Question

Two forces, each of magnitude $F$ have a resultant of the same magnitude $F$. The angle between the two forces is:
A) $45^\circ $
B) $120^\circ $
C) $150^\circ $
D) $60^\circ $

Answer 1

Hint: Already we know that force is a vector quantity. Also, when two forces are acting on a body, then in that case the net force acting on the body is given by its resultant. In this question, we have to find the angle between the forces. For that we will use the Resultant formula and get the required angle.

Complete step by step solution:
The value of first force,$P$ = the value of second force, $Q$ = $F$
Also the magnitude of each force is $F$, then the value of the resultant will also be $F$
We know that the resultant of two forces is given by
$R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } $
We need to put the values in the above equation.
Now, putting the respective values we get,
$F = \sqrt {{F^2} + {F^2} + 2F \times F\cos \theta } $
$ \Rightarrow $${F^2} = {F^2} + {F^2} + 2{F^2}\cos \theta $
$ \Rightarrow $${F^2} - {F^2} - {F^2} = 2{F^2}\cos \theta $
$ \Rightarrow $$ - {F^2} = 2{F^2}\cos \theta $
$ \Rightarrow $$\dfrac{{ - {F^2}}}{{2{F^2}}} = \cos \theta $
$ \Rightarrow $$\dfrac{{ - 1}}{2} = \cos \theta $
Or, ${\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = \theta $
$\therefore $ $ \theta = 120^\circ $

Hence, according to the given options, option (B) i.e. $120^\circ $ is correct.

Note: As in the given question it is being asked for the angle between the two forces. So, we need to use the resultant formula, i.e. $R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } $. We should not be confused with the formula used for finding the direction of the resultant, i.e. $\tan \alpha = \dfrac{{Q\sin \theta }}{{P + Q\cos \theta }}$.