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(B) The resultant $R$ of two vectors $A$ and $B$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) The resultant $R$ of two vectors $A$ and $B$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given that the magnitudes of the two forces are $A = F$ and $B = F$,and their resultant magnitude is $R = F$.Substituting these values into the formula: $F = \sqrt{F^2 + F^2 + 2(F)(F) \cos 	heta}$.Squaring both sides: $F^2 = F^2 + F^2 + 2F^2 \cos 	heta$.$F^2 = 2F^2 + 2F^2 \cos 	heta$.Dividing by $F^2$: $1 = 2 + 2 \cos 	heta$.$-1 = 2 \cos 	heta$.$\cos 	heta = -\frac{1}{2}$.Since $\cos 	heta = -\frac{1}{2}$,the angle $	heta = 120^\circ$.

Two forces,each of magnitude $F$,have a resultant of the same magnitude $F$. The angle between the two forces is....... $^o$

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