Two forces of magnitude P and Q acting at a p | vedclass

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Question

(B) Let the angle between the forces $P$ and $Q$ be $	heta$.The resultant $R$ is given by the vector sum $\vec{R} = \vec{P} + \vec{Q}$.The resolved p

Vedclass · Accepted Answer

$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$

Vedclass · Answer

(B) Let the angle between the forces $P$ and $Q$ be $	heta$.The resultant $R$ is given by the vector sum $\vec{R} = \vec{P} + \vec{Q}$.The resolved part of $R$ in the direction of $P$ is given by $P + Q \cos 	heta$.According to the problem,this resolved part is equal to $Q$.Therefore,$P + Q \cos 	heta = Q$.Rearranging the terms,we get $Q \cos 	heta = Q - P$.Thus,$\cos 	heta = \frac{Q - P}{Q} = 1 - \frac{P}{Q}$.Using the trigonometric identity $\cos 	heta = 1 - 2 \sin^2 \left(\frac{	heta}{2}ight)$,we have:$1 - 2 \sin^2 \left(\frac{	heta}{2}ight) = 1 - \frac{P}{Q}$.Subtracting $1$ from both sides,we get $-2 \sin^2 \left(\frac{	heta}{2}ight) = -\frac{P}{Q}$.Therefore,$\sin^2 \left(\frac{	heta}{2}ight) = \frac{P}{2Q}$.Taking the square root,we get $\sin \left(\frac{	heta}{2}ight) = \left(\frac{P}{2Q}ight)^{1/2}$.Finally,the angle $	heta = 2 \sin^{-1} \left(\frac{P}{2Q}ight)^{1/2}$.

Two forces of magnitude $P$ and $Q$ acting at a point have a resultant $R$. The resolved part of $R$ in the direction of $P$ is of magnitude $Q$. The angle between the forces is:

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