Overcoming Friction Sliding Box.

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Homework Help Overview

The problem involves a box sliding down a ramp, focusing on calculating the coefficient of kinetic friction and the acceleration of a different mass box. The context includes forces acting on the box and the effects of friction on its motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the direction of forces and the importance of coordinate systems. There are attempts to clarify the relationships between forces acting on the box and the ramp.

Discussion Status

Some participants have provided guidance on adjusting the coordinate system to simplify the analysis. Others have identified mistakes in reasoning related to the application of trigonometric functions and the arrangement of forces. There is an ongoing exploration of these concepts without a clear consensus on the final approach.

Contextual Notes

Participants note the challenge of visualizing the problem and the potential impact of geometry on their calculations. There is mention of specific values and assumptions, such as the gravitational force and the angles involved, which are critical to the problem but not fully resolved in the discussion.

PEZenfuego
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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
 
Last edited by a moderator:
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
 
Last edited by a moderator:
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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