Question

The maximum and minimum of the resultant of two forces are $F$ and $G$ (the angle between these forces is $2\alpha$), Then, the resultant of $F$ and $G$ is

• A. $(F^2{\cos }^2\alpha +G^2{\sin }^2\alpha {)}^{\frac{1}{2}}$
• B. $(F^2{\sin }^2\alpha +G^2{\cos }^2\alpha {)}^{\frac{1}{2}}$
• C. $(F^2\sin \alpha +G^2{\cos }^2\alpha {)}^{\frac{1}{2}}$
• D. $(F^2\sin \alpha +G^2\cos 2\alpha {)}^{\frac{1}{2}}$

Answer 1

Answer: (A)

Let two forces be $F_1$ and $F_2$.
Given, $F$ and $G$ are maximum and minimum
resultant of two forces $F_1$ and $F_2$ respectively and angle between two forces $F_1$ and $F_2$ is $2\alpha$.
$\therefore F=F_1+F_2$ and $G=F_1-F_2$
$\Rightarrow F_1=\frac{F+G}{2}$
and $F_2=\frac{F-G}{2}$
Resultant
$R=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos 2\alpha }$
$=\sqrt{{\left(\frac{F+G}{2}\right)}^2+{\left(\frac{F-G}{2}\right)}^2+2\left(\frac{F+G}{2}\right)\left(\frac{F-G}{2}\right)\cos 2\alpha }$
$=\sqrt{\frac{1}{4}\left[2\left(F^2+G^2\right)+2\left(F^2-G^2\right)\cos 2\alpha \right]}$
$=\sqrt{\frac{1}{2}\left[F^2(1+\cos 2\alpha )+G^2(1-\cos 2\alpha )\right]}$
$=\sqrt{\frac{1}{2}\left[F^2\left(2{\cos }^2\alpha \right)+G^2\left(2{\sin }^2\alpha \right)\right]}$
$=\sqrt{F^2{\cos }^2\alpha +G^2{\sin }^2\alpha }$
$={\left(F^2{\cos }^2\alpha +G^2{\sin }^2\alpha \right)}^{1/2}$