Two forces, while acting on a particle in opposite directions, have the resultant of 10 N. If they act at right angles to each other, the resultant is found to be 50 N. Find the two forces.

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$B y p a r a l l e \log r a m l a w o f v e c t o r a d d i t i o n . F_{n e t_{1}} = \sqrt{{F_{1}}^{2} + {F_{2}}^{2} + 2 F_{1} F_{2} C o s θ} 10 = \sqrt{{F_{1}}^{2} + {F_{2}}^{2} + 2 F_{1} F_{2} C o s 180} 10 = \sqrt{{F_{1}}^{2} + {F_{2}}^{2} - 2 F_{1} F_{2}} 10 = \sqrt{{(F_{1} - F_{2})}^{2}} F_{1} - F_{2} = 10 F_{n e t_{2}} = \sqrt{{F_{1}}^{2} + {F_{2}}^{2} + 2 F_{1} F_{2} C o s θ} 50 = \sqrt{{F_{1}}^{2} + {F_{2}}^{2} + 2 F_{1} F_{2} C o s 90} 50 = \sqrt{{F_{1}}^{2} + {F_{2}}^{2} + 0} 2500 = {F_{1}}^{2} + {F_{2}}^{2} 2500 = {(10 + F_{2})}^{2} + {F_{2}}^{2} 2500 = 2 {F_{2}}^{2} + 100 + 20 F_{2} {F_{2}}^{2} + 10 F_{2} - 1200 = 0 {F_{2}}^{2} - 30 F_{2} + 40 F_{2} - 1200 = 0 F_{2} (F_{2} - 30) + 40 (F_{2} - 30) = 0 (F_{2} - 30) (F_{1} + 40) = 0 F_{2} = 30 N F_{1} = 30 + 10 = 40 N$

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