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How to determine the force of friction from a graph?

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I had to a lab on the Newton's second law of motion, and one of the direction was to "determine the force of friction" from a graph I constructed. The experiment is similar to this: http://people.sc.fsu.edu/~dduke/manual/gp1.htm [Broken]
    This image of the data is a lot cleaner than my representation of it. wmo5k5.jpg

    Data:
    Trial Mass Weight Hanger (g) Time Average (sec) Acceleration (m/s^2) Force(N) Theoretical Acceleration (m/s^2)
    1 15 3.320 0.0998 0.1472 0.0934
    2 25 2.519 0.1734 0.2453 0.1556
    3 35 2.261 0.2152 0.3434 0.2179
    4 45 1.904 0.3034 0.4415 0.2801
    5 55 1.761 0.3547 0.5396 0.3424
    6 65 1.602 0.4286 0.6377 0.4047
    7 75 1.496 0.4915 0.7358 0.4669
    8 85 1.283 0.6683 0.8339 0.5292

    Experimental Slope 1.4018
    Theoretical Slope 1.5758
    Total mass of system 1.5758 kg


    2. Relevant equations
    F = uN = umg ( But we weren't given a coefficient of friction, so I wonder if calculating frictional force is even possible)
    Fk = -F (Fk = kinetic force of friction)


    3. The attempt at a solution
    I think you would subtract 1.5758 - 1.4018 to get the difference in kg, and then multiply that by an acceleration of 1 to get the coefficient of friction 0.174. Then I guess you multiply that by the 1.5758 to get the force of friction 0.2742 N.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 29, 2012 #2

    TSny

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    You want to get the force of friction from your graph rather than by calculation. So, you need to consider how friction affects the look of the graph. To do this, write out Newton's 2nd law (symbolically) for the horizontal motion of the cart including friction. Solve the equation for the applied force and interpret in terms of what the equation would look like if you graphed the applied force vs acceleration when friction is present.
     
  4. Sep 29, 2012 #3
    What values would I use for the mass and acceleration to find the applied force? The total mass of the system is 1.5758 kg. So then F = 1.5758a, which gets me back to the equation of the line for theoretical acceleration. Do you take the average of all the accelerations?
     
    Last edited: Sep 29, 2012
  5. Sep 29, 2012 #4

    TSny

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    Don't plug in any numbers. Just consider the equation (in symbols) that would express the applied force in terms of the mass, acceleration and the friction force. Then you can tell how the friction force can be obtained from the graph without any calculation. To get the equation, set up Newton's second law and solve for the applied force.
     
  6. Sep 29, 2012 #5
    F = ma?
     
  7. Sep 29, 2012 #6

    TSny

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    Yes, but the left hand side must represent the net force. How would you write the net force in terms of the applied force and the friction force?
     
  8. Sep 29, 2012 #7
    applied force - friction force = net force
    net force = applied force - friction force
    applied force - friction force = ma ?
     
  9. Sep 29, 2012 #8

    TSny

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    Now solve for the applied force. Your graph is a plot of the applied force on the "y" axis and acceleration on the "x" axis. Does your equation for the applied force have the form y = mx+b? If so, how does the force of friction affect the graph?
     
  10. Sep 29, 2012 #9
    y = mx + b
    applied force = ma + friction force
    It causes an increase in force?
     
  11. Sep 29, 2012 #10

    TSny

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    What does b represent in y = mx + b?
     
  12. Sep 29, 2012 #11
    friction force, wouldn't that value have to be negative though because it opposes the applied force?
     
  13. Sep 29, 2012 #12

    TSny

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    You took care of that when you wrote the net force as Fapplied minus Ffriction. Rearranging makes the friction force appear as a positive contribution to the right hand side.
     
  14. Sep 29, 2012 #13
    Ah, so the frictional force is just -0.174?
     
  15. Sep 29, 2012 #14

    TSny

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    Where did you get that number?
     
  16. Sep 29, 2012 #15

    TSny

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    Also, when graphing the data, hopefully you didn't "force" the graph to go through the origin.
     
  17. Sep 29, 2012 #16
    1.5758 - 1.4048
    and I did; I suspected that something was fishy there, too. My equation of the experimental without "the force" was y = 1.2729x + 0.056 and the theoretical was the same as before, y = 1.5758x.
    This was my normal graph:

    nmjklt.jpg
     
    Last edited: Sep 29, 2012
  18. Sep 29, 2012 #17

    TSny

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    Ok. Well, we still need to see the connection between Fapplied = ma + Ffriction and y = mx + b. What does the force of friction correspond to in the equation y = mx + b?
     
  19. Sep 29, 2012 #18
    the y-intercept
     
  20. Sep 29, 2012 #19

    TSny

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    Good. So, you should be able to get the friction force from your graph if you don't force the graph to go through the origin.
     
  21. Sep 29, 2012 #20
    Why wouldn't the experimental acceleration line go through the origin though? I thought an acceleration of zero corresponds to a force of zero. Or is that what corresponds to the force of friction? Also to determine the percent error of the graph, what value x would I subtract from the theoretical value over the theoretical. ((x-1.5758)/1.5758). The addition of the y-intercept to the equation confuses me. And I presume the force of friction is just +0.0556?

    Sorry for the torrent of questions.
     
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