Two forces P + Q and P - Q make an angle 2 al | vedclass

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Question

(A) Let the two forces be $F_1 = P + Q$ and $F_2 = P - Q$. The angle between them is $2 \alpha$. The bisector divides this into two angles of $\alpha$

Vedclass · Accepted Answer

$P 	an 	heta = Q 	an \alpha$

Vedclass · Answer

(A) Let the two forces be $F_1 = P + Q$ and $F_2 = P - Q$. The angle between them is $2 \alpha$. The bisector divides this into two angles of $\alpha$ each.If the resultant makes an angle $	heta$ with the bisector,then the angle between the resultant and $F_1$ is $(\alpha - 	heta)$ and the angle between the resultant and $F_2$ is $(\alpha + 	heta)$.Resolving the forces perpendicular to the resultant,the components must cancel out:$(P + Q) \sin (\alpha - 	heta) = (P - Q) \sin (\alpha + 	heta)$Expanding the terms:$P \sin (\alpha - 	heta) + Q \sin (\alpha - 	heta) = P \sin (\alpha + 	heta) - Q \sin (\alpha + 	heta)$Rearranging terms to group $P$ and $Q$:$Q [\sin (\alpha + 	heta) + \sin (\alpha - 	heta)] = P [\sin (\alpha + 	heta) - \sin (\alpha - 	heta)]$Using trigonometric identities $\sin(A+B) + \sin(A-B) = 2 \sin A \cos B$ and $\sin(A+B) - \sin(A-B) = 2 \cos A \sin B$:$Q [2 \sin \alpha \cos 	heta] = P [2 \cos \alpha \sin 	heta]$Dividing both sides by $2 \cos \alpha \cos 	heta$:$Q 	an \alpha = P 	an 	heta$Thus,$P 	an 	heta = Q 	an \alpha$.

Two forces $P + Q$ and $P - Q$ make an angle $2 \alpha$ with each other,and their resultant makes an angle $\theta$ with the bisector of the angle between them. Then:

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