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Puck Force Question

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    A puck of m = 0.20 kg has F1 of 1N directed -30° and F2 of 2N directed to the West. What is the acceleration?

    I've dealt with these problems before by finding the vector components of the angled force and adding it to the other forces. This meant i took into account both x and y directions. However in this book example, they did (1N*cos30 - 2N)/0.20 kg = -5.7 m/s^2. Why did they only take the x direction into account when the force also had a y component?

    2. Relevant equations
    F = ma

    3. The attempt at a solution
  2. jcsd
  3. Feb 6, 2012 #2


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    Hi Supernejihh! :smile:

    (without seeing the full question :redface: …) my guess is that F1 is diagonally downward, and y is constant because the puck can't go through the ice :wink:
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