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(B) The resultant $R$ of two forces $F_1$ and $F_2$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \c

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) The resultant $R$ of two forces $F_1$ and $F_2$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos 	heta}$.Given $F_1 = F$,$F_2 = F$,and the resultant $R = F$.Substituting these values into the formula: $F = \sqrt{F^2 + F^2 + 2F^2 \cos 	heta}$.Squaring both sides: $F^2 = 2F^2 + 2F^2 \cos 	heta$.Dividing by $F^2$: $1 = 2 + 2 \cos 	heta$.Rearranging the terms: $2 \cos 	heta = 1 - 2 = -1$.Thus,$\cos 	heta = -1/2$.Since $\cos(120^\circ) = -1/2$,the angle between the two forces is $	heta = 120^\circ$.

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

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