Problem 3-25b - Vector difference - Part 4

Determine the vector that must be added to the sum of $A$ and $B$ in the figure to give a net displacement of $4.0\; km\; \text{W}$ .

diagram indicating directions north and east as well as vectors A and B

Accumulated Solution

diagram indicating C at the desired 4 km West

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)	$-5.1 \sin 71$	$6.8 \cos 52$	$C_y$	$0$

$5.1 \cos71 + 6.8 \sin52 + C_x = -4 \\ -5.1 \sin71 + 6.8 \cos52 + C_y = 0$

You should get:
$C_x = -11\; km \\ C_y = 0.63 \;km$

The vector $C$ has a length:

(A) $-11 + 0.63 = -10.4\; km$

(B) $[(-11)^2 + (0.63)^2]^{1/2} = 11\; km$

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