Question

Vector equation. Please help

Answer 1

Three coplanar forces \(A, B, C\) have magnitudes of 1000 N, 1300 N and 1800 N respectively. They act on a body such that the resultant force is zero. Find the angle between and \(A \text{ & } C\)

Answer 2

dont you just have a triangle?

Answer 3

if the resultant forces are 0 then the vectors end up where they started

Answer 4

use .... law of cosines maybe

Answer 5

c^2 = a^2 + b^2 -2ab cos(C)

we know abc and that leaves us with one unknown

Answer 6

I thought of that asweel, but apparently we are not meant to solve that way? Is there any other way?

Answer 7

We normally find angle using the dot product

Answer 8

you might be able to do it another way, but as far as a solution goes .... finding it one way should be sufficient.

Answer 9

the formula:

cos(a) = (u.v)/(|u||v|) can be created from the law of cosines, since that is its proof

Answer 10

thats why i suggested law of cosines to start with, thought it would be simpler to work

Answer 11

Could you tell me how I could do with the dot product formula?

Answer 12

not really, i think i would bring in two many variables trying to solve a general u.v

Answer 13

:( Not sure how to do this question then

Answer 14

why are you not allowed to use law of cosines? or is that just an assumption on your part?

Answer 15

Because I have to do they way they "teach me". I asked if we can use Law of Sine/Cosine, and she said no, we won't get any marks on the test, since we didn't "learn" it from this Maths Class (Maths C)

Answer 16

hmm, i had a teacher like that once. really stiffles creativety if you ask me

Answer 17

define the vectors A and C generically

A = k1(ax,ay)
C = k2(cx,cy)

A.C = k1k2(ax*cx + ay*cy)

since k1 and k2 are the magnitudes of A and C, dividing them off just gives us

cos(t) = ax*cx + ay*cy

which i have no idea what we could possibly do from there

Answer 18

Hmmm

Answer 19

Could we simply add the vectors \(A+B+C=0\), find the appropriate i and j, then solve?

Answer 20

Set A , B, or C as the x-axis?

Answer 21

C^2 = A^2 + B^2 -2AB cos(t)

C^2 -(A^2 + B^2) = -2AB cos(t)

A^2 + B^2 - C^2/(2AB) = cos(t)

Answer 22

i dont think we can determine the vectors beforehand ... if we could work the dot product into the end results there then maybe

Answer 23

forgot some ()s :)

Answer 24

(A^2 + B^2 - C^2)/(2AB) = cos(t)


A^2 + B^2 - C^2 A.C
---------------- = ------
2AB AC



A^2 + B^2 - C^2 A.C
---------------- = ------
2B C



CA^2 + CB^2 - C^3
---------------- = A.C
2B


maybe this proves another way to do A.C ?

Answer 25

since all we have are 3 magnitudes, defining the individual vectors is not going to be a real effecient alternative in my mind

Answer 26

thats all i got, srry :)

Answer 27

@amistre64, couldn't we say that vector A is straight right? Then, draw the other vecotrs attached to the tail of A?Then, we know that A only has an I component, and the other two have i and j components. Then, simply state that sinds the whole vectors = 0, all the i = 0 and all the j = 0. Then, solve for each individual one?

Answer 28

It doesn't matter what direction they are facing, we only care about the angles between them, right? And this should be dictated by their magnitudes, since they equal 0

Answer 29

the only way youll know where they end up at is using the laws tho

Answer 30

the end of B to attach C to that is

Answer 31

Want me to try, and see what I get?

Answer 32

if you have an idea, work with it :) id love to see what you get

Answer 33

:D
BRB, doing some actual work now

Answer 34

a circle centered on the x axis at x=A would be:

circle: (x-A)^2 + y^2 = B^2

lets assume the top half of the circle ... therefore: y = sqrt(B^2 - (x-A)^2)

now the (distance)^2 from point on the halfcircle to the origin has to equate to C^2

C^2 = x^2 + y^2

C^2 = x^2 + B^2 - (x-A)^2

C^2 = x^2 + B^2 - (x^2-2Ax+A^2)

C^2 = x^2 + B^2 - x^2+2Ax-A^2

C^2 = B^2 +2Ax -A^2, solve for x

x = (C^2+A^2-B^2)/(2A), subit in and solve for y to get your B abd C vectors

Answer 35

but that seems to be going thru a lot of effort just to define the vector parts

Answer 36

Yeah, thorugh. I will post my way when i finish, @amistre64

Answer 37

its 4am where im at, i might be asleep by the time youre done ;)

Answer 38

lol, Then you have something to look up to when you wake up then ;)

Answer 39

fair enough lol, good luck

Answer 40

A + B + C = 0
B = -A - C (1)

projection of A and C on B should equal B ie dot product
B.B + A.B + C.B = 0
B.B = - B . (A + C) (2)

sub in from (1)

B.B = - (-A-C).(A+C) = (A+C).(A+C) = A.A + C.C + 2 A.C

|B||B| = |A||A| + |C||C| + 2 |A||C|cos (π-b)
|B||B| = |A||A| + |C||C| - 2 |A||C|cos (b)

law of cosines...