Lami's Theorem - Statement, Formula, Proof, Problems

By Aina Parasher|Updated : August 3rd, 2022

Lami's Theorem associates the magnitudes of Coplanar, Concurrent and Non-Collinear forces that maintain an object in equilibrium. It can be used to solve a lot of real-life problems. Lami’s Theorem is named after Bernard Lamy, a French Mathematician.

Lami’s Theorem comes in handy in analyzing most mechanical and structural systems. In this article, we will discuss in detail Lami’s Theorem Proof, Formula, its derivation, use, and some examples.

State Lami’s Theorem

Lami’s Theorem states that if three coplanar forces acting at a point be in equilibrium, then each force is proportional to the sine of the angle between the other two. According to Lami's theorem, if forces are in equilibrium, as shown in the image, then they will be related to each other with the sin of the angle formed between them.

Lami's Theorem

where P, Q & R are the Coplanar forces acting at a point.

Lami’s Theorem Formula

Lami’s Theorem is related to magnitudes of concurrent, coplanar, and non-collinear forces that help to keep a body or an object in static equilibrium. The theorem is very useful in analyzing most mechanical and structural systems. Lami's theorem formula for three forces is:

P/sinα = Q/sinβ = R/sinγ

Where, P, Q, R are the Coplanar Forces acting at a point.

Lami’s Theorem Proof

Now, let us see how we can state and prove Lami’s Theorem. This theorem can be obtained by applying the sine rule for triangles. Consider three Coplanar Forces P, Q, and R in equilibrium acting at a point ‘O’ as shown. Now, refer to the figure below.

Lami's Theorem

  • Complete the parallelogram OACB with OA and OB as shown.
  • Diagonal OC represents the resultant of two forces, P and Q.

Since these forces are in equilibrium, the resultant of the forces P and Q must be in line with OD and equal to R, but in the opposite direction.

OA = P

OB = Q

OC = R

Now, α + β + γ= 360°

Angle AOC = 180° – β

Angle COB = 180° – α

Angle ACO = 180° – α (As AC is parallel to OB)

In Triangle AOC,

∠A + ∠B + ∠C = 180°

∠A + (180°-β) + (180°-α) =180°

∠A = α + β -180° ……………………………. (i)

We know that,

α + β + γ= 360°

α + β = 360° – γ

Now, in equation (i), replacing α + β by 360° –γ, we get,

∠A = 360° – γ-180°

∠A = 180° – γ

Lami's Theorem

Consider Triangle AOC,

Applying Sine Rule, we get,

Lami's Theorem

Limitations of Lami’s Theorem

Lami’s Theorem, though very useful, has its own limitations. As we all know, this theorem has been derived from Sine Rule for triangles; all the limitations of the Sine Rule are automatically liable to this theorem. A few limitations of Lami’s Theorem are listed below:

  • It is applicable only if the body is in the State of Equilibrium.
  • It is not applicable for more or less than Three Concurrent Forces.
  • It is not applicable for Parallel or Non- Concurrent Force systems.
  • It is not applicable for Non- Coplanar Forces.

Lami’s Theorem Questions

Let us see a few Lami's theorem questions to understand how to apply Lami's theorem in problems:

  1. A vehicle is pulled by means of two ropes, as shown in the figure below. If the resultant pull is 1200N, find the angle and the force F. (Ans - angle = 30°, F = 1.98 kN)
    • Lami's Theorem
  2. If the resultant of Force F and 70N is R = 50kN, as shown in the figure below, find force ‘F’. (Ans – 80N)

Lami's Theorem

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FAQs on Lami's Theorem

  • Lami’s Theorem is derived on the basis of the Sine rule in a triangle that relates angles and opposite sides to those angles. Therefore, it is not applicable for systems having a force less than or greater than 3.

  • Lami’s Theorem states that “If three coplanar forces acting at a point be in equilibrium, then each force is proportional to the sine of the angle between the other two.”

  • ∠AOB = 180°– (60°+ 40°) = 80°

    According to Lami’s Theorem,

    500/ sin sin 80° = TOA/ sin sin 150°

    TOA = 253.85 ≅254N

  • Tension in String 1 -

    T/sin sin 150 = 5/ sin sin 70

    T=2.87 kN

    Tension in String 2 -

    T/ sin sin 140 = 5/ sin sin 70

    T=3.42 kN

  • According to Lami’s Theorem:

    R1/ sin sin 135 = R2/ sin sin 120 = 600/ sin sin 105

    Reaction 1 –

    R1/ sin sin 135 = 600/ sin sin 105

    R1 = 439.2≅440 N

    Reaction 2 –

    R2/ sin sin 120 = 600/ sin sin 105

    R1 = 537.9≅538 N

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