Question

two forces are in the ratio √5:√7 acting at angle of 2π have their resultant equal to 1453 N. Find the two forces​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{F_{1}=\frac{1453\sqrt{5}}{\sqrt{5}+\sqrt{7}}\:N}}}$

$\green{\tt{\therefore{F_{2}=\frac{1453\sqrt{7}}{\sqrt{5}+\sqrt{7}}\:N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Ratio \: of \: forces = \sqrt{5} : \sqrt{7} \\ \\ \tt: \implies Angle \: between \: vectors = 2\pi \\ \\ \tt: \implies resultant \: vector = 1453 \: N \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies forces =?$

• According to given question :

$\tt \circ \: \frac{ F_{1} }{ F_{2}} = \frac{ \sqrt{5} }{ \sqrt{7} } \\ \\ \tt:\implies \theta = 2\pi \\ \\ \tt \circ \: Angle \: between \: two \: vector \: is \: 2\pi \: it \: means \\ \\ \tt \: \: \: that \:both \: vector \: acting \: in \: same \: direction \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies F_{1} + F_{2} = 1453\:\:\:\:(Same\:direction) \\ \\ \tt: \implies \sqrt{5}x + \sqrt{7}x = 1453 \\ \\ \tt: \implies x( \sqrt{5} + \sqrt{7} ) = 1453 \\ \\ \green{\tt: \implies x = \frac{1453}{ \sqrt{5} + \sqrt{7} } } \\ \\ \green{\tt \therefore F_{1} = \sqrt{5} \times \frac{1453}{ \sqrt{5} + \sqrt{7} } = \frac{1453 \sqrt{5} }{ \sqrt{5} + \sqrt{7} }\:N } \\ \\ \green{\tt \therefore F_{2} = \sqrt{7} \times \frac{1453}{ \sqrt{5} + \sqrt{7} } = \frac{1453 \sqrt{7} }{ \sqrt{5} + \sqrt{7} }\:N}$