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Forces on an object

  1. Feb 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Three forces acting on an object are given by F1 = ( 2.20 j ) N, F2 = ( 3.45 j ) N, and F3 = ( - 40.0 i ) N. The object experiences an acceleration of magnitude 3.95 m/s2.

    (a) What is the direction of the acceleration?

    (b) What is the mass of the object?

    (c) If the object is initially at rest, what is its speed after 13.0 s?

    (d) What are the velocity components of the object after 13.0 s?


    2. Relevant equations
    f=ma pythagorean theorem


    3. The attempt at a solution

    (a)I drew this triangle to find both the resultant force and angle.
    [​IMG]

    I calculated theta to be 171.9602 deg from positive x-axis using tan theta=(40/5.65)
    and calculated R to be 40.3971N.

    (b)Taking R from (a), I have f=ma --> m=f/a m=40.3971/3.95=10.2271kg

    (c) Using vf=v0t+at we have vf=(0)(13)+(3.95)(13)=51.35m/s

    (d)This is the one I'm lost on. I have no idea how to convert to velocity vector.


    Anyone have any insight on (d)? Also, does the rest of the work look to be correct? Thanks in advance.
     
  2. jcsd
  3. Feb 8, 2008 #2

    PhanthomJay

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    You should first determine the direction of the acceleration vector in part (a), using Newton's 2nd law. That should help to understand the direction of the velocity vector. The rest looks OK. You're almost there.
     
  4. Feb 8, 2008 #3
    How would I relate newton's second law to the direction of the acceleration vector? Isn't it just f=ma?
     
  5. Feb 8, 2008 #4

    PhanthomJay

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    Yes, it's f_net = ma, where f and a are vector quantities. If f acts in a certain direction, in what direction must a act?
     
  6. Feb 8, 2008 #5
    The same direction as the force vector, correct?

    I still don't think I'm understanding. I take it I drew the triangle wrong initially? I took the three forces to be (2.2N@90deg), (3.45N@90deg), and (-40N@180deg), so theta in the triangle is 81.9602 deg which is 171.9602deg from the positive x-axis.
     
    Last edited: Feb 8, 2008
  7. Feb 8, 2008 #6

    PhanthomJay

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    Why do you think you drew it incorrectly? The resultant force acts 171.9 degrees from the +x axis, this is correct. So what's the direction of the acceleration with respect to the +x axis?
     
  8. Feb 8, 2008 #7
    Also 171.9 degrees. I guess I'm struggling to relate this angle with the velocity components asked for in part (d).
     
  9. Feb 8, 2008 #8

    PhanthomJay

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    Yes, and since the object starts from rest, v=at (as you noted), and since a and v are vectors, and you now know the direction of a, what's the direction of v? Then break up v, which you have already calculated, into its x and y components using basic trig, or by comparison to the force components using similar triangles.
     
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