Question

For the vectors shown in the figure (A = 56.0 and θ = 57.0°) determine the following.

Answer 1

Please help me!

Answer 2

How did you get 64...?

Answer 3

I tried doing the problem but got confused.

Answer 4

I did b-a

Answer 5

Your question asks for \[2 \vec B - 3 \vec A\]

Answer 6

yes, and when I tried to do that i got confused on the values.

Answer 7

The magnitude of B is 26.5 and the magnitude for A is 56.0 correct?

Answer 8

yes

Answer 9

Vector A: 49.4450652i +26.29040752j

Answer 10

Then \[2 \vec B - 3 \vec A = 2(26.5)-3(56.0)\]

Answer 11

The sign only indicates direction

Answer 12

The magnitude is -115?

Answer 13

and is the direction then arctan( 168/53)?

Answer 14

I tried looking through yahoo answers but it is confusing
there

Answer 15

Sorry ignore what I said above, we should be using components yeah?

Answer 16

We just need the magnitude and direction

Answer 17

Yes but we require components for that (x and y components)

Answer 18

correct

Answer 19

So we must find the x and y components for all A, B, and C

Answer 20

Any idea how to find the components?

Answer 21

yes, for example Vector A would be 56cos28 + 56sin28

Answer 22

Yes exactly!

Answer 23

So for part a we only need vector B and A

Answer 24

So we have \[A_x = 56 \cos (28°) = 49.4450652\]\(A_y = 56 \sin (28 °) = 26.29040752\)

\[B_x = -26.5\cos(57.0 °) = -14.43293443\] \(B_y = 26.5\sin(57.0°) = 22.22477005\)

Answer 25

I am confused what we need to do next

Answer 26

Oh, we would need to combine b(x) with a(x) and b(y) with a(y)

Answer 27

So we have \[\vec A = A_x+A_y = 49.4 \hat i +26.2 \hat j\]
and

\[\vec B = B_x+B_y = -14.4 \hat i + 22.2 \hat j\] maybe this is easier to see, now we want \[2 \vec B - 3 \vec A \] so first multiply the components by the "coefficients"

Answer 28

We will get to the magnitude after, lets take this slow so you understand each step

Answer 29

so 2(-14.4i +22.2j) = -28.8i + 44.4j

Answer 30

Yes good now do it for vector A

Answer 31

Vector A: 148.2i + 78.6j

Answer 32

\[2 \vec B = -28.86586886 \hat i + 44.4495401 \hat j\]
and we have \[3 \vec A = 148.3351956 \hat i + 78.87122256 \hat j\] looks good, I'm leaving all the digits until the end just in case

Answer 33

Now we can go ahead and subtract the components

Answer 34

So now we will have -177.2010644i - 34.42168246j

Answer 35

did I do the math correctly?

Answer 36

\[2 \vec B - 3 \vec A = -177.2010645 \hat i - 34.42168246 \hat j\] looks good to me!

Answer 37

so now we will square both of the components and take a square root?

Answer 38

is the magnitude 180.5133503?

Answer 39

Now we find the magnitude \[R= \sqrt{(-177.2...)^2+(-34.4...)^2} = 180.5133499\]

Answer 40

Our decimals are a bit off but maybe they just want a round answer in any case that is our magnitude

Answer 41

for the direction it would be: 10.99292254

Answer 42

That looks good!

Answer 43

\[\theta = arc \tan ( \frac{ 34.4... }{ 177.2... }) = 10.99292254\]

Answer 44

Part B looks different though.

Answer 45

Oh I see you mean counter clockwise

Answer 46

do we add 180?

Answer 47

You can just do 180°-10.99...° = 169.0070775 °

Answer 48

it still says it is wrong

Answer 49

Can you take a screen shot

Answer 50

it is still wrong

Answer 51

I have only one try left before the answer is marked wrong

Answer 52

Oh crap! It should be 180 degrees + 10.99...

Answer 53

Because of the resultant components bleh

Answer 54

But now since you know how to do (a) try (b) on your own and see how it goes, if you need further help I can check it for you if you like.