Closed Triangle of Forces in Equilibrium
When three coplanar, concurrent, non‑collinear forces act on a body in static equilibrium, Lami’s Theorem may be used.
Lami’s Theorem
\[
\frac{A}{\sin \alpha}
=
\frac{B}{\sin \beta}
=
\frac{C}{\sin \gamma}
\]
Derivation of Lami’s Theorem
Using geometric properties:
giving alternate angle
then
Putting everything together:
so
which is the vector addition B + A + C
This also represents the vector addition:
\[
\vec{A} + \vec{B} + \vec{C} = 0
\]
Applying the Sine Rule
\[
\frac{a}{\sin\alpha}
=
\frac{b}{\sin\beta}
=
\frac{c}{\sin\gamma}
\]
\[
\frac{A}{\sin(180^\circ - \alpha)}
=
\frac{B}{\sin(180^\circ - \beta)}
=
\frac{C}{\sin(180^\circ - \gamma)}
\]
But:
\[ \sin(180^\circ - \theta) = \sin\theta \]
\[
\alpha + \beta + \gamma = 180^\circ
\]
\[ \frac{A}{\sin\alpha} = \frac{B}{\sin\beta} = \frac{C}{\sin\gamma} \]
Example
A baby suspended from a doorway jumper is in static equilibrium. Calculate forces \(x\) and \(y\).
\[
\frac{A}{\sin\alpha}
=
\frac{B}{\sin\beta}
=
\frac{C}{\sin\gamma}
\]
\[
\frac{x}{\sin120^\circ}
=
\frac{7.6}{\sin100^\circ}
\]
\[
x
=
\frac{7.6}{\sin100^\circ}
\;\sin120^\circ
\]
\[
x = 6.683\ \text{N}
\]
\[
\frac{A}{\sin\alpha}
=
\frac{B}{\sin\beta}
=
\frac{C}{\sin\gamma}
\]
\[
\frac{y}{\sin140^\circ}
=
\frac{7.6}{\sin100^\circ}
\]
\[
y
=
\frac{7.6}{\sin100^\circ}
\;\sin140^\circ
\]
\[
y = 4.960\ \text{N}
\]
