Solving Puck Force Question: Find Acceleration

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The discussion revolves around calculating the acceleration of a puck subjected to two forces, F1 and F2. The confusion arises from the book's solution, which only considers the x-direction component of the forces, despite F1 having both x and y components. Participants suggest that the y-component may be negligible due to the puck's movement being constrained on a horizontal surface. The key equation used is F = ma, emphasizing the need to resolve forces into their vector components. The final consensus indicates that the y-component's impact on acceleration is minimal in this scenario.
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Homework Statement


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?


Homework Equations


F = ma


The Attempt at a Solution

 
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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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