Two forces $F_{1}$ acting om a body have a resultant $R_{1}$ is doubled the new resultant is $R_{2}$ forms are right angle with $F_{1}$ . prove that $R_{1}$ and $R_{2}$ have the same magnitude.

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Solution

Let $a$ be the angle between $F_{1}$ and $F_{2}$
$F_{1} + F_{2} = r_{1}$ -------------------- $(1)$
$F_{1} + 2 F_{2} = r_{2}$ ------------------ $(2)$
$R_{1} \times R_{1} = F_{1} \times F_{1} + F_{1} \times F_{2} + F_{2} \times F_{1} + F_{2} + F_{2}$
$= {R_{1}}^{2} = {F_{1}}^{2} + {F_{2}}^{2} + 2 F_{1} \times F_{2} \times a$ --------------
- $(3)$
$F_{1} \times (F_{1} + 2 F_{2}) = F_{1} \times R_{2}$ --------------- $(4)$
Since $F_{1}$ and $R_{2}$ perpendicular $F_{1} \times R_{2} = 0.$
${F_{1}}^{2} + 2 F_{1} \times a = 0.$
by equation $(3)$ and $(4)$
${R_{1}}^{2} = {F_{1}}^{2} + 2 F_{1} \times F_{2} a + {F_{2}}^{2} = 0 + {F_{2}}^{2}$
$R_{1} = F_{2}$

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Two forces F1 acting om a body have a resultant R1 is doubled the new resultant is R2 forms are right angle with F1. prove that R1 and R2 have the same magnitude.

Two forces $F_{1}$ acting om a body have a resultant $R_{1}$ is doubled the new resultant is $R_{2}$ forms are right angle with $F_{1}$ . prove that $R_{1}$ and $R_{2}$ have the same magnitude.