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Vector Forces

  1. Jan 31, 2008 #1
    1. The problem statement, all variables and given/known data
    Three forces, given by 1 = ( -2.35 i - 2.10 j ) N, 2 = ( 4.90 i + 3.05 j ) N, and 3 = (49.0 j ) N, act on an object to give it an acceleration of magnitude 3.65 m/s2.

    (a) What is the direction of the acceleration?
    86.1° (from the positive x axis)

    (b) What is the mass of the object?
    _____kg

    (c) If the object is initially at rest, what is its speed after 10.0 s?
    _____m/s

    (d) What are the velocity components of the object after 10.0 s?
    (____ i + ____ j) m/s

    2. Relevant equations
    F = ma
    Vf = Vi + at

    3. The attempt at a solution
    F = ma
    m = F/a
    To find mass (kg), I divided 2.55 by 3.65 and got 0.7 kg as the answer. That should be the correct answer but it isn't. Can someone please tell me what I did wrong? Thanks.
     
  2. jcsd
  3. Jan 31, 2008 #2

    mjsd

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    where did you get 2.55 for magnitude of force?
     
  4. Jan 31, 2008 #3
    2.55i + 49.9j were the magnitudes.

    I found I was supposed to use 49.9 as the F, and that resulted in a mass of 3.68 kg, which is the correct answer.

    Now I just need to solve for the i component for part d.). I got 36.42j for the j component, which is right, and 2.48i as the i component, which is wrong.
    Help?
     
  5. Jan 31, 2008 #4

    mjsd

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    the magnitude of this vectorial quantity (2.55, 49.95) is given by

    [tex]\sqrt{2.55^2 + 49.95^2} \approx 50[/tex] and NOT 49.9, although in this case it turned out that it is close to 49.9
     
  6. Jan 31, 2008 #5
    Ok that is right.

    However I wasn't able to find the i factor of velocity components after 10 second
     
  7. Jan 31, 2008 #6
    For part d
     
    Last edited: Jan 31, 2008
  8. Feb 2, 2008 #7

    mjsd

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    did you mean to say mass is about m= 13.7 Kg?
    then 2.55/m x 10 = 1.86
    49.95/m x 10 = 36.4
     
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