# Vector A=125 m/s, 40 degrees north of west. Vector B is 185 m/s, 30 degrees south of west and vector C is 175 m/s 50 east of south. How do you find A+B-C by vector resolution method?

Jan 8, 2017

The resultant vector will be $402.7 \frac{m}{s}$ at a standard angle of 165.6°

#### Explanation:

First, you will resolve each vector (given here in standard form) into rectangular components ($x$ and $y$).

Then, you will add together the $x -$components and add together the $y -$components. This will given you the answer you seek, but in rectangular form.

Finally, convert the resultant into standard form.

Here's how:

Resolve into rectangular components

A_x = 125 cos 140° = 125 (-0.766) = -95.76 m/s

A_y = 125 sin 140° = 125 (0.643) = 80.35 m/s

B_x = 185 cos (-150°) = 185 (-0.866) = -160.21 m/s

B_y = 185 sin (-150°) = 185 (-0.5) = -92.50 m/s

C_x = 175 cos (-40°) = 175 (0.766) = 134.06 m/s

C_y = 175 sin (-40°) = 175 (-0.643) = -112.49 m/s

Note that all given angles have been changed to standard angles (counterclockwise rotation from the $x$-axis).

${R}_{x} = {A}_{x} + {B}_{x} - {C}_{x} = - 95.76 - 160.21 - 134.06 = - 390.03 \frac{m}{s}$

and

R_y = A_y+B_y-C_y = 80.35-92.50+112.49=100.34m/s

This is the resultant velocity in rectangular form. With a negative $x$-component and a positive $y$-component, this vector points into the 2nd quadrant. Remember this for later!

Now, convert to standard form:

$R = \sqrt{{\left({R}_{x}\right)}^{2} + {\left({R}_{y}\right)}^{2}} = \sqrt{{\left(- 390.03\right)}^{2} + {100.34}^{2}} = 402.7 \frac{m}{s}$

theta=tan^(-1)(100.34/(-390.03)) = -14.4°#

This angle looks a bit strange! Remember, the vector was stated to point into the second quadrant. Our calculator has lost track of this when we used the ${\tan}^{- 1}$ function. It noted that the argument $\left(\frac{100.34}{- 390.03}\right)$ has a negative value, but gave us the angle of the portion of a line with that slope that would point into quadrant 4. We need to be careful not to put too much faith in our calculator in a case like this. We want the portion of the line that points into quadrant 2.

To find this angle, add 180° to the (incorrect) result above. The angle we want is 165.6°.

If you get into the habit of always drawing a reasonably accurate diagram to go along with your vector addition, you will always catch this problem when it occurs.