Overcoming Friction Sliding Box.

  • Thread starter PEZenfuego
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  • #1
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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.
 

Answers and Replies

  • #2
71
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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 [Broken]

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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  • #3
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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.
 
  • #4
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take a look
[PLAIN]http://img718.imageshack.us/img718/8605/blockvg.png [Broken]

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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  • #5
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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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