Stuck for a long time :(..........Please help me with vector addition! Illustration below. I need to find the direction and magnitude of the sum vector. The correct answer is 41.62, 42 degrees from x to -y

- anonymous

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- anonymous

|dw:1381109606282:dw| (Construct c = a + b)

- anonymous

|dw:1381109717758:dw|

- wolfe8

I believe you are supposed to add them like this: |dw:1381110194843:dw|

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## More answers

- anonymous

?

- anonymous

I was trying to use the component method (x and y components)

- mathstudent55

What are the magnitudes of the two vectors?

- anonymous

Oh! I forgot to write that! One second...

- anonymous

|dw:1381111020150:dw|

- mathstudent55

|dw:1381111017373:dw|

- anonymous

255 or 75?

- mathstudent55

\(R_x = A_x + B_x = 5 \cos20 + 5 \cos255 \)
\(R_y = A_y + B_y = 5 \sin20 + 5 \sin255 \)
\(R = \sqrt{R_x^2 + R_y^2} \)
\(\theta = \tan^{-1} \dfrac{R_y}{R_x} \)

- mathstudent55

An angle of 75 deg down form the negative y-axis is an angle of 255 deg from the positive x-axis. The advantage of using the 255 deg angle is that the sin and cos will automatically be negative numbers since it's in the third quadrant.

- anonymous

Oh!! Okay That makes sense!

- anonymous

@mathstudent55 do you mind If I calculate that out to see if I get the correct answer? I've been half an hour, so please don't leave :)

- mathstudent55

No problem.

- anonymous

Hmmm I get 4.61 for magnitude?

- anonymous

7 degrees?

- mathstudent55

I got 4.62 magnitude, and -42.5 deg.

- mathstudent55

What did you get for Rx and Ry?

- anonymous

Rx: \(\ 5cos20+5cos255 \) \(\ = \) \(\ -0.13... \)
Rx: \(\ 5sin20 + 5sin255 \) \(\ = \) \(\ 0.416 \)

- anonymous

\(\ \huge \text{The latter should be } Ry. \)

- mathstudent55

This is what I get for the components of the resultant.
\(R_x = 5 \cos 20 + 5 \cos 255 = 4.6985 + (-1.29409) = 3.4044 \)
\(R_y = 5 \sin 20 + 5 \sin 255 = 1.7101 + (-4.82963) = -3.11952 \)

- mathstudent55

\( R^2 = R_x^2 +R_y^2 \)
\( R = \sqrt{(3.4044)^2 + (-3.11952)^2} \)
\( R = 4.6175 \)
\( \theta = \tan^{-1}\dfrac{R_y}{R_x} \)
\( \theta = \tan^{-1} \dfrac{-3.11952}{3.4044} \)
\(\theta = 42.5^o\)

- anonymous

Hmmm So perhaps the answer key is incorrect?

- anonymous

I hope this not too much, but there is one other problem I am stuck on... this time with vector subtraction. Do you mind helping me with that?

- anonymous

|dw:1381113129371:dw|

- mathstudent55

a - b = a + (-b)

- mathstudent55

The new vector is -b.
Now find x and y components of vector a and vector -b.
Add the components together and find the magnitude of the resultant.
Then use the inverse tangent to find the angle of the resultant.
|dw:1381113881005:dw|

- mathstudent55

|dw:1381114004562:dw|

- anonymous

Okay, and I would still add the components except add the OPPOSITE of the b components?

- anonymous

would x still be cos and y still be sin in this case? I always forget when those are switched

- mathstudent55

\(A_x = 8 \cos 135\)
\(A_y = 8 \sin 135 \)
\(B_x = 4.5 \cos 240 \)
\( B_y = 4.5 \sin 240 \)
\(R_x = A_x + B_x = 8 \cos 135 + 4.5 \cos 240 = -7.90685\)
\(R_y = A_y + B_y = 8 \sin 135 + 4.5 \sin 240 = 1.75974\)

- anonymous

Where do the 135 and 240 come from?

- anonymous

It's subtraction, so shouldn't it be - 4.5cos240?

- mathstudent55

Let me explain that again.

- mathstudent55

|dw:1381114977171:dw|

- mathstudent55

These are the original vectors A and B, ok?

- anonymous

k

- mathstudent55

|dw:1381115051999:dw|

- mathstudent55

Instead of A + B, they want A - B, right?

- anonymous

Yup

- mathstudent55

Mathematically speaking, A + B is the same as A + (-B).
For example, 4 - 3 = 4 + (-3)

- anonymous

Yes

- mathstudent55

So in order to perform the subtraction of vectors A - B, we can instead add vectors A and -B.

- anonymous

Okay

- mathstudent55

The first step now is to find what the vector -B is.

- anonymous

k

- mathstudent55

If vector \(B = B_x + B_y\), then \(-B = -( B_x + B_y) = - B_x + (- B_y)\)

- mathstudent55

If B is the vector in the figure below,
|dw:1381115308130:dw|

- anonymous

uhuh

- mathstudent55

|dw:1381115353688:dw|

- mathstudent55

Those are the components of B.
Then this is -B:
|dw:1381115399994:dw|

- anonymous

ok

- mathstudent55

Since vector B goes up to the right, vector -B goes down to the left.

- mathstudent55

Now remember that we need to add vectors A and -B to subtract A - B.

- mathstudent55

|dw:1381115515926:dw|

- mathstudent55

There you have vector A and vector -B.
Now we need to add them.

- mathstudent55

First, let's translate the angles of the vectors to angles starting at the positive x axis.

- mathstudent55

|dw:1381115630429:dw|

- anonymous

okay

- mathstudent55

Now we need to add a vector with magnitude 8 at 135 degrees and a vector of magnitude 4.5 at 240 degrees.

- anonymous

k

- mathstudent55

Now we get x and y components of both vectors.

- anonymous

ok

- mathstudent55

\(A_x = 8 \cos 135 = -5.65685\)
\(A_y = 8 \sin 135 = 5.65685\)
\(B_x = 4.5 \cos 240 = -2.25\)
\(B_y = 4.5 \sin 240 = -3.89711 \)

- anonymous

Ahhh! That makes more sense!

- mathstudent55

Now to find the resultant we add the x components to find the x component of the resultant, and we do the same for the y components.

- mathstudent55

\( R_x = A_x + B_x = -5.65685 + (-2.25) = -7.90685 \)
\(R_y = A_y + B_y = 5.65685 + (-3.89711) = 1.75974 \)

- mathstudent55

Now that we have the x and y components of the resultant, we can use the Pythagorean theorem to find the magnitude of the resultant.
\(R = \sqrt{R_x^2 + R_y^2} \)
\(R = 8.1\)
We find the angle by using the inverse tangent:
\(\theta = \tan^{-1} \dfrac{R_y}{R_x} \)
\( \theta = \tan^{-1} \dfrac{1.75974}{-7.90685} \)
\(\theta = -12.5^o\)

- mathstudent55

|dw:1381116568714:dw|

- anonymous

Thank you so much @mathstudent55! You're help is invaluable! I was stuck on these for soooo long, thank you thank you!

- mathstudent55

You're welcome.
BTW, notice how I drew the direction in the last picture.
The direction is 12.5 deg up from the negative x-axis, or 167.5 deg measured counterclockwise from the positive x-axis.
The answer is 8.1 at 167.5 deg.

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