Two forces whose magnitudes are in the ratio | vedclass

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(C) Let the two forces be $F_1 = 5x$ and $F_2 = 3x$.Given the resultant $R = 35 \ N$ and the angle $	heta = 60^{\circ}$.The formula for the resultant

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$25 \ N, 15 \ N$

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(C) Let the two forces be $F_1 = 5x$ and $F_2 = 3x$.Given the resultant $R = 35 \ N$ and the angle $	heta = 60^{\circ}$.The formula for the resultant of two vectors is $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos 	heta}$.Substituting the values: $35 = \sqrt{(5x)^2 + (3x)^2 + 2(5x)(3x) \cos 60^{\circ}}$.Since $\cos 60^{\circ} = 0.5$,we have $35 = \sqrt{25x^2 + 9x^2 + 30x^2(0.5)}$.$35 = \sqrt{25x^2 + 9x^2 + 15x^2} = \sqrt{49x^2} = 7x$.Solving for $x$: $x = 35 / 7 = 5$.Therefore,$F_1 = 5 	imes 5 = 25 \ N$ and $F_2 = 3 	imes 5 = 15 \ N$.

Two forces whose magnitudes are in the ratio $5:3$ are acting at a point at an angle $60^{\circ}$ simultaneously. If the resultant of the two forces is $35 \ N$,then the magnitudes of the two forces respectively are

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