Question

Two equal forces have the square of the resultant equal two three times their product. find the angle between them.

Answer 1

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Answer 2

Answer:  

The angle between the two forces is $\bold{60^{\circ}}$.

Solution:

Let the two forces be $\mathrm{F}_{1} \text { and } \mathrm{F}_{2}$ and the resultant is represented as R. As it is stated that both the forces are equal to each other, then $\mathrm{F}_{1}=\mathrm{F}_{2}=\mathrm{F}$. It is also stated that resultant $\mathrm{R}=3\left(\mathrm{F}_{1} \times \mathrm{F}_{2}\right)$. So if we put all the conditions together we will get the following equations.

It is given that

$R=3\left(F_{1} \times F_{2}\right) \rightarrow(1)$

And

$F_{1}=F_{2}=F \rightarrow(2)$

If we substitute the condition (2) in (1), we will get

$R=3 F^{2} \rightarrow(3)$

We know that the resultant R original formula is  

$R=\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta}$

On squaring on both sides, we get

$R^{2}=F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta \rightarrow(4)$

Substitute the condition of (2) and (3) in (4), we get

$3 F^{2}=F^{2}+F^{2}+2 F^{2} \cos \theta$

$3 F^{2}=2 F^{2}+2 F^{2} \cos \theta$

$F^{2}=2 F^{2} \cos \theta$

$\cos \theta=\frac{1}{2}$

$\theta=\cos ^{-1}\left(\frac{1}{2}\right)$

Thus,

$\theta=60^{\circ}$

The angle between the two forces is $\bold{60^{\circ}}$.