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

• osob
In summary: ChetIn summary, Chestermiller needed steps and guidance on how to solve questions like this, but was not able to complete the task. He attempted to solve the problem by resolving vectors into components in the x and y directions, but ran into trouble with the y component. He was then able to correctly solve for the angle, cos(23 x 8), and tanθ.
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 ..

Last edited:
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

Chestermiller said:
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?

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

Chestermiller said:
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?

osob said:
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?

Chestermiller said:
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

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

Chestermiller said:
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?

OR

tan theta = 2.87 / 7.36
= 21 degrees

osob said:
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

osob said:
OR

tan theta = 2.87 / 7.36
= 21 degrees
Yes.

Chestermiller said:
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.

osob said:
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

osob

## 1. What is a resultant vector?

A resultant vector is the sum of two or more individual vectors. It represents the overall displacement or force resulting from the combination of these vectors.

## 2. How do I find the components of the resultant vector?

To find the components of the resultant vector, you can use basic trigonometry. The x-component can be determined by adding the x-components of the individual vectors, and the y-component can be determined by adding the y-components. You can then use the Pythagorean theorem to find the magnitude of the resultant vector.

## 3. Can the resultant vector be negative?

Yes, the resultant vector can have a negative magnitude or direction. This can occur when the individual vectors have opposite directions or when they cancel each other out.

## 4. How do I determine the angle of the resultant vector?

The angle of the resultant vector can be found using inverse trigonometric functions. For example, if you have the x-component and y-component of the resultant vector, you can use the arctangent function to find the angle.

## 5. Can I use the parallelogram method to find the resultant vector?

Yes, the parallelogram method is one way to visually determine the resultant vector. It involves drawing the individual vectors as sides of a parallelogram and then drawing the diagonal, which represents the resultant vector. This method can also be used for more than two vectors.

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