1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating coefficient of friction given m, applied force, and a

  1. Jun 30, 2012 #1
    1. The problem statement, all variables and given/known data
    From http://library.thinkquest.org/10796/index.html (#6)

    [tex]g=9.80m/s^2[/tex]

    2. Relevant equations
    [tex]F=ma[/tex]
    [tex]F_f=\mu F_N[/tex]
    [tex]F_N=mg[/tex] (The site actually states the normal force to be equal to negative mass times gravitational acceleration, but with a negative value for gravitational acceleration. I'm going with Wikipedia, though.)
    [tex]n\textrm{g}=\frac{n}{1000}\textrm{kg}[/tex]

    3. The attempt at a solution
    First off, I'd like to say that this site was made by high school seniors, so I'm put in the uncomfortable position of not being able to readily accept everything that's there.

    Next, why is the applied force positive but the acceleration negative? I'll just assume that that was a mistake and that the applied force should actually be -5N.

    [tex]-5\textrm{N}+F_f=F\implies F_f=F+5\textrm{N}[/tex]
    (Right? It seems right to me....)
    [tex]F=0.4\textrm{kg} \times -1.5\textrm{m/s}^2[/tex]
    [tex]F_f=\mu \times 0.4\textrm{kg} \times 9.80 \textrm{m/s}^2[/tex]
    [tex]\mu=\frac{F_f}{0.4\textrm{kg} \times 9.80 \textrm{m/s}^2}=\frac{0.4\textrm{kg}\times -1.5\textrm{m/s}^2+5\textrm{N}}{0.4\textrm{kg} \times 9.80 \textrm{m/s}^2}[/tex]
    [tex]\mu=1.122[/tex]
    I get the same answer when I keep the applied force positive, make the acceleration positive, and use [tex]5\textrm{N}-F_f=F\implies F_f=5\textrm{N}-F.[/tex]

    Yet, the site's answer is 0.15.

    I even tried using a positive applied force with a negative acceleration (pretending that friction could make an object go in the opposite direction of the applied force).

    [tex]5\textrm{N}-F_f=F\implies F_f=5\textrm{N}-F[/tex]
    [tex]F=0.4\textrm{kg} \times -1.5\textrm{m/s}^2[/tex]
    [tex]F_f=\mu \times 0.4\textrm{kg} \times 9.80 \textrm{m/s}^2[/tex]
    [tex]\mu=\frac{F_f}{0.4\textrm{kg} \times 9.80 \textrm{m/s}^2}=\frac{5\textrm{N}-0.4\textrm{kg}\times -1.5\textrm{m/s}^2}{0.4\textrm{kg} \times 9.80 \textrm{m/s}^2}[/tex]
    [tex]\mu=1.429[/tex]

    Then I realized that, using 400 instead of 0.4, you get -0.152, -0.152, and 0.154, respectively, for the three attempts described above.

    Somebody, please, what is going on here?? I'd really appreciate some help, and it'd be great if you simply told me that the site was really wrong. :tongue:

    Thank you!!
     
  2. jcsd
  3. Jun 30, 2012 #2
    From data given, maximum frictional force=mg<5N(applied force)
    The direction of acceleration should be positive.
     
  4. Jun 30, 2012 #3

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    As aziziwi states, the acceleration given in the problem should be in the same direction as the applied force ... so it should have the same sign.

    The site says that the normal force is -mg is correct if g=-9.8m.s-2 ... i.e. the normal force is equal and opposite to gravity, which you'd normally give as mg. Take care interpreting equations ... a minus sign in front of a thingy does not have to imply that you take the negative of that thingy.

    But the site looks somewhat confused to me. It says at the top that this is by students for students, and I see no way to edit it or submit corrections, so I'm not really all that surprised.
     
  5. Jun 30, 2012 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Site is pretty bad. I'm surprised it is not first proofread and checked by a teacher before posting.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook