Question

The magnitudes of two forces X and Y are in the ratio
2: 3. If the resultant of two forces is 25 N and angle
between them is 60°, find the magnitude of each force,
close to one place of decimal.​

Answer 1

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

$\green{\tt{\therefore{x=18.2\:N}}}$

$\green{\tt{\therefore{y=12.1\:N}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Ratio \: of \: two \: forces = 2 : 3 \\ \\ \tt: \implies Resultant \: force(R) = 25 \: n \\ \\ \tt: \implies Angle \: between \: vectors = 60 \degree \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Magnitude \: of \: two \: forces =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \frac{ x}{y} = \frac{2}{3} \\ \\ \tt: \implies x = \frac{2y }{3} - - - - - (1) \\ \\ \bold{As \: we \: know \:that} \\ \tt: \implies R = \sqrt{ {a}^{2} + {b}^{2} + 2ab \: cos \theta } \\ \\ \tt: \implies 25 = \sqrt{ {x}^{2} + {y}^{2} + 2x \times \times y \: cos \: 60 \degree} \\ \\ \tt: \implies 25 = \sqrt{ (\frac{2y}{3})^{2} + {y}^{2} + 2 \times \frac{2y}{3} \times y \times cos \: 60 \degree } \\ \\ \tt: \implies {25}^{2} = \frac{4 {y}^{2} }{9} + {y}^{2} + \frac{4 {y}^{2} }{3} \times \frac{1}{2} \\ \\ \tt: \implies 625 = \frac{4 {y}^{2} + 9 {y}^{2} + 6{y}^{2} }{9} \\ \\ \tt: \implies 625 \times 9 = 17 {y}^{2} \\ \\ \tt: \implies \frac{625 \times 9}{17} = {y}^{2} \\ \\ \tt: \implies {y}^{2} = 330.9 \\ \\ \tt: \implies y = \sqrt{330.9} \\ \\ \green{\tt: \implies y = 18.2 \: N} \\ \\ \text{Putting \: value \: of \: y \: in \: (1)} \\ \tt: \implies x = \frac{2 \times 18.2}{3} \\ \\ \green{\tt: \implies x = 12.1 \: N}$