Q. Two equal forces are acting at a point with an angle of 60$^{\circ}$ between them. If the resultant force is equal to $40\sqrt{3}$, the magnitude of each force is

Solution:

Let equal forces $F_1 = F_2 = F$ Newton.
Angle between the forces $(\theta ) = 60^\circ$
Resultant force $ \ \ R = \sqrt{F^2_1 + F^2_2 + 2F_1F_2 \\cos \ \theta}$
$\therefore \ \ \ \ \ \ \ \ \ 40\sqrt{3} = \sqrt{F^2 + F^2 + 2F \cdot F \ \cos \ 60^\circ}$
$or \ \ \ \ \ \ \ \ \ \ \ F = 40 \ N$

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