Question

Two forces, each of magnitude F have a resultant of the same magnitude F. The angle between the two forces is
459
1200
150°
60°​

Answer 1

Answer:

Ans is 120⁰.

Force is vector quantity . Resultant of two vectors A and B is given by

R = √{A² + B² + 2A.BcosФ} , where Ф is angle between A and B

Now, magnitude of each force is F and resutant is also F

Then, F = √{F² + F² + 2.F.FcosФ}

⇒F² = F² + F² + 2F².cosФ

⇒- F² = 2F²cosФ

⇒ -1/2 = cosФ

⇒cosФ = -1/2 = cos120°

⇒Ф = 120°

Hence angle between them is 120°

Answer 2

${ \bold{ \huge{ \underline{ \underline{ \red{Question:-}}}}}}$



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



${ \huge{ \bold{ \underline{ \underline{ \red{ Solution:-}}}}}}$



${ \bold{ \purple{GIVEN- }}}$



▪ magnitude of two forces = equal = F



▪magnitude of the Resultant force = F



${ \bold{ \purple{TO \: FIND- }}}$



▪ The angle between the two forces???



${ \bold{ \underline{ \pink{resultant \: vector}}}}$



${ \boxed{ \bold{ \red{R = \sqrt{ {A}^{2} + {B}^{2} + 2AB \cos( \alpha )}}}} }$



here,


R = magnitude of Resultant vector



A = magnitude of vector 1


B = magnitude of vector 2



a = alpha= angle between the two vectors


▪ substituting the information given in the formula....



${ \bold{F = \sqrt{ {F}^{2} + {F}^{2} + 2F\times F \cos( \alpha ) } }}$



${ \bold{ \implies{ {F}^{2} = 2 {F}^{2} + 2 {F}^{2} \cos( \alpha ) }}}$



${ \bold{ \implies{ {F}^{2} - 2 {F}^{2} = 2 {F}^{2} \cos( \alpha ) }}}$



${ \bold{ \implies{2 {F}^{2} \cos( \alpha ) = - {F}^{2} }}}$



${ \bold{ \implies{2 \cos( \alpha ) = - 1}}}$



${ \bold{ \implies{ \cos( \alpha ) = - \frac{1}{2} }}}$



▪ we know that....



${ \underline{ \pink { \bold{ \cos(120) = - \frac{1}{2}}}}}$



${ \underline{ \huge{ \boxed{ \bold{ \red{ \alpha = 120}}}}}}$