What is the coefficient of kinetic friction on a 30 degree ramp?

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The discussion revolves around calculating the coefficient of kinetic friction for a box sliding down a 30-degree ramp with an acceleration of 1.20 m/s². The correct coefficient of kinetic friction is determined to be 0.436. The calculations involve using gravitational forces, where the component of gravity acting down the ramp is found using the sine of the angle. The participant expresses confusion about the use of sine versus cosine in the calculations but ultimately arrives at the correct answer. The conversation highlights collaboration among peers to clarify the problem-solving process.
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Homework Statement



A box slides down a 30.0 degree ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box and the ramp.

Homework Equations





The Attempt at a Solution


Ok so what i did was different from what my teacher has, but I can't explain it.
(the answer is supposed to be .436, which is what i got)

1) F=m(9.81)(sin30.0) ...divide by m... a=4.9m/s^2 that would be the ramp with no friction right?

2) so then... F_k=(a_withoutfriction-a_with friction)m --> F_k (4.9-1.2)m= (3.7m/s^2)(m) and that would be the force of kinetic friction (I think).

3) \mu=F_k/F_n --> (3.7)(m)/(9.81)(cos30)(m)= .436

Ok so i got the right answer but I can't explain it. Just for starters, why in step 1, did it have to be sin30? would it be cos30 to find the accleration in the x direction?
 
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why in step 1, did it have to be sin30?

If you translate your coordinate system, where the X axis is on the incline you will get mgsin(t) for the forces of gravity in the X direction.

[PLAIN]http://img524.imageshack.us/img524/4726/gradianthw.png
 
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Ok me a two of my freinds got together tonight and figured out most the theorizing and problems with the problem. And yeah, that diagram was basically what we came up with.

But I really appreciate you taking your time to respond to my question. Thank you!
 
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