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Force magnitude and direction question help

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Three charged particles are placed at the corners of an equilateral triangle of side 1.2m. The charges are +4 macroC, -8macroC and -6macroC. Calculate the magnitude and direction of the net force on each due to the other two.
    [​IMG]



    2. Relevant equations
    F=k (Q1Q2)/r^2



    3. The attempt at a solution
    I started buy using the force formular using Q1 and Q2 got .2N dont know if that right
    then i did it using Q1 and Q3 and got .15N now im so stuck and have no idea what to do with these numbers. Any help will be appreciated!
     
  2. jcsd
  3. Sep 8, 2007 #2

    learningphysics

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    Can you show your calculations? The forces are vectors so you need to take magnitude and direction into account.
     
  4. Sep 8, 2007 #3
    ok for the one i did ill try to type what i did:

    F12=9 X 10^9Nm^2/C^2(4CX10^-6 X 8CX10^-6/1.2m^2)=.2N
    F13=9 X 10^9Nm^2/C^2(4CX10^-6 X 6CX10^-6/1.2m^2)=.15N

    then i did this to break the .2N into x and y components:
    .2cos60=.1
    .2sin60=(aprx.) .173

    and the same for the .15N

    .15cos60=.075
    .15sin60=.13

    if that is right what do i do from there?
     
  5. Sep 8, 2007 #4

    learningphysics

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    Ok. That looks good to me, except for signs... you want to calculate the total force on 1...

    What is the x component of the force of 2 on 1? What is the y component of the force of 2 on 1?

    What is the x component of the force of 3 on 1? What is the y component of the force of 3 on 1?

    You've already calculated the magnitudes... but you haven't taken the signs into account... take negative for left and down. positive for right and up.

    So once you know the x-components and y-components taking signs into account... just add the x-components... add the y-components... Then you can write the force in vector form:

    [tex]\vec{F} = x\ihat{i} + y\ihat{j}[/tex]

    So then get the magnitude of this vector... and the direction. So that would be the force on charge 1.

    Same way you need to force on charge 2... then on 3.
     
  6. Sep 8, 2007 #5
    ok the x component of the force of 2 on 1 is .1 isnt it? and the y is .173
    and for 3 on 1 the x is .075 and the y is .13. Q1 and Q2 attract (-,+) and and same for Q1 and Q3. im not to sure what your telling me ughh sorry.. but ill try to continue
    since Q2 is attracting Q1 downward the x comp is negative(to the left) and the y is negative(down). As for Q3 the x comp is positive(right) and y comp is negative(down). ok then i add the total x's:
    -.1+.075=-.025
    and same for the y's
    -.173+-.13=-.303

    then use the resultant forumula

    and i get: .30N!

    check the back of box for answer and...

    IM RIGHT HAHAHA TY MAN!! now lately how do i get the angle?
    i know it has something to do with tan(y/x) or something please just help with that last thing, THANKS AGAIN!!
     
  7. Sep 8, 2007 #6
    ok ok ok before u respond i think i got it i did:

    -.303/-.025 and got the tan inverse of that number and the result was about 85 degress. That is my reference angle so i drew a diagram and the resultant pointed in the 3rd quadrant so i added the 85 to 180 and for 265 which is the same answer as in back! if that is the right way to do it please let me know! im so excited was never more happy to do physics!
     
  8. Sep 8, 2007 #7

    learningphysics

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    yup. tan^-1 (y/x)... but remember that more than 1 angle have the same tan... so check to make sure the angle makes sense.
     
  9. Sep 8, 2007 #8

    learningphysics

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    Yup. exactly right.
     
  10. Sep 8, 2007 #9
    ok im sorry to say im last again..smh
    now for F21 i know its the same number except this time both the x and y comp are postive so:
    X21=.1, and Y21=.173
    but now for F23
    i did the Force formula:
    9x10^9Nm^2(8X10^-6 X 6X10^-6/1.2m^2) and i got .3N
    break that down into x and y:
    .3cos60=.15
    .3sin60=.26
    now this is where im stuck since both 2 and 3 are negative charges they move away from each other making the x component negative, but there is no up or down movement so how do i know what sign the y component gets?
     
  11. Sep 8, 2007 #10

    learningphysics

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    F23 has no y component ie: the y-component is 0.
     
  12. Sep 8, 2007 #11
    if thats the case then i add the numbers and i get:
    total x is -.05
    and total y is .173

    then i do resultant formula and i get .18 i think its right but according to book the answer is .26N??
     
  13. Sep 8, 2007 #12

    learningphysics

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    No, you don't need to break down F23... the angle between particle 2 and particle 3 is 0 (as measured from the positive x axis).

    So it should just be F23 = F23x = 0.3N.
     
  14. Sep 8, 2007 #13
    god bless you my friend you just saved me i can sleep easy tonite, since there was the angle that meant all the force went to the x-component...smh at myself for not realizing that.. Thanks again!
     
  15. Sep 8, 2007 #14

    learningphysics

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    no prob.
     
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