Two forces,each numerically equal to 10 dyne, | vedclass

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(A) To find the resultant of two vectors,we must use the angle between them when they are placed tail-to-tail.In the given figure,the angle between th

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$10$

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(A) To find the resultant of two vectors,we must use the angle between them when they are placed tail-to-tail.In the given figure,the angle between the two vectors is $120^{\circ}$.The magnitude of the resultant $R$ is given by the formula:$R = \sqrt{P^2 + Q^2 + 2PQ \cos 	heta}$Here,$P = 10$ $dyne$,$Q = 10$ $dyne$,and $	heta = 120^{\circ}$.Substituting these values:$R = \sqrt{(10)^2 + (10)^2 + 2(10)(10) \cos 120^{\circ}}$Since $\cos 120^{\circ} = -1/2$:$R = \sqrt{100 + 100 + 200(-1/2)}$$R = \sqrt{100 + 100 - 100}$$R = \sqrt{100} = 10$ $dyne$.

Two forces,each numerically equal to $10$ $dyne$,are acting as shown in the adjoining figure. The magnitude of the resultant force is ......... $dyne$.

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