Overcoming Friction Sliding Box.

AI Thread Summary
A 75 kg box sliding down a 25-degree ramp with an acceleration of 3.60 m/s² requires the calculation of the coefficient of kinetic friction, resulting in a value of 0.061. The discussion highlights the complexity of the problem, particularly noting that the forces in the y-direction do not equal zero, which is a deviation from previous problems. The user initially miscalculated due to confusion between sine and cosine functions and the handling of forces. After correcting these mistakes, they were able to arrive at the correct solution. The importance of proper geometric understanding in physics problems is emphasized.
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Homework Statement



A 75 kg box slides down a 25 degree ramp with an acceleration of 3.60 m/s/s

Find the coefficient of kinetic friction (I'll use "uk") between the box and the ramp.
Find the acceleration for a 175 kg box.

Homework Equations



F=ma

Fg= mXg (g=9.81 m/s/s in my class)

uk= Fk/Fn

Solution: 0.061 and 3.61 m/s/s

This is kind of a curveball question compared to what we've had. The big difference is that the sum of the forces in the y direction does not equal 0...I am also pretty sure Fapp is 0 since there is no applied force.

The Attempt at a Solution



I wish I could include a drawing because that could be where the problem is...here was my go:

Fnet=Fk+Fp

Fnet=m X a

m X a=Fk-Fp

(Fk=(m X a)+(Fp)

Fp=cos(theta)m X g

Fn=sin(theta)m X g

uk=Fk/Fn

uk=((m X a)+(cos(theta)m X g))/(sin(theta) X m X g)

uk=((75X3.6)-(cos(25)X75X9.81))/sin(25)X75X9.81

uk=3.01 which is obviously way off...

I haven't tried to find the answer to the second part yet.

Any help would be very much appreciated.
 
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when you do these problems, put your x-axis flush with the ramp, that way the sum of the forces in the Y direction equals zero. It simplifies the problem.[PLAIN]http://img710.imageshack.us/img710/6459/blockp.png

If you do that than the forces in the Y = 0

Forces in Y = N - mg = 0

therefor N just equals mg, the x direction will than compensate for the coordinate transformation
 
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I know what I am doing wrong, but I do not know why I am doing it wrong.

I had two problems. Firstly, I should have subtracted Fk from Fp. I figured this out and it was a stupid mistake in the first place. My second problem was that I mixed up sin and cos in the problem. When I fixed those, I got the right answer. The problem is that I do not understand why that happened.
 
take a look
[PLAIN]http://img718.imageshack.us/img718/8605/blockvg.png

Fy = Fn - Fgcos(t) = 0
Fx = Fgsin(t) - Fk = ma

Fk = (uk)Fn = (uk)Fgcos(t)

Fx = Fgsin(t) - (uk)Fgcos(t) = ma
 
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Thank you so much. I need to pay more attention to my geometry, I switched those angles. Thanks for the diagram, that's exactly what I needed.
 
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