Vectors

Let East direction be the positive x axis and North the positive y axis

Component is the horizontal (x) direction = (Magnitude) x (Cos phi)
Component is the vertical (y) direction = (Magnitude) x (Sin phi)

a) 70 LBS AT AN ANGLE 90 DEGREES N OF WEST
90 degrees N of West is also 90 degrees from the positive x direction

Magnitude = 70 LBS
phi = 90 degrees
x component = 0 LBS =70 Cos90
y component = 70 LBS =70 Sin90

b) 55LBS AT AN ANGLE OF 63 DEGREES S OF WEST
63 degrees S of West is 180+ 63 = 243 degrees from the positive x direction

Magnitude = 55 LBS
phi = 243 degrees
x component = -24.97 LBS =55 Cos243
y component = -49.01 LBS =55 Sin243

c) 42 LBS AT AN ANGLE OF 158 DEGREES N OF EAST
Magnitude = 42 LBS
phi = 158 degrees
x component = -38.94 LBS =42 Cos158
y component = 15.73 LBS =42 Sin158

Total force in x direction= -63.91 LBS =0+-24.97+-38.94
Total force in y direction= 36.72 LBS =70+-49.01+15.73

Resultant force = square root of {(x component)^2 + (y component)^2}= 73.71 LBS =square root of (-63.91^2+36.72^2)

Angle between the resultant and x axis = tan inverse ( y component / x component)

-29.88 degrees =tan inverse (36.72/-63.91)
or = 150.12 degrees =180-29.88 (expressing it as a positive angle)
Answer:
Resultant displacement = 73.71 LBS at an angle of 150.12 degrees from x axis
0r (73.71LBS,150.12degrees)

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