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Dynamics: Forces at 90 and 45 degrees

  1. Apr 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi, I have a system of forces which is showed in the following picture:

    http://imageshack.us/photo/my-images/197/fuerzas.JPG/

    The exercise says: A force F produces an acceleration of 5 m/s^2 when it acts on a particle of mass m. Calculate the acceleration of the particle when subjected to the forces as shown in the image, for the following cases: alpha: 90, alpha: 45.

    As you can see, I did not give the values ​​of both F or m.

    Results: The acceleration must be 7.1 m/s^2 to 90 ° and 9.2 m/s^2 to 45 °, that is, the smaller the angle the bigger the acceleration.


    2. Relevant equations

    F= m.a

    3. The attempt at a solution


    I think I have to use Pitagoras, using the 2 Fs as the legs and the resultant force will be the hypotenuse?

    Thanks.
     
  2. jcsd
  3. Apr 20, 2013 #2

    Redbelly98

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    Welcome to PF!
    Pythagoras only works when the two Fs are at 90°, since it only applies to a right triangle. So yes, you can use that for the alpha=90° case.

    When alpha=45°, you should draw yourself a free-body diagram, and look at components of the individual forces.
     
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