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. $(F^2\sin 2\alpha +G^2\cos 2\alpha {)}^{\frac{1}{2}}$

Answer 1

The correct option is A $(F^2{\cos }^2\alpha +G^2{\sin }^2\alpha {)}^{\frac{1}{2}}$

Let the two forces be $\frac{F+G}{2}$ and $\frac{F-G}{2}$
Therefore resultant of these forces
$=\sqrt{{\left(\frac{F+G}{2}\right)}^2+{\left(\frac{F-G}{2}\right)}^2+2\left(\frac{F+G}{2}\right)\left(\frac{F-G}{2}\right)\cos 2\alpha }$
$=\sqrt{\left(\frac{F^2+G^2+2FG}{4}\right)+\left(\frac{F^2+G^2-2FG}{4}\right)+\left(\frac{2F^2-2G^2}{4}\right)\cos 2\alpha }$
$=\sqrt{\left(\frac{F^2+G^2}{2}\right)({\cos }^2\alpha +{\sin }^2\alpha )+\left(\frac{F^2-G^2}{2}\right)({\cos }^2\alpha -{\sin }^2\alpha )}$
$=\sqrt{(F^2{\cos }^2\alpha +G^2{\sin }^2\alpha )}$