Problem 3-25b - Vector difference - Part 6 - (B)
Determine the vector that must be added to the sum of \(A\) and \(B\) in the figure to give a net displacement of \(4.0\; km\; \text{W}\).
Accumulated Solution
| Solution | \(A_x\) | \(B_x\) | \(C_x\) | \(R_x\) |
|---|---|---|---|---|
| (A) | 5.1 cos 71 | 6.8 sin 52 | \(C_x\) | -4 |
| Solution | \(A_y\) | \(B_y\) | \(C_y\) | Ry |
|---|---|---|---|---|
| (B) | \(-5.1 \sin 71\) | \(6.8 \cos 52\) | \(C_y\) | \(0\) |
\(5.1 \cos71 + 6.8 \sin52 + C_x = -4 \\ -5.1 \sin71 + 6.8 \cos52 + C_y = 0\)
\(C_x = -11 \;km \\ C_y = 0.63 \;km\)
\(|C| = 11 \;km\)
\(\theta = \tan-1C_y/C_x = \tan-1 0.63/11 = 3.3^\circ\)
No. \(C_x\) is west \((-)\) and \(C_y\) is north \((+)\)