IIT JEE , AIEEE Forum - Mathematics , Vectors

neeraj_agarwal_1990

The forces P and Q have a resultant R and the resolved part of R in the direction of P is of magnitude Q.The angle between the forces is?

ans - $2{sin}^{-1}\sqrt{\frac{P}{2Q}}$

feynmann

simple

resolved part of resultant along any dirn.= sum of resolved part of component forces along this dirn.

This theorem gives

Q = P + Q cos $\alpha$

so 2Q sin^2 $\alpha/2$ = P

so $\alpha$ = $2 {sin}^{-1}( P/2Q)$

sriram.a

$R=\sqrt{P^2+Q^2+2PQcos}\theta$

Rcos $\alpha$ =lQl,cos $\alpha$ = $\frac{P+Qcos\theta}{\sqrt{P^2+Q^2+2PQcosP+Qcos\theta}}$

$P=Q(1-cos\theta)\Rightarrow$ $sin^2{\theta/2}=\frac{P}{Q}$

$\theta=2\sqrt{\frac{P}{2Q}}$

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