(a) An object is subjected to two displacements. The displacement vector of the first, V1 , has a magnitude of 186 m at an angle 23° North of East. The second displacement vector, V2 , has a magnitude of 327 m at an angle 43° West of North. find the magnitude and direction of the resultant vector.

Question

i got 502.02 m and 38.25 degrees but have no idea if this is anywhere near correct ! Help I split each into x and y components and added but had to use 47 degrees for the angle for vector two as subracted the 43 degrees from 90 because it said it is west of north... have i done this correctly ?

Answer 1

In terms of x & y,

186@E23°N = 171.21,72.68
327@N43°W = -223.01,239.15
Add them up: -51.80,311.83

That is 361.10 at 99.43°
or, as a heading, [email protected]°W

Answer 2

Vr = 186m[23o] + 327m[133o]

X = 186*cos23 + 327*cos133 = -51.8 m/s.
Y = 186*sin23 + 327*sin133 = 311.8 m/s.

Tan Ar = Y/X = 311.8/-51.8 = -6.01986
Ar = -80.6o = Reference angle.
A = -80.6 + 180 = 99.4o CCW = 9.4o W of
N.

Vr = Y/sin A = 311.8/sin99.4)=316 m/s
[99.4o] = 316 m/s[9.4o]W of N.

Answer 3

thank you ! got it now