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Given :

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Force F₁ magnitude = 47 N

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Force F₁ inclined angle θ₁ =  18º  ( with positive X-axis in counter clockwise direction )

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Force F₂ magnitude = 74 N

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Force F₂ inclined angle θ₂ =  280º   ( with positive X-axis in counter clockwise direction )

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θ₂ =  360 - 280 = 80º   ( with positive X-axis in clockwise direction )

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Resultant Force magnitude F =?

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Resultant Force angle θ = ?

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From above figure

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Force F₁ magnitude OA = BC = 47 N

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Force F₂ magnitude OB = AC = 74 N

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Resultant Force magnitude F = OC

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Force F₁ in X-Axis direction F_1x = OA cos18

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F_1x = 47 cos18

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F_1x = 44.7

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Force F₁ in Y-Axis direction F_1y = OA sin18

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F_1y = 47 sin18

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F_1y = 14.52

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Force F₂ in X-Axis direction F_2x = OB cos(80)

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F_2x = 74 cos(80)

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F_2x = 12.8

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Force F₂ in Y-Axis direction F_2y = OB sin(80)

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F_2y = 74 sin80

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F_2y = 72.88

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F_2y direction is downwards.So

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F_2y = - 72.88

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Sum of all forces in X-Axis direction ΣF_x = F_1x+F_2x

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ΣF_x = 44.7+12.8

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ΣF_x = 57.5

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Sum of all forces in Y-Axis direction ΣF_y = F_1y+F_2y

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ΣF_y = 14.52-72.88

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ΣF_y = -58.36

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F = √ [( ΣF_x )2+ ( ΣF_y )²]

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F = √ [ 57.5² + (-58.36)² ]

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F = √ [6712.1396]

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F = 81.93 N

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Resultant Force magnitude is 81.93 N

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Θ = tan^-1( ΣF_y / ΣF_x )

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Θ = tan^-1(-58.36/57.5)

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Θ = - 45.43^o

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Negative symbol indicates clockwise direction (downwards) with positive X-axis.

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Resultant Force angle is 45.43º with positive X-axis in clockwise direction

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