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Coplanar Forces help!

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data

    I Have to work out the following using mathematical and graphical methods

    i have to work out the resultant force of

    Vertical = 8kN

    Horizontal = 6kN

    we were given an example which i understood but with that we were given an extra component of 3kN at 45 degrees, so really not sure how to work this one out!

    where would i draw my lines etc
     
  2. jcsd
  3. Jan 15, 2013 #2

    CAF123

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    Set up a coordinate system and draw a vertical and horizontal vector (force is a vector) coinciding with the y and x axis respectively. Would you know how to compute the resultant of these two vectors?
     
  4. Jan 15, 2013 #3
    Thats what i have done mate, horizontal is 6kN and vertical is 8kN , but this is where i am stuck as this is my first time doing it

    the example i did was with 3 vectors and i joined the end of the 3'rd to the beginning and measured and got my answer if that makes sense

    i have 6 of these but if i can understand this first one the others will be fine!

    cheers dude
     
  5. Jan 15, 2013 #4

    CAF123

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    Do you mean you are stuck after this or are you stuck in finding the resultant of these two vectors?
    Joined the end of the third vector to the beginning of what?
     
  6. Jan 15, 2013 #5
    the resultant mate.. to do it graphically you have to draw another line and measure it dont you?

    the start point mate (4kN vertical, 4kN horizontal then 3kN at 45 degrees from that, then joined the end of that to the very start of the vertical)
     
  7. Jan 15, 2013 #6

    CAF123

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    Yes, find the resultant of the vertical and horizontal vectors and (vectorially) add this to the 3kN force at 45 degrees. In this case, it is probably easier to do a proper scale drawing and simply measure the magnitude of the resultant and it's angle from +ve x. You could also use various trig techniques (such as sine/cosine rules).
    EDIT: did you notice your vectors have changed from having magnitudes 8 and 6 to 4 and 4?
     
  8. Jan 15, 2013 #7
    thats what i have to do mate, i have to work out using a scaled drawing and using trig, and the vectors 4,4,3 was the one i did at college.. the one im struggling with at the moment is the 8 and 6 :/

    i have drawn the lines on graph paper but have no idea how to get the resultant from it..
     
  9. Jan 15, 2013 #8

    CAF123

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    Have you done vector addition at all?
    See for example
    http://www.physicsclassroom.com/class/vectors/u3l1b.cfm
     
  10. Jan 15, 2013 #9
    my scenario look like this picture

    u3l1b3.gif

    but with one vector 6kN and one 8kN , would my resultant be the red line?
     
  11. Jan 15, 2013 #10

    CAF123

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    Yes.
     
  12. Jan 15, 2013 #11
    thanks mate :) thats what i thought but didn't seem right..

    what do i do after 6(2)+8(2) = 100?

    do i square root it?
     
  13. Jan 15, 2013 #12

    CAF123

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    Yes, but do you know why?
     
  14. Jan 15, 2013 #13
    i dont mate.. care to enlighten me?
     
  15. Jan 15, 2013 #14

    CAF123

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    You have two vectors that are perpendicular and when you add them together you get a vector going from tail of one to the head of another. You can now think of this as a vector triangle. If the resultant you think of as the hypotenuse of a right angled triangle, you would use Pythagoras to find the length.
    Remember your intial problem is a special case (the vectors being orthogonal). You can't always use Pythagoras to find the length.
     
  16. Jan 15, 2013 #15
    cheers mate :)

    how would i work out this one

    its a horizontal line with a 45 degree vector going left and a 45 degree vector going right?
     
  17. Jan 15, 2013 #16

    CAF123

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    I can't make much sense of that. Can you post a picture?
     
  18. Jan 15, 2013 #17
    number 4 mate

    IMAG0273_zps7ac069e0.jpg
     
  19. Jan 15, 2013 #18

    CAF123

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    At this point you have to provide an attempt (forum rules) before I can help. What have you learnt in the previous problem that can be applied here?
     
  20. Jan 15, 2013 #19
    Well i would draw the first 12kN 45 degree vector, then on top of that i will do the other one, and draw a line from the start to the tip of the 2nd vector - then i would measure that line..

    to do it mathematically im not sure.. as i cant use trig? 12(2)+12(2) and square it?
     
    Last edited: Jan 15, 2013
  21. Jan 15, 2013 #20

    CAF123

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    What does that line look like? (i.e is it vertical, horizontal or inclined at some angle)
    Are the vectors perpendicular in this example?
     
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