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A lamp hangs from two strings tied at a knot. You could resolve every tension into horizontal and vertical parts and solve two simultaneous equations. Or you could write one line of ratios and read off both answers. That line is Lami's theorem. This post shows where it comes from, works two examples with checks, and points out the angle error that most wrong answers share.
Is Lami's theorem on the syllabus?
The JEE (Main) 2026 syllabus lists "equilibrium of concurrent forces" under Unit 3, Laws of Motion. The NCERT Class 11 chapter on laws of motion covers this in Section 4.8, Equilibrium of a particle, but does not name Lami's theorem. It gives the idea the theorem rests on: three forces in equilibrium "can be represented by the sides of a triangle with the vector arrows taken in the same sense." Lami's theorem is that triangle plus the sine rule, so it is a shortcut, not new physics. Check the syllabus for your exam year.
What does the theorem say?
If three coplanar forces F₁, F₂ and F₃ act at one point and keep it in equilibrium, then
F₁/sin α = F₂/sin β = F₃/sin γ
where α is the angle between F₂ and F₃, β is the angle between F₃ and F₁, and γ is the angle between F₁ and F₂. Each force goes with the angle between the other two, the angle it does not touch.
Why is it true?
Put the three force arrows head to tail. Because they add to zero, they close into a triangle. The sine rule says each side divided by the sine of the opposite angle is the same for all three sides. The interior angle opposite F₁ in that triangle is 180° − α, and sin(180° − α) = sin α. So the ratios become exactly Lami's form.
Worked example 1: a lamp on two strings
A 10 kg lamp hangs from a knot. One string goes to the ceiling at 30° to the horizontal, the other at 60° to the horizontal. Take g = 10 m/s², so the weight W = 100 N. Find the tensions T₁ (the 30° string) and T₂ (the 60° string).
Angles between forces. T₁ is 60° from the upward vertical, T₂ is 30° from it, on the other side. So the angle between T₁ and T₂ is 60° + 30° = 90°.
The angle between T₁ and W (straight down) is 180° − 60° = 120°. The angle between T₂ and W is 180° − 30° = 150°.
Lami: T₁/sin 150° = T₂/sin 120° = W/sin 90°.
So T₁ = 100 × 0.5 = 50 N and T₂ = 100 × 0.866 ≈ 86.6 N.
Check with components. Horizontal: T₁ cos 30° = 50 × 0.866 = 43.3 N and T₂ cos 60° = 86.6 × 0.5 = 43.3 N, which balance. Vertical: T₁ sin 30° + T₂ sin 60° = 25 + 75 = 100 N, equal to the weight. The steeper string carries more of the load.
OpenStax University Physics solves the same kind of problem, a 15.0 kg traffic light on two wires, by components. Solve one problem both ways to trust the shortcut.
Worked example 2: why a nearly straight wire is dangerous
A 20 N picture hangs from the midpoint of a wire. Each half of the wire makes 10° with the horizontal. Find the tension.
By symmetry both tensions are equal, T.
Angle between the two tensions: 180° − 2 × 10° = 160°. Angle between each tension and the weight: 90° + 10° = 100°.
Lami: T/sin 100° = 20/sin 160°.
T = 20 × 0.985/0.342 ≈ 57.6 N.
The same answer comes from 2T sin 10° = 20, so T = 20/(2 × 0.1736) ≈ 57.6 N. The tension is nearly three times the weight. The flatter the wire, the larger the tension, OpenStax's tightrope example makes the same point: a wire that sags by only 5° under a 686 N walker carries a tension almost six times the walker's weight. In the limit of a perfectly straight wire, no finite tension could hold any weight.
When can you use Lami's theorem?
All of these must hold:
Exactly three forces. A block on a rough incline has weight, normal force and friction (three), but add a pull and you have four, so resolve instead.
Concurrent: all three act through one point, such as a knot, a small ring or a particle.
Coplanar and in equilibrium: no acceleration.
A rigid body such as a ladder also needs torque balance. Our post on ladders and torque equilibrium covers that case.
Common mistakes and how to avoid them
Using angles with the horizontal or vertical instead of angles between forces. Draw all three forces from the knot first.
Pairing a force with an angle it touches. It must be the angle between the other two.
Using it with four forces. Combine two of them into one first, or resolve.
Forgetting that strings only pull. Every tension points away from the knot along its string.
Ignoring a heavy rope. Lami's theorem treats each string as massless. Our post on tension in a rope with mass shows what changes when it is not.
A practice routine
Day 1: Solve Worked example 1 by components without looking, then by Lami. Time both.
Day 2: Three knot problems with angles given from the vertical. Convert to angles between forces before writing ratios.
Day 3: Two symmetric sagging-wire problems at 5°, 10° and 30°. Tabulate how T changes.
Day 4: Two mixed problems where Lami does not apply, so you practise spotting the fourth force.
If you set up the angles but keep getting the wrong ratio, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare your force diagram with a complete solution. For the free-body diagram habit itself, see our guide to Newton's second law and free-body diagrams.
Tonight, redo Worked example 1 with strings at 45° and 45°, and check that each tension is about 70.7 N.
