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Resultant force

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known dat
    the ans given is 52.8N .
    I tried to let Fx and F y =0.
    for Fx , I have 120-20(12/13) -170(8/17)=21.5N (to the right)
    for Fy , I have 90-170(15/17) = -60N( in upwards direction )
    but my ans is incorrect = 63.7 N ( by using resultant force formula)
    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 21, 2015 #2

    billy_joule

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    It's best to type out the problem statement in full. We have no idea what you are trying to do.
     
  4. Sep 21, 2015 #3
    Sorry, I have forgotten about the problem. The problem is find the resultant force
     
  5. Sep 21, 2015 #4

    billy_joule

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    How many forces have a Y component?
     
  6. Sep 21, 2015 #5
    90, 170, 20
     
  7. Sep 21, 2015 #6

    billy_joule

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    ...And do they all appear in your ∑Fy calculation?
     
  8. Sep 21, 2015 #7
    So my new fy =90-170(15/17)-20(5/13)(15/17)= -66.7n
    I made 20(5/13)(15/17) becoz I resolve 20n to become force which is parallel with 170, then I resolve it to y axis
     
  9. Sep 21, 2015 #8

    billy_joule

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    That's not right. You can find the Y component of the 20N force directly from the info shown on the vector itself, no need for any info from the 170N vector.
    Also, you have an incorrect sign in that expression.
     
  10. Sep 21, 2015 #9
    You mean 20(5/13) ? But , when I resolve the force in that way, it's not exactly the y axis , right? It's slanted.
     
  11. Sep 22, 2015 #10

    billy_joule

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    I think it's safe to assume the the sides of the three triangles given are referenced to the x and y axis, otherwise, what use would they be?
    (that is, the lines with length 12, 12 and 8 are horizontal, lines 15, 5 & 5 are vertical)
    Your previous working implies you've already made that assumption. So I'm not quite sure why you didn't make it for the 20N vector?
     
  12. Sep 22, 2015 #11
    What do u mean by12, 12 and 8 are horizontal, lines 15, 5 & 5 are vertical)
    I m confused now...
     
  13. Sep 22, 2015 #12

    billy_joule

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    There are 3 small right triangles in your image, One on the incline, one on the 20N vector and one on the 170N vector.
    the length of two of each triangles sides are labelled. These labelled sides are either horizontal or vertical.
    The lines labelled 12, 12 and 8 are horizontal, lines 15, 5 & 5 are vertical. If this were not true, then you could not solve the question at all! It is the only information you have to find the relative direction of the force vectors.


    This post implies you do not believe the side of a triangle with length 5 is vertical:

    I am saying ; both lines labelled with length 5 are vertical. So 20(5/13) is the 20N vector resolved on the Y axis.
     
  14. Sep 22, 2015 #13
    for Fx , I have 120-20(12/13) -170(8/17)=21.5N (tothe right)
    So, which of the sign is incorrect? I have no clue at all....
     
  15. Sep 22, 2015 #14

    billy_joule

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    Fx looks correct.

    Which direction (up or down) does the Y component of the 20N vector point?
     
  16. Sep 22, 2015 #15
    Downwards
     
  17. Sep 22, 2015 #16

    billy_joule

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    Correct...post #7 suggests otherwise..
     
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