The resultant of two forces F1 and F2 is F. If F2 is reversed then the resultant is 2F. If F2 is doubled then also the resultant is also doubled. Find the ratio F1 : F2 : F .

F2=F12+F22+2F1F2cos⁡θ4F2=F12+F22−2F1F2cos⁡θIf F2 is doubled then resultant is 2F4F2=F12+4F22+4F1F2cos⁡θsolving the equations, we getF1:F2:F=2:3:2

The resultant of two forces F1 and F2 is F. If F2 is reversed then the resultant is 2F. If F2 is doubled then also the resultant is also doubled. Find the ratio F1 : F2 : F .

F2=F12+F22+2F1F2cos⁡θ4F2=F12+F22−2F1F2cos⁡θIf F2 is doubled then resultant is 2F4F2=F12+4F22+4F1F2cos⁡θsolving the equations, we getF1:F2:F=2:3:2

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The resultant of two forces F₁and F₂ is F. If F₂is reversed then the resultant is 2F. If F₂is doubled then also the resultant is also doubled. Find the ratio F₁: F₂ : F .

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$\sqrt{3} : \sqrt{2} : 1$

$\sqrt{2} : \sqrt{3} : \sqrt{2}$

$1 : \sqrt{2} : \sqrt{3}$

$\sqrt{3} : \sqrt{2} : \sqrt{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} F^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ \\ 4 F^{2} = F_{1}^{2} + F_{2}^{2} - 2 F_{1} F_{2} \cos θ \end{array}$
If F₂ is doubled then resultant is 2F
$4 F^{2} = F_{1}^{2} + 4 F_{2}^{2} + 4 F_{1} F_{2} \cos θ$
solving the equations, we get
$F_{1} : F_{2} : F = \sqrt{2} : \sqrt{3} : \sqrt{2}$