anonymous
  • anonymous
last vector question - application
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
Find the unit vectors of each rope. Scale them by the pounds of force they have. Add the vectors up, then find the magnitude of that vector.
anonymous
  • anonymous
dude can plz go step by step ... i m confused

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anonymous
  • anonymous
Can you find the unit vectors for the ropes?
anonymous
  • anonymous
Everything you have been doing for has lead up to this point.
amistre64
  • amistre64
im thinking more along the lines of calculating the trig stuff involved
amistre64
  • amistre64
|dw:1378221930615:dw|
amistre64
  • amistre64
im assuming it wants a resultant vectors maybe?
ganeshie8
  • ganeshie8
interesting.. . yes it wants resultnat of T1 and T2
anonymous
  • anonymous
yes ,it is asking for the force exerted on point A
amistre64
  • amistre64
(8,10,6) @ 420 (-10,10,6) @ 650
amistre64
  • amistre64
find the magnitude of each vector, and scale it to obtain the stated Tension ... then add the results and find the magnitude/direction
amistre64
  • amistre64
im assuming the point A to be associated with the x axis, and positive to be towards us .... just the way i saw it initially is all
anonymous
  • anonymous
alright the magnitude would be 10 sqrt(2) and 2sqrt(59)
anonymous
  • anonymous
wait am i lost somwhere
amistre64
  • amistre64
good, so multiply the vector parts by say: 420/(10sqrt2) and 650/(2sqrt59)
amistre64
  • amistre64
this should represent the appropriate forces instead of just distances
anonymous
  • anonymous
42/sqrt(2) and 325/sqrt(59)
amistre64
  • amistre64
if simplifying as you go is what you want to do ... then thats fine scale out the vectors now
anonymous
  • anonymous
how?
amistre64
  • amistre64
(8a/b,10a/b,6a/b) (-10n/m,10n/m,6n/m)
amistre64
  • amistre64
\[k(x,y)=(kx,ky)\]
amistre64
  • amistre64
then its just adding the parts ... x = 8a/b-10n/m y = 10a/b+10n/m z = 6a/b+6n/m
anonymous
  • anonymous
ok i m confused where r u getting this a,b and n,m
amistre64
  • amistre64
im not going to sit there and write out all the specifics that we found all over again .... so i just represented them in some generic form
anonymous
  • anonymous
is there a easy way to figure out the solution
anonymous
  • anonymous
lol
amistre64
  • amistre64
a/b represents the 420/10sqrt2 n/m represents the 620/2sqrt59
amistre64
  • amistre64
using a calculator is the easiest way ....
anonymous
  • anonymous
haha so whats the answer
amistre64
  • amistre64
x = 8a/b-10n/m y = 10a/b+10n/m z = 6a/b+6n/m
amistre64
  • amistre64
that gives us the direction; the magnitude of that is the "applied Force"
anonymous
  • anonymous
did we use unit vector
amistre64
  • amistre64
in a way yes. we divided by the original magnitudes (unit them), then scaled them by their respective forces.
anonymous
  • anonymous
isn't the force scalar
amistre64
  • amistre64
magnitiude and force are scalar quantities
anonymous
  • anonymous
alright thx
amistre64
  • amistre64
youre welcome, and good luck
anonymous
  • anonymous
alright so here we r finding vector ab and ac and them their magnitude
anonymous
  • anonymous
then whats the step where we r dividing 420 / ab and 650/ac @ganeshie8
ganeshie8
  • ganeshie8
yes ! first find the magnitude, then divide it by the tension
anonymous
  • anonymous
ya what is that step called, like is it \[F_{1}\]
anonymous
  • anonymous
F1 = tension/ magnitude of ab
ganeshie8
  • ganeshie8
T1 is already given as 420 T2 is already given as 650 F1 is same as T1 F2 is same as T2
ganeshie8
  • ganeshie8
but they're just magnitudes, to add the forces you need directions also so we find the coordinates for points A, B and C
ganeshie8
  • ganeshie8
lets fix A at (0,0,0)
anonymous
  • anonymous
ok so i have tension of ab coordinates and magnitude of ab....
ganeshie8
  • ganeshie8
can you find B and C relatice to A ?
anonymous
  • anonymous
(8,10,6) and c = (-10,10,6)
ganeshie8
  • ganeshie8
now multiply your scaled forces(scalar) wid these coordinates, that gives you direction also to the forces then you can add them
anonymous
  • anonymous
so plzzzz hurry, just 7 minutes....add what and what ab and ac???
ganeshie8
  • ganeshie8
ab = magnitude of AB = ||<8,10,6|| ac = magnitude of AC = ||<-10,10,6||
anonymous
  • anonymous
ya i got ||ab|| = 10 sqrt(2) and bc = 2 sqrt(59)
ganeshie8
  • ganeshie8
scale the given forces/tensions nw 420 / ab and 650/ac
anonymous
  • anonymous
why
ganeshie8
  • ganeshie8
good q, cuz you're given the magnitude of forces. to account for the direction of resultant force in space, you need to scale the given force along the direction of rope (il give better explanation ltr)
anonymous
  • anonymous
thx got that part 2 min whats last part
ganeshie8
  • ganeshie8
add both the forces
ganeshie8
  • ganeshie8
420/ab <8,10,6> + 650/ac <-10,10,6>
ganeshie8
  • ganeshie8
that gives u a force vector
ganeshie8
  • ganeshie8
take its magnitude to get the force in pounds
anonymous
  • anonymous
72 pounds and at last i will get x y and z right?
anonymous
  • anonymous
yes or no
ganeshie8
  • ganeshie8
dint get u, force is just a number
ganeshie8
  • ganeshie8
just give them the number, you dont need to given them components
anonymous
  • anonymous
alright thx bye
ganeshie8
  • ganeshie8
np :)

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