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Homework Help: 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?

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


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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.
  5. Jan 31, 2008 #4


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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


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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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