Two forces F1 and F2 are acting on a body . One force is double that of the other force and the resultant is equal to the greater force. Then the angle between the two forces is ?

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Solution

$L e t F_{1} = F F_{2} = 2 F F_{R} = 2 F$

$w e k n o w F_{R} = \sqrt{{F_{1}}^{2} + {F_{2}}^{2} + 2 F_{1} F_{2} \cos θ} 2 F = \sqrt{{(F)}^{2} + {(2 F)}^{2} + 2 F (2 F) \cos θ} \cos θ = - \frac{1}{4} θ = π + \cos$ ^-1(1/4)

which is the angle between them.

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Two forces F1 and F2 are acting on a body . One force is double that of the other force and the resultant is equal to the greater force. Then the angle between the two forces is ?

Let F1 = FF2 = 2FFR = 2F we knowFR = F12 + F22 + 2F1F2cos θ2F = F2 + 2F2 + 2F(2F)cos θcos θ = -14 θ = π + cos-1(1/4) which is the angle between them.

$L e t F_{1} = F F_{2} = 2 F F_{R} = 2 F$

$w e k n o w F_{R} = \sqrt{{F_{1}}^{2} + {F_{2}}^{2} + 2 F_{1} F_{2} \cos θ} 2 F = \sqrt{{(F)}^{2} + {(2 F)}^{2} + 2 F (2 F) \cos θ} \cos θ = - \frac{1}{4} θ = π + \cos$ ^-1(1/4)

which is the angle between them.