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Homework Help: What did I do wrong? (Newton's Second Law)

  1. Nov 7, 2005 #1
    Only two horizontal forces act on a 3.0 kg body. One force is 9.0N, acting due east, and the othr is 8N, acting 62 degrees north of west. What is the magnitude of the body's acceleration?
    Fnet= ma ............so I took that to mean F1x + F2x = ma
    F1x = 9 cos 0 = 9
    f2x = 8 cos 62 = 3.76
    a = (9 + 3.76)/3 = 4.25
    The solution says 2.9
     
  2. jcsd
  3. Nov 7, 2005 #2

    arildno

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    East and west are opposite directions. What does that tell you?
     
  4. Nov 7, 2005 #3
    I think it tells me that it should be F1 - F2 instead. If yes then

    a = (9 - 3.76)/3 = 1.75

    It still doesn't come to 2.9
     
  5. Nov 7, 2005 #4

    arildno

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    Oops, sorry to mislead you a bit.
    The MAGNITUDE of the acceleration evidently includes the contribution from the VERTICAL acceleration the body also expriences.
    That is, you are to find [itex]a=\sqrt{a_{x}^{2}+a_{y}^{2}}[/itex]
     
  6. Nov 7, 2005 #5
    You're right because I got the answer using it, but I don't know why. The problem says that only 2 'horizontal' forces act on the object. Why is the vertical involved? I know F2 is making an angle, but then what do they mean by horizontal forces? Shouldn't it just say 'forces'? I have a sample problem in the book that specifies there is one-dimensional motion yet 1 of the forces is making an angle.
     
  7. Nov 7, 2005 #6

    arildno

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    The phrasing in the book is just dumb, that's all there is to it.
     
  8. Nov 7, 2005 #7
    thanks for your help
     
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