# How do I determine the resultant vector R from R = A + B ?

1. Apr 8, 2015

### osob

Hi everyone,
I need steps and some guidance on how to solve questions like this. I know how to do right triangle questions using the Pythagorean method, but when it comes to these non right angle triangles I get so confused. Help!

1. The problem statement, all variables and given/known data

2. Relevant equations
a) The magnitude of the component "R" along the x-axis?
b) The magnitude of the component "R" along the y-axis?
c) The magnitude of the resultant "R" ?
d) The direction of the resultant "R" represented by the angle it makes with the x-axis?

3. The attempt at a solution
I don't know if i'm on the right track..

a) 8cos23= 7.36 m
b) 6sin23= 2.34 m
c) R^2 = a^2 + b^2 - 2abcos23 = 12.24 m
d) I have no clue where to begin ..

Last edited: Apr 8, 2015
2. Apr 8, 2015

### Staff: Mentor

Start by resolving A and B into components in the x and y directions. Even though the magnitudes of the vectors don't add algebraically, their components in each direction do. So when you are done adding their components, you will have the components of the resultant.

Chet

3. Apr 8, 2015

### osob

Hmm I'm not quite sure how to do that... Do I have to solve for the x and y axis separately?

for x axis do i have to do this:
8cos23 =
6cos23 =

and y axis:
8sin23 =
6sin23 =

And I'm also confusing myself because am I supposed to use 23 degrees or 67degrees? and why?

4. Apr 8, 2015

### Staff: Mentor

Well, first of all, the vector A, length 6, is pointing exclusively in the y direction. It doesn't have a component in the x direction. So, what are its components in the x and y directions?

Your components for vector B are more on target. But, there is something a little amiss with your y component. Which way is the y component of B pointing, up or down? What does that tell you about its sign?

Chet

5. Apr 8, 2015

### osob

magnitude of R on the x axis:
ax = 6cos90 = 0
bx = 8cos23 = 7.36

rx = ax + bx
rx = 7.36

along the y axis:
ay = 6sin90 = 6
by = -8sin23 = - 3.13

ry = ay + by
ry = 2.87

B is pointing down, meaning its negative... but I still don't know what I'm supposed to do with that information .. Am i going in the right direction?

6. Apr 8, 2015

### Staff: Mentor

Yes. You've already correctly answered parts a and b. For part c, the resultant R is obtained from the pythagorean theorem using rx and ry. Do you see how rx and ry are the components of the vector R in the figure?

7. Apr 8, 2015

### osob

Yes now I do. Phew I went a long way from my original thread, I appreciate the help Chestermiller.

So for part c,

R = rx^2 + ry^2
= (7.36)^2 + (2.87)^2
=7.89

As for part d,
I would have to use the sine law? Im totally stuck here.

R = 7.89
angle = 23 degrees

B = 8.0
angle = ?

angle R makes with the x-axis=
cos23/7.89 = cos(theta)/8
( cos23 x 8 )/ 7.89
= 21 degrees

8. Apr 8, 2015

### Staff: Mentor

Drop a normal from the tip of the vector R to the x axis. This produces a right triangle. One side of this triangle is rx, a second side is ry, and the hypotenuse is R. From this right triangle, what is tanθ?

Chet

9. Apr 8, 2015

### osob

ok i created a right angle triangle...
hypotenuse is 7.89
rx = 7.36
ry = 2.87

tan theta = 7.89 / 7.36
= 46.99 degrees

Does this look right?

10. Apr 8, 2015

### osob

OR

tan theta = 2.87 / 7.36
= 21 degrees

11. Apr 8, 2015

### Staff: Mentor

No. Opposite = 2.87, Adjacent = 7.36,

tanθ = (2.87)/(7.36)

Chet

12. Apr 8, 2015

### Staff: Mentor

Yes.

13. Apr 8, 2015

### osob

I believe I can now do these questions on my own. Through all the errors and trials, I appreciate your help so much Chestermiller.

14. Apr 8, 2015

### Staff: Mentor

I like your agressiveness, persistence, and determination. Great qualities.

Chet