Three forces of magnitude 30N, 60N and PN that are acting at a point are in equilibrium. If the angle between the first two is 60o60o , then what is the value of P?

 Three forces of magnitude 30N, 60N and PN that are acting at a point are in equilibrium. If the angle between the first two is 

60o${60}^o$, then what is the value of P?

SOLUTION

*A2A

⋆$\star$ Use Lami's theorm

⋆$\star$ The angles opposite toP$P$, 30N$30N$ and 60N$60N$ force will be, 60o${60}^o$, θ$\theta$ and (300−θ)$(300-\theta )$ respectively. So, now we have :-

⟹Psin60o=30sin(300−θ)=60sinθ$⟹\boxed{{\displaystyle {\displaystyle \frac{P}{\sin {60}^o}}={\displaystyle \frac{30}{\sin (300-\theta )}}={\displaystyle \frac{60}{\sin \theta }}}}$

⋆$\star$ Now just solve for P$P$ and θ$\theta$ :-

⟹30sin(300−θ)=60sinθ$⟹{\displaystyle \frac{30}{\sin (300-\theta )}}={\displaystyle \frac{60}{\sin \theta }}$

⟹2sin(300−θ)=sinθ$⟹2\sin (300-\theta )=\sin \theta$

⟹2sinθ=−3–√cosθ$⟹2\sin \theta =-\sqrt{3}\cos \theta$

⟹tanθ=−3–√2$⟹\tan \theta =-{\displaystyle \frac{\sqrt{3}}{2}}$

⟹sinθ=37−−√$⟹\boxed{{\displaystyle \sin \theta =\sqrt{{\displaystyle \frac{3}{7}}}}}$

Now,

⟹Psin60o=60sinθ$⟹{\displaystyle \frac{P}{\sin {60}^o}}={\displaystyle \frac{60}{\sin \theta }}$

⟹P=307–√N