Three forces vec{P}, vec{Q}, and vec{R} act a | vedclass

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(D) For a system of three forces in equilibrium,Lami's theorem states that $\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$,

Vedclass · Accepted Answer

$\sqrt{3} : 2 : 1$

Vedclass · Answer

(D) For a system of three forces in equilibrium,Lami's theorem states that $\frac{P}{\sin \alpha} = \frac{Q}{\sin \beta} = \frac{R}{\sin \gamma}$,where $\alpha, \beta, \gamma$ are the angles opposite to the forces $P, Q, R$ respectively.
Let $	heta_{PQ} = 150^{\circ}$ be the angle between $\vec{P}$ and $\vec{Q}$.
Let $	heta_{QR} = 120^{\circ}$ be the angle between $\vec{Q}$ and $\vec{R}$.
The angle between $\vec{R}$ and $\vec{P}$ is $	heta_{RP} = 360^{\circ} - (150^{\circ} + 120^{\circ}) = 90^{\circ}$.
The angles opposite to the forces are:
$\alpha = 	heta_{QR} = 120^{\circ}$
$\beta = 	heta_{RP} = 90^{\circ}$
$\gamma = 	heta_{PQ} = 150^{\circ}$
Applying Lami's theorem:
$P : Q : R = \sin(120^{\circ}) : \sin(90^{\circ}) : \sin(150^{\circ})$
$P : Q : R = \frac{\sqrt{3}}{2} : 1 : \frac{1}{2}$
Multiplying by $2$,we get:
$P : Q : R = \sqrt{3} : 2 : 1$.

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