Question

The greatest and least magnitude of the resultant of two forces of constant magnitude is$F$and $G$. When the forces act at an angle$2\alpha$, the resultant in magnitudes is equal to

• A. $\sqrt{F^2{\cos }^2\alpha +G^2{\sin }^2\alpha }$
• B. $\sqrt{F^2{\sin }^2\alpha +G^2{\cos }^2\alpha }$
• C. $\sqrt{F^2+G^2}$
• D. $\sqrt{F^2-G^2}$

Answer 1

The correct option is A

$\sqrt{F^2{\cos }^2\alpha +G^2{\sin }^2\alpha }$

Step 1: Given that:

Draw the required diagram,

The greatest and least magnitude of the resultant of two forces of constant magnitude is$F$and $G$.

An angle$=2\alpha$

Step 2: Formula used:

Suppose that the forces are A and B, resultant will be defined as-

$\overrightarrow{R}=\sqrt{A^2+B^2+2AB\cos \theta }$

The maximum resultant will be when $\theta =0\Rightarrow \cos 0=1$

$$\begin{gathered} \overrightarrow{F}=\sqrt{A^2+B^2+2AB} \\ F=A+B \end{gathered}$$

The least resultant will be when $\theta =180°\Rightarrow \cos 180=-1$

$\begin{array}{rcl}\overrightarrow{G}& =& \sqrt{A^2+B^2-2AB}\\ & =& A-B\end{array}$

Step 3: Finding the relationship between $F$ and $G$

The greatest resultant $F=A+B\dots ..\left(1\right)$

The least resultant $G=A-B\dots \dots .\left(2\right)$

By adding equations (1) and (2) we get-

$A=\frac{\left[F+G\right]}{2}$

On subtracting equation (1) to equation (2) we get-

$B=\frac{\left[F-G\right]}{2}$

Where A and B act an angle $2\alpha$

Therefore the resultants,

$\overrightarrow{R}=\sqrt{A^2+B^2+2AB\cos \theta }$……(3)

Step 4: Calculation of resultant magnitude when the forces act at an angle $2\alpha$

Now substitute the value of A and B in the equation (3)

Therefore,

$\begin{array}{rcl}R& =& \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 }\\ \overrightarrow{R}& =& \sqrt{\frac{F^2+G^2+2FG+F^2+G^2-2FG}{4}+\frac{1}{2}\left(F^2-G^2\right)\cos 2\alpha }\\ & =& \sqrt{\frac{F^2+G^2}{2}+\frac{1}{2}\left(F^2-G^2\right)\cos 2\alpha }\\ & =& \sqrt{{\frac{F}{2}}^2\left(1+\cos 2\alpha \right)+{\frac{G}{2}}^2\left(1-\cos 2\alpha \right)}\\ & =& \sqrt{F^2{\cos }^2\alpha +G^2{\sin }^2\alpha }\end{array}$$$\begin{gathered} \{\cos 2\theta =2{\cos }^2\theta -1 \\ \cos 2\theta =1-2{\sin }^2\theta \} \end{gathered}$$

Therefore the resultant magnitude is$\sqrt{F^2{\cos }^2\alpha +G^2{\sin }^2\alpha }$.

Hence, option A is the correct answer.