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Proper way to find forces

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0N, acting due east, and the other is 8.0N acting 62° north of west. What is the magnitude of the body's acceleration.

    Okay I know the answer is 2.9m/s^2.

    It is from F1 = 9i + oj

    F2 = -8cos62 + 8sin62

    Fnet = 5.2i + 7.1j

    Resultant vector = sqrt [(5.2^2) + (7.1^2)]

    =8.8
    F = ma

    8.8/3 = a

    a = 2.9.


    Now in the book, they had a similar question asking: to find the acceleration of a puck when there was F1 of magnitude 1N pointing 30 degrees below the horizontal and F2 of magnitude 2N pointing west of the puck. The puck weighs .20 kg.

    They only used (1Ncos30 - 2N)/0.20 = a

    a = -5.7 m/s^2

    Now im asking which way is correct? because in the first example i got the correct answer with both x and y directions taken in account for. but in the book's example they only found the x direction.


    2. Relevant equations

    F = ma


    3. The attempt at a solution
     
    Last edited: Feb 5, 2012
  2. jcsd
  3. Feb 5, 2012 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    This might mean pointing into the ground at 30 degrees, and with F2 exactly opposing the horizontal component of this, making the relevant components of F1 and F2 to be co-linear.
     
  4. Feb 5, 2012 #3
    So i don't take into account the y direction when it's pointing into the ground, but I do if it's pointing out of the ground?
     
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