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Can someone explain this equation?

  1. Nov 3, 2012 #1
    F= (mg/sin θ) + K

    I don't understand how this equation came about. I have a physics practical to finish and I need to use this to finish it.

    I plotted a graph of F against 1/sin θ
    I got a straight line passing through the origin.

    So i tried to see if i could get the formula from

    y=mx + c

    Could someone tell me where mg comes from in this?
     
  2. jcsd
  3. Nov 3, 2012 #2

    Simon Bridge

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    Take your equation written at the top - erase the ##F## and put a ##y## in it's place. Erase the ##1/\sin(\theta)## part and put an ##x## there instead. Now compare what you have with the equation ##y=mx+c##.

    m=?
    c=?

    which part of mx+c is the slope of the graph and which the y-intercept?
     
  4. Nov 3, 2012 #3
    m is the slope c is the intercept.

    but how would mg be the gradient?
     
  5. Nov 3, 2012 #4
    from my graph c = 0

    and for m I got 1.92, but how is that related to mg?
     
  6. Nov 3, 2012 #5

    SammyS

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    mg = 1.92
     
  7. Nov 3, 2012 #6
    Oh, I'm sorry I still quite fully understand. so Force/sin X is equal to mass x gravity?
     
  8. Nov 4, 2012 #7

    haruspex

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    You determined that K (=c) = 0. Put that in your original equation.
     
  9. Nov 4, 2012 #8

    SammyS

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    How about [itex]\ \ F\times\sin(\theta)=mg\ ?[/itex]
     
  10. Nov 4, 2012 #9

    Simon Bridge

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    How can it be anything else?

    You know from the theoretical relation that:
    F=mg/sin(θ)+K

    (Presumably you started with Fsin(θ)=mg+Ksin(θ) or something - then rearranged it so you can get a graph that is a straight line.)

    Since x=1/sin(θ) and y=F, this is the same as:

    y = mgx + K

    This has to be the same as y=μx+c
    (using μ for the gradient because m is already taken for "mass".)

    So what do μ and c have to be to make these two lines the same?

    hint - if the lines are the same then: mgx + K = μx+c

    ---------------------

    Note: starting from Fsin(θ)=mg+Ksin(θ) ... you could have plotted y=Fsin(θ) vs x=Ksin(θ): then the slope would have been 1 and the y-intercept would have been mg. But it is best practice to find the thing you want in the gradient of something.
     
  11. Nov 4, 2012 #10
    mu would have to be 1.92 and c 0.
     
  12. Nov 4, 2012 #11

    Simon Bridge

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    No - using only mgx+K = μx+c.
     
  13. Nov 5, 2012 #12
    mgx+ k = μx+c

    1.92/sin + 0 = 1.92x + 0

    right?
     
  14. Nov 5, 2012 #13

    Simon Bridge

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    You did one step too many.

    If Ωx+ζ=μx+c
    then μ=Ω
    and c=ζ

    If <carrot>x+<banana>=μx+c
    then μ=<carrot>
    and c=<banana>

    If (a+b)x+qwerty=μx+c
    then μ=(a+b)
    and c=qwerty

    If (a/b)x+A=μx+c
    then μ=(a/b)
    and c=A

    If mgx+K=μx+c
    then μ=
    and c=
     
  15. Nov 5, 2012 #14
    μ= mg
    c=K
     
  16. Nov 5, 2012 #15

    Simon Bridge

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    Great - in relation to your question... "how is mg the slope":

    You have a theory that predicts that if you plot y=F vs x=1/sin(θ) you will get a line whose slope is mg - namely: y=(mg)x+K. You did the experiment and came up with a line ... the slope of which you found to be 1.92units (is that correct? what are the units?): namely: y=(1.92)x+0.

    Therefore:
    1. The theory has successfully predicted the line.
    2. If the theory is correct then (mg)x+K=(1.92)x+0 ... which means that mg=?, K=?

    This is a common technique: you rearrange the theory so that some experiment - if the theory is right - will produces a line. Then you measure the line and compare with the theoretical prediction.
     
  17. Nov 5, 2012 #16
    1.92 Newtons
     
  18. Nov 5, 2012 #17
    Oh I think I understand now.
     
  19. Nov 6, 2012 #18

    Simon Bridge

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    Great. 1.92N suggests to me a mass something like 200g.
    Has your question been answered?
     
  20. Nov 6, 2012 #19
    Yes! Cause when I worked out the mass I got .20kg. The mass would be equal to

    m= gradient/9.81 right?
     
  21. Nov 6, 2012 #20
    Well thank you for the help, I finished everything.
     
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