1.8 Equilibrium of a particle and Lami's theorem

Engineering Mechanics Chapter 1: Force systems and equilibrium Reviewed 25 Oct 2026 Author: Indiacon

Learning objectives

After this lesson you should be able to:

Write and solve the equilibrium equations of a particle in a plane.
State and derive Lami's theorem and use it for three concurrent forces.
Choose between resolution of forces and Lami's theorem for a given problem.

Theory

A particle is in equilibrium when the vector sum of all forces on it is zero. In a plane this gives two scalar equations, ΣFx = 0 and ΣFy = 0, so at most two unknowns can be found. Typical civil engineering cases are a ring or hook where slings meet, a pin joint of a truss, and a weight hanging from two cables.

Graphically, forces in equilibrium form a closed polygon when drawn tip to tail. For exactly three forces the polygon is a triangle. That observation gives Lami's theorem: if three coplanar concurrent forces keep a particle in equilibrium, each force is proportional to the sine of the angle between the other two.

Figure 1.8: A weight W held by two cables at 30° and 45° to the horizontal, with Lami's relation between the three forces. Indiacon original, not to scale.

Lami's theorem is quick when the three angles between the forces are easy to find, but it applies only to exactly three forces, all concurrent and coplanar, and the angles must be the angles between the forces themselves (measured between their directions drawn away from the point), not the angles they make with some axis. With more than three forces, or when angles are awkward, resolution into components is safer.

A useful consequence: if three non-parallel forces hold a rigid body in equilibrium, their lines of action must meet at one point. This three-force principle helps locate an unknown reaction direction, for example at the pin of a ladder or a strut.

Derivation

Let P, Q and R act on a particle in equilibrium, and let α be the angle between Q and R, β between R and P, γ between P and Q.
Since P + Q + R = 0, the three vectors drawn tip to tail form a closed triangle.
The interior angle of the triangle opposite side P is 180° − α, opposite Q is 180° − β, opposite R is 180° − γ.
By the sine rule: P / sin(180° − α) = Q / sin(180° − β) = R / sin(180° − γ).
Since sin(180° − θ) = sin θ: P / sin α = Q / sin β = R / sin γ, which is Lami's theorem.

Formulas, symbols and SI units

Particle equilibrium (2D): ΣFx = 0, ΣFy = 0
Lami: P / sin α = Q / sin β = R / sin γ (α opposite P, i.e. between Q and R)
Two equal cables at angle θ each to the vertical carrying W: T = W / (2 cos θ)

Assumptions and limits

The body is a particle: all forces pass through one point.
Lami's theorem: exactly three coplanar, concurrent, non-parallel forces.
Cables and strings are weightless and carry tension only.

Codes and standards

Indian Standards are cited by number and clause only; no clause text, tables or figures are reproduced. Other standards are listed for comparison, not ranked.

No design code governs this lesson; it is pure mechanics. In lifting plans, sling leg forces computed this way are compared with the rated capacities stated by the sling manufacturer for the actual leg angle.

Worked examples

Example 1 (exam style)

Problem. A 2 kN weight hangs from two cables that make 30° and 45° with the horizontal. Find the cable tensions by resolution, then check with Lami's theorem.

Horizontal: T₁ cos 30° = T₂ cos 45°. Vertical: T₁ sin 30° + T₂ sin 45° = 2.
From the first, T₂ = T₁ cos 30° / cos 45°. Substituting: T₁ (sin 30° + cos 30° tan 45°) = 2, so T₁ = 2 / (0.5 + 0.8660) = 1.4641 kN.
T₂ = 1.4641 × 0.8660 / 0.7071 = 1.7932 kN.
Lami check: the angle between T₂ and W is 135°, between T₁ and W is 120°, between T₁ and T₂ is 105°. T₁ = 2 sin 135° / sin 105° = 1.4641 kN; T₂ = 2 sin 120° / sin 105° = 1.7932 kN. Both agree.

Answer. T₁ = 1.464 kN (30° cable); T₂ = 1.793 kN (45° cable)

Example 2 (site style)

Problem. A smooth sphere of weight 500 N and radius 0.2 m hangs against a smooth vertical wall from a string 0.3 m long attached to the wall. Find the string tension and the wall reaction.

The string, extended, passes through the sphere centre (three-force principle). From the attachment point to the centre is 0.3 + 0.2 = 0.5 m.
The angle θ of the string with the wall: sin θ = 0.2 / 0.5 = 0.4, so θ = 23.58° and cos θ = 0.9165.
Vertical: T cos θ = 500, so T = 500 / 0.9165 = 545.5 N.
Horizontal: N = T sin θ = 500 tan θ = 218.2 N.

Answer. T = 545.5 N; wall reaction 218.2 N

Step-by-step procedure

Isolate the particle and draw its FBD.
Count unknowns: no more than two can be solved for a particle in a plane.
For three forces with known angles between them, apply Lami directly; otherwise resolve.
Solve and check that all cable tensions come out positive.
Check with the other method or a force-triangle sketch.

Site notes

Sling tension grows quickly as the slings flatten: with two legs at 60° to the vertical each leg carries the full load. Lifting plans limit the included angle for this reason.
Rated capacities of chain and wire-rope slings are given for specific leg angles; recompute leg forces whenever the geometry on site differs from the plan.
Taglines used to steady a load add a third force and change the leg tensions; keep them light and nearly horizontal.

Common mistakes

Using the angle a force makes with an axis in Lami's formula instead of the angle between the other two forces.
Applying Lami to four forces or to non-concurrent forces.
Assuming the two cable tensions are equal when the angles are different.
Forgetting that a string can only pull: a negative tension means the assumed configuration is impossible.
Mixing degrees and radians on the calculator.

Quick check

For two symmetric slings at angle θ to the vertical, each carries W / (2 cos θ): 0.5 W at 0°, 0.58 W at 30°, 0.71 W at 45°, 1.0 W at 60°. Above 60° the leg force exceeds the load itself.

Summary

Particle equilibrium: ΣFx = 0, ΣFy = 0; two unknowns at most.
Three forces in equilibrium form a closed triangle.
Lami: P / sin α = Q / sin β = R / sin γ, angles between the other two forces.
Three non-parallel forces in equilibrium must be concurrent.

Practice questions

Tap a question to see the answer.

1A 100 N weight hangs from two strings, each at 60° to the vertical. Find the tension in each.

T = 100 / (2 cos 60°) = 100 N.

2Three equal forces keep a particle in equilibrium. What are the angles between them?

120° each.

3A 10 kN load hangs from a cable pulled aside by a horizontal force P until the cable is at 30° to the vertical. Find P and the cable tension.

P = 10 tan 30° = 5.774 kN; T = 10 / cos 30° = 11.547 kN.

4Two strings at right angles to each other support a 50 N weight; one string is at 30° to the vertical. Find both tensions.

T₁ = 50 cos 30° = 43.30 N (the 30° string); T₂ = 50 sin 30° = 25.0 N.

5A particle is held by forces of 50 kN at 0° and 30 kN at 120° and a third unknown force F. Find F.

Fx = −(50 + 30 cos 120°) = −35 kN; Fy = −30 sin 120° = −25.98 kN; F = √(35² + 25.98²) = 43.59 kN at 216.59° from the x-axis.

References

Standard texts, listed in no order of merit. Our explanations, examples and diagrams are original; nothing is copied from these books.

F. P. Beer, E. R. Johnston, D. F. Mazurek, P. J. Cornwell, B. P. Self, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, 12th edition, 2019.
R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, 15th edition, 2022 (published 2021).
J. L. Meriam, L. G. Kraige, J. N. Bolton, Engineering Mechanics: Statics, Wiley, 9th edition, 2018.
S. Timoshenko, D. H. Young, J. V. Rao, S. Pati, Engineering Mechanics (in SI units), McGraw Hill Education (India), 5th edition, 2013.