Two forces P and Q,of magnitude 2F and 3F,res | vedclass

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(A) Let the magnitudes be $P = 2F$ and $Q = 3F$. The resultant $R_1$ of two vectors $\vec{P}$ and $\vec{Q}$ is given by $R_1^2 = P^2 + Q^2 + 2PQ \cos

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(A) Let the magnitudes be $P = 2F$ and $Q = 3F$. The resultant $R_1$ of two vectors $\vec{P}$ and $\vec{Q}$ is given by $R_1^2 = P^2 + Q^2 + 2PQ \cos 	heta$.Substituting the values: $R_1^2 = (2F)^2 + (3F)^2 + 2(2F)(3F) \cos 	heta = 4F^2 + 9F^2 + 12F^2 \cos 	heta = F^2(13 + 12 \cos 	heta)$.When force $Q$ is doubled,the new force is $Q' = 2Q = 6F$. The new resultant $R_2 = 2R_1$,so $R_2^2 = 4R_1^2$.The new resultant $R_2$ is given by $R_2^2 = P^2 + (Q')^2 + 2P(Q') \cos 	heta$.Substituting the values: $R_2^2 = (2F)^2 + (6F)^2 + 2(2F)(6F) \cos 	heta = 4F^2 + 36F^2 + 24F^2 \cos 	heta = F^2(40 + 24 \cos 	heta)$.Equating $R_2^2 = 4R_1^2$: $F^2(40 + 24 \cos 	heta) = 4 	imes F^2(13 + 12 \cos 	heta)$.Dividing by $4F^2$: $10 + 6 \cos 	heta = 13 + 12 \cos 	heta$.$-3 = 6 \cos 	heta \Rightarrow \cos 	heta = -1/2$.Therefore,$	heta = 120^o$.

Two forces $P$ and $Q$,of magnitude $2F$ and $3F$,respectively,are at an angle $\theta$ with each other. If the force $Q$ is doubled,then their resultant also gets doubled. Then,the angle $\theta$ is ....... $^o$

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