Math Help - four displacement vectors problem.

1. four displacement vectors problem.

heres another physics displacement problem i'm having trouble with. I also included a drawing diagram to make it easier to see what is going on.

The magnitudes of the four displacement vectors shown in the drawing are A = 11.0 m, B = 11.0 m, C = 12.0 m, and D = 32.0 m.
what is the magnitude and directional angle for the resultant that occurs when these vectors are added together.

magnitude is:
and the directional angle is (in degrees)
(from the +x axis)

http://img339.imageshack.us/img339/2716/p142xd0.gif

thanks for any help you can offer here.

2. Originally Posted by rcmango
heres another physics displacement problem i'm having trouble with. I also included a drawing diagram to make it easier to see what is going on.

The magnitudes of the four displacement vectors shown in the drawing are A = 11.0 m, B = 11.0 m, C = 12.0 m, and D = 32.0 m.
what is the magnitude and directional angle for the resultant that occurs when these vectors are added together.

magnitude is:
and the directional angle is (in degrees)
(from the +x axis)

http://img339.imageshack.us/img339/2716/p142xd0.gif

thanks for any help you can offer here.
You have the given coordinate axes, so break the vectors down into their components:
$A_x = -11.0 \cdot cos(20^o)$ and $A_y = 11.0 \cdot sin(20^o)$

$B_x = 11.0 \cdot cos(90^o)$ and $B_y = 11.0 \cdot sin(90^o)$

etc.

Then add all the x components and y components together.
$R_x = A_x + B_x + C_x + D_x = 0.40276$
and
$R_y = A_y + B_y + C_y + D_y = -16.6341$

The magnitude of the resultant vector is
$R = \sqrt{R_x^2 + R_y^2} = 16.639$

The reference angle is
$\theta = tan^{-1} \left ( \left | \frac{R_y}{R_x} \right | \right ) = 88.613^o$
and is in the fourth quadrant (as $R_x$ is positive and $R_y$ is negative.)

So your vector is 16.6 m at 88.6 degrees below the +x axis.

-Dan