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Force and Motion Question

  1. Sep 24, 2009 #1
    I keep getting the wrong answers for these two questions no matter what way i seem to structure it.
    1. Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 62` north of west.What is the magnitude of the body's acceleration?

    If follow the guideline example in my book this is how i try to answer it.
    [tex] F = m a \Rightarrow F_1 - F_2 = m a [/tex]

    [tex] 9 - 8 \cos 62 = (3) a \Rightarrow \frac {9 - 3.75}{3} = a [/tex]

    [tex] \frac {5.2}{3} = a \Rightarrow 1.733... = a [/tex]

    But my book says the answer is [tex] 2.9[/tex], what am I missing?
     
  2. jcsd
  3. Sep 24, 2009 #2

    ehild

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    The resultant force has a component to north, too, and you need to calculate the magnitude of the total force. Moreover, I do not know what does 62´ north of west mean.

    ehild
     
  4. Sep 24, 2009 #3
    From the question I assumed you only had to take the horizontal motion;
    And I tried (to the best of my ability) using separate [tex]x_i[/tex] and [tex] y_j[/tex] coordinates and failed. I can't get that elusive [tex] 2.9 [/tex]
    .
    This would mean what? To take [tex] |a| \sqrt(x^2 i+ y^2 j ) [/tex] accelerations?

    62 Degrees North of the west axis. I assume it's like 118 degrees counterclockwise from [tex] 1,0[/tex] on a unit circle,
     
  5. Sep 24, 2009 #4

    ehild

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    You either go to west or north, you move horizontally still you stay on the floor and do not rise up, or sink down...

    You have got the x component of the acceleration, using the x component of the north-west force. What is the y (north) component of this force?

    And yes, you have to use the magnitude (absolute value) of the acceleration at the end, sqrt(ax2+ay2) .

    ehild
     
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