The resultant of two forces 3P and 2P is R. I | vedclass

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(B) Let the angle between the two forces be $	heta$.Using the law of parallelogram of vectors,the resultant $R$ is given by:$R^2 = (3P)^2 + (2P)^2 +

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$120$

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(B) Let the angle between the two forces be $	heta$.Using the law of parallelogram of vectors,the resultant $R$ is given by:$R^2 = (3P)^2 + (2P)^2 + 2(3P)(2P) \cos 	heta$$R^2 = 9P^2 + 4P^2 + 12P^2 \cos 	heta = 13P^2 + 12P^2 \cos 	heta$ ... $(i)$When the first force is doubled,it becomes $6P$. The new resultant is $2R$:$(2R)^2 = (6P)^2 + (2P)^2 + 2(6P)(2P) \cos 	heta$$4R^2 = 36P^2 + 4P^2 + 24P^2 \cos 	heta = 40P^2 + 24P^2 \cos 	heta$ ... (ii)Divide equation (ii) by $4$:$R^2 = 10P^2 + 6P^2 \cos 	heta$ ... (iii)Equating $(i)$ and (iii):$13P^2 + 12P^2 \cos 	heta = 10P^2 + 6P^2 \cos 	heta$$3P^2 = -6P^2 \cos 	heta$$\cos 	heta = -3/6 = -1/2$Therefore,$	heta = 120^\circ$.

The resultant of two forces $3P$ and $2P$ is $R$. If the first force is doubled,then the resultant is also doubled. The angle between the two forces is ........... $^o$.

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