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

[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. Homework Statement

Homework 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?

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 ..

 

[Chestermiller]

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

 

[osob]

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

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?

 

[Chestermiller]

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

 

[osob]

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

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?

 

[Chestermiller]

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?

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?

 

[osob]

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?

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? I am 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

 

[Chestermiller]

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

 

[osob]

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

ok i created a right angle triangle...
tan(theta) = opp /adj
hypotenuse is 7.89
rx = 7.36
ry = 2.87

tan theta = 7.89 / 7.36
= 46.99 degrees

Does this look right?

 

[osob]

OR

tan theta = 2.87 / 7.36
= 21 degrees

 

[Chestermiller]

ok i created a right angle triangle...
tan(theta) = opp /adj
hypotenuse is 7.89
rx = 7.36
ry = 2.87

tan theta = 7.89 / 7.36
= 46.99 degrees

Does this look right?

No. Opposite = 2.87, Adjacent = 7.36,

tanθ = (2.87)/(7.36)

Chet

 

[Chestermiller]

OR

tan theta = 2.87 / 7.36
= 21 degrees

Yes.

 

[osob]

Yes.

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

 

[Chestermiller]

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

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

Chet