in DSSSB PGT and KVS.

Lami's Supplementary Angle Property

$$\alpha + \beta + \gamma = 360^\circ$$

Rule: Use the property that the angles between the lines of action satisfy $\alpha + \beta + \gamma = 360^\circ$ when all forces radiate outward from a point.

Application: Saves time in geometry-heavy statics problems.

Triangle Law Vector Closure

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

Rule: Use the law of cosines directly for non-equilibrium resultant force magnitude calculations.

Application: Appears in foundational mechanics numericals.

Rapid Elimination via Symmetry

$$P = Q \implies \text{Resultant bisects } \theta$$

Rule: Symmetrical force systems always have their resultant along the angle bisector, allowing immediate elimination of asymmetric options.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Interior vs Exterior Angle Confusion in Lami's Theorem

Common Pitfall: Using the angle between the force vectors themselves instead of the angle between the lines of action of the *other two* forces.

Counterexample: Using $\theta$ instead of $180^\circ - \theta$

Why it fails: Lami's theorem requires the angle between the *other two* forces meeting at the point, not the angle between the force and its component.

🛡️ Exam Defense: Always extend lines of action and measure angles between the opposing forces meeting at the junction.

⚠️ COMMON ERROR #02

Applying Lami's Theorem to Non-Concurrent Systems

Common Pitfall: Applying Lami's theorem when forces are parallel or acting along different lines on a rigid body.

Counterexample: Two parallel forces acting on opposite ends of a rod

Why it fails: Non-concurrent forces produce couples and moments; Lami's theorem only applies to point-concurrent equilibrium.

🛡️ Exam Defense: Verify point concurrency before invoking $\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta}$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Triangle Law of Forces	3 Coplanar forces in equilibrium	$\|\vec{P} + \vec{Q} + \vec{R}\| = 0$	Closed triangle of vectors	Forces represented by sides in order
Lami's Theorem	3 Concurrent forces in equilibrium	$P/\sin\alpha = Q/\sin\beta = R/\sin\gamma$	Sine rule applied to force polygon	Angles between lines of action given
Polygon Law of Forces	$n$ Coplanar forces in equilibrium	$\sum \vec{F}_i = 0$	Closed multi-sided polygon	More than 3 concurrent forces
Converse of Triangle Law	Sides proportional to forces	Triangle is similar to force triangle	Geometric similarity	Proving equilibrium or finding angles

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Three forces $P, Q, R$ acting at a point are in equilibrium. If the angle between $P$ and $Q$ is $90^\circ$ and between $Q$ and $R$ is $120^\circ$, find the ratio $P : Q : R$.

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

📚 Method A: Formal Textbook Derivation

Step 1: Sum of angles around a point is $360^\circ$. Angle between $R$ and $P$ is $360^\circ - (90^\circ + 120^\circ) = 150^\circ$. Step 2: Apply Lami's theorem: $P / \sin(120^\circ) = Q / \sin(150^\circ) = R / \sin(90^\circ)$. Step 3: Substitute values: $\sin(120^\circ) = \frac{\sqrt{3}}{2}$, $\sin(150^\circ) = \frac{1}{2}$, $\sin(90^\circ) = 1$. Step 4: Ratio becomes $\frac{\sqrt{3}}{2} : \frac{1}{2} : 1 = \sqrt{3} : 1 : 2$.

⚡ Method B: 25-Second Exam Shortcut

Note that the angle opposite to $R$ is $90^\circ$ ($R=2$ units if scale is 2). Check options matching $\sin 120^\circ : \sin 150^\circ : \sin 90^\circ$. Option (B) matches immediately.

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

CANONICAL PROBLEM 02 Dual-Method Breakdown

A weight $W$ is suspended by two strings of length $L$ from two points at the same horizontal level at distance $L$ apart. The tension in each string is:

(A) $W$(B) $\sqrt{3}W$(C) $\frac{W}{\sqrt{3}}$(D) $\frac{2W}{\sqrt{3}}$

📚 Method A: Formal Textbook Derivation

Step 1: The strings and the distance between supports form an equilateral triangle of side $L$. Step 2: Angles of the equilateral triangle are $60^\circ$. Step 3: By Lami's theorem at the weight junction, the angle between the strings is $60^\circ$, and angle between each string and vertical downward weight $W$ is $90^\circ + 30^\circ = 120^\circ$. Step 4: $T / \sin(120^\circ) = W / \sin(60^\circ) \implies T = W \frac{\sqrt{3}/2}{\sqrt{3}/2}$? Wait, angle between string and vertical is $30^\circ$, so angle between string and opposite tension is $120^\circ$, angle between tension and $W$ is $90^\circ + 30^\circ = 120^\circ$. Thus $T / \sin 120^\circ = W / \sin 60^\circ$ gives $T = W$. Let's re-verify: equilateral triangle implies string makes $60^\circ$ with horizontal, so $30^\circ$ with vertical. Vertical equilibrium: $2 T \cos 30^\circ = W \implies 2 T (\sqrt{3}/2) = W \implies T = \frac{W}{\sqrt{3}}$? Wait, if angle with vertical is $30^\circ$, angle between strings is $60^\circ$. By Lami's: $T / \sin(90^\circ + 30^\circ) = W / \sin(60^\circ) \implies T / \cos 30^\circ = W / \sin 60^\circ \implies T = W$. Let's check option C: $\frac{W}{\sqrt{3}}$.

⚡ Method B: 25-Second Exam Shortcut

Resolve vertically: $2 T \cos 30^\circ = W \implies T \sqrt{3} = W \implies T = W / \sqrt{3}$.

💡 Tip: Equilateral triangle geometry combined with vertical resolution yields the answer.

Practice questions

Choose an answer to check it and see the solution:

If three forces acting in one plane are in equilibrium, then:

A They must be parallel

B They must be concurrent or parallel

C They must be equal in magnitude

D They must act at right angles

Derivation & Analysis:

By principles of statics, three coplanar forces in equilibrium must either meet at a single point (concurrent) or be parallel to each other.

Lami's theorem is applicable only when the number of forces in equilibrium is:

A Two

B Three

C Four

D Any number

Derivation & Analysis:

Lami's theorem is specifically formulated for exactly three concurrent coplanar forces in equilibrium.

Two forces $P$ and $Q$ act at an angle $\theta$. Their resultant $R$ is maximum when $\theta$ is:

A $0^\circ$

B $90^\circ$

C $180^\circ$

D $45^\circ$

Derivation & Analysis:

Resultant $R = \sqrt{P^2 + Q^2 + 2PQ \cos \theta}$. Maximum value occurs when $\cos \theta = 1$, i.e., $\theta = 0^\circ$.

If the resultant of two equal forces $P$ has magnitude equal to $P$, the angle between them is:

A $60^\circ$

B $90^\circ$

C $120^\circ$

D $180^\circ$

Derivation & Analysis:

$P^2 = P^2 + P^2 + 2P^2 \cos \theta \implies -P^2 = 2P^2 \cos \theta \implies \cos \theta = -1/2 \implies \theta = 120^\circ$.

According to the triangle law of forces, if three forces are represented in magnitude and direction by the sides of a triangle taken in order, their resultant is:

A Zero

B Maximum

C Equal to the third side

D Undefined

Derivation & Analysis:

A closed vector polygon in order represents a zero net resultant, hence the system is in equilibrium.

If $P, Q, R$ are three forces in equilibrium such that $P = Q = R$, the angles between their lines of action are:

A $90^\circ, 90^\circ, 180^\circ$

B $60^\circ, 60^\circ, 120^\circ$

C $120^\circ, 120^\circ, 120^\circ$

D $45^\circ, 45^\circ, 90^\circ$

Derivation & Analysis:

By Lami's theorem, equal forces must have equal sine ratios of opposite angles, hence $\sin \alpha = \sin \beta = \sin \gamma$, giving $120^\circ$ each.

The angle between two forces of equal magnitude $P$ when their resultant is also of magnitude $P$ is:

A $60^\circ$

B $90^\circ$

C $120^\circ$

D $150^\circ$

Derivation & Analysis:

As derived earlier, $R = P$ for equal forces $P$ at $120^\circ$.

If three forces $P, Q, R$ at a point are in equilibrium and $P \perp Q$, then the relation between $P, Q, R$ is:

A $R^2 = P^2 + Q^2$

B $R = P + Q$

C $R^2 = P^2 - Q^2$

D $R = \sqrt{P} + \sqrt{Q}$

Derivation & Analysis:

Since $P$ and $Q$ are perpendicular, their resultant must be equal and opposite to $R$, hence $R^2 = P^2 + Q^2$.

Which of the following is a necessary condition for Lami's theorem?

A Forces must be parallel

B Forces must be concurrent and coplanar

C Forces must be equal

D Forces must be non-coplanar

Derivation & Analysis:

Lami's theorem strictly requires coplanar and concurrent force systems.

Q10

If the magnitudes of two forces are 3 N and 4 N, their minimum possible resultant force magnitude is:

A 7 N

B 5 N

C 1 N

D 12 N

Derivation & Analysis:

Minimum resultant occurs at $180^\circ$, which is $|4 - 3| = 1\text{ N}$.

Common questions

No. Lami's theorem is strictly restricted to exactly three concurrent coplanar forces in equilibrium. For four or more forces, one must use the polygon law or resolve forces into rectangular Cartesian components ($ \sum X = 0, \sum Y = 0 $).

Students frequently make sign errors or take the wrong interior/exterior angles when setting up Lami's equations. Always ensure the angles used are between the lines of action of the opposing forces.

Memorize standard symmetric angle configurations (like $120^\circ$ for equal forces and $90^\circ-120^\circ-150^\circ$ triangles) and directly utilize proportional ratio shortcuts without full derivations.

← Previous Topic

Resolution into two perpendicular forces

Next Topic →

Trigonometrical theorem