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Bizzare question on friction constant

  • Thread starter pfssassin
  • Start date
  • #1
16
0

Homework Statement



Create a lab experiment where you can find the friction constant of the tissue box with the floor (no surface tilting or accelerometer allowed).

Materials:
  • Various Weights
  • Tissue Box
  • Ruler
  • Timer
  • spring scales

So since I couldn't vary the inclination of the floor or use a accelerometer and advance devices, came up with the quickest thing I could do in the time I had. I pulled the tissue box on the floor using a constant force eg 400g and time how long it took to go 1 meter. So I have velocity and force. And I did the same thing with different weights.

I was thinking at the time I could use the .5mv2 to get the energy then use the F=Ed where d is 1 meter, to get Force and use that and the F=muN to get the friction coefficient.

So my questions:

1. Is that allowed?
2. I used different horizontal forces to pull each tissue box weight for example I pulled the 700g tissue box with a force of 200g and got a velocity of .199 m/s and I pulled the 1200g tissue box with a force of 400g with a velocity of 0.151285930408472 how do I convert them so they are the same?


Homework Equations


F=[tex]\mu[/tex]N
F=ma
KE=.5mv2
F=Ed


The Attempt at a Solution


Explained above
 

Answers and Replies

  • #2
157
0
What you need to be able to do is apply a force and measure the resultant acceleration of the system. This poses two problems that you have noticed:
1. How do we know how much force has been applied?
2. How do we measure the resultant acceleration?

To measure the force applied you can use the spring scales by placing it horizontal between the block and the applied force. Because of Newtons 2nd law the force you apply to the scale will be the force applied to the block.
Next apply the force such that the box just barley moves along at constant velocity. Thus you have no need to worry about acceleration since its equal to zero. Likewise the net force acting on the system will be zero so you can then sum forces in the x direction to get:

Fscale-Ffriction=0

If you analyze it doing an energy analysis your just doing one extra step that will reduce to the same equation as above since :
work=change in kinetic energy
Fscale*d-Ffriction*d=change in kinetic energy
change in kinetic energy=0 ---> constant velocity
Fscale-Ffriction=0
 
  • #3
16
0
Thanks a lot chief but I dont have that stuff at home! wish I had posted yesterday.

Any other thoughts on what I can do with the data I have? I cant really redo this experiment (its do tmrw...)
 
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