Vectors
What is the resultant force of an object if there are forces of 70 lbs at an angle 90 degrees N of west, 55lbs at an angle of 63 degrees S of west, and 42 lbs at an angle of 158 degrees N of east?
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SOLUTION This solution is FREE courtesy of BrainMass!
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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