Solving Puck Force Question: Find Acceleration

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SUMMARY

The discussion centers on calculating the acceleration of a puck with a mass of 0.20 kg subjected to two forces: F1 of 1N at -30° and F2 of 2N directed west. The correct approach involves resolving the angled force into its vector components, specifically addressing both x and y directions. The book's solution, which only considers the x direction, yields an acceleration of -5.7 m/s², prompting questions about the omission of the y component in the calculation.

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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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