Problem 3-25b - Vector difference - Part 4 - (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) | \(-5.1 \sin 71\) | \(6.8 \cos 52\) | \(C_y \) | \(0\) |
\(5.1 \cos71 + 6.8 \sin52 + C_x = -4 \\ -5.1 \sin71 + 6.8 \cos52 + C_y = 0\)
\(C_x = -11\; km \\ C_y = 0.63\; km\)
Correct
\(|C| = 11\; km\)
The direction of \(C\) is given by:
(A) \(\cos^{-1} C_y/C_x\)
(B) \(\sin^{-1} C_y/C_x\)
(C) \(\tan^{-1} C_y/C_x\)