The magnitudes of the four displacement vectors shown in the drawing are A = 14.

Question

The magnitudes of the four displacement vectors shown in the drawing are A = 14.0 m, B = 10.0 m, C = 12.0 m, and D = 26.0 m. Determine the (a) magnitude and (b) direction for the resultant that occurs when these vectors are added together. Specify the direction as a positive (counterclockwise) angle from the +x axis.

See the image below. For part b, I keep coming up with 62.395 and it will not accept it. What am I doing wrong!?

My calculations: Ax= -13.16, Ay= 4.79, Bx= 0, By= 10, Cx= -9.83, Cy = -6.88, Dx= 16.71 Dy= -19.92

Explanation / Answer

Break down the resultant into its x- and y-components, using trigonometry to add the vectors:

R_x = Acos(a) + Bcos(b) + Ccos(c) + Dcos(d)

R_x = 14cos(160) + 10cos(90) + 12cos(215) + 26cos(310) = -6.27
R_y = Asin(a) + Bsin(b) + Csin(c) + Dsin(d)

R_y= 14sin(160) + 10sin(90) + 12sin(215) + 26sin(310) = -12.01

where
R_x = x-component of the resultant
R_y = y-component of the resultant
A, B, C, D = magnitudes of vectors A, B, C, D
a, b, c, d = angle each vector makes with the +ve x axis

Now you can find the magnitude and direction of the resultant:

R = [(R_x)² + (R_y)²] = 13.54 m
= tan^-1(R_y/R/x) = 62.43°

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