Physics help?

Question

Consider two vectors 
~ 
F1 
with magnitude 47 N 
inclined at 18◦ and 
~ 
F2 
with magnitude 74 N 
inclined at 280◦ 
measured from the positive 
x axis with counterclockwise positive. 
What is the magnitude of the resultant vector 
~F1 + ~ F2? Draw the vectors to scale on a 
graph to determine the answer.

Answer 1

Given :

Force F₁ magnitude = 47 N

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

Force F₂ magnitude = 74 N

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

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

Resultant Force magnitude F =?

Resultant Force angle θ = ?

From above figure

Force F₁ magnitude OA = BC = 47 N

Force F₂ magnitude OB = AC = 74 N

Resultant Force magnitude F = OC

Force F₁ in X-Axis direction F_1x = OA cos18

F_1x = 47 cos18

F_1x = 44.7

Force F₁ in Y-Axis direction F_1y = OA sin18

F_1y = 47 sin18

F_1y = 14.52

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

F_2x = 74 cos(80)

F_2x = 12.8

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

F_2y = 74 sin80

F_2y = 72.88

F_2y direction is downwards.So

F_2y = - 72.88

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

ΣF_x = 44.7+12.8

ΣF_x = 57.5

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

ΣF_y = 14.52-72.88

ΣF_y = -58.36

F = √ [( ΣF_x )2+ ( ΣF_y )²]

F = √ [ 57.5² + (-58.36)² ]

F = √ [6712.1396]

F = 81.93 N

Resultant Force magnitude is 81.93 N

Θ = tan^-1( ΣF_y / ΣF_x )

Θ = tan^-1(-58.36/57.5)

Θ = - 45.43^o

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

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