Question

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?

Answer 1

The resultant vector will be `402.7m/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).

Now, add the one-dimensional components

`R_x = A_x+B_x-C_x = -95.76-160.21-134.06 = -390.03m/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((R_x)^2+(R_y)^2) = sqrt((-390.03)^2+100.34^2) = 402.7m/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 `(100.34/(-390.03))` 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.