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Finding coefficient of friction and the mass

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Ben pushes a box along a floor against a constant force of friction. When Ben pushes with a horizontal force of 75N the acceleration of the box is 0.5 m/s/s; when he increases the force to 81N the acceleration is 0.75m/s/s.


    2. Relevant equations

    Find the mass of the box and the coefficient of friction between the box and the floor.


    3. The attempt at a solution

    i used the formula ma = uk*mg and that allows to cancel out mass and get a/g = uk
    with that i get .051 but i really dont think that is right im not really sure where the 2 different accelerations and forces come into play...
     
  2. jcsd
  3. Mar 1, 2009 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    You are pushing with a constant force. Where does that force go to?

    Write an equation for how the 75N is used up.
     
  4. Mar 1, 2009 #3
    well not really sure what your saying by that but with uk*F = m i can get the mass but im really not sure how to use the 75N to first get uk or the mass.... thats what im missing
     
  5. Mar 1, 2009 #4

    rock.freak667

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    So we know that Fnetmanet.

    If the horizontal force is F' and the force of friction is F, then the net force is given by

    Fnet=F'-F
    so that manet=F'-F

    In the question you are given that the horizontal force is 75N and the net acceleration is 0.5ms-2. You can now form one equation. Use the other information given to form the other equation.

    You will now have two equations in m and F, solve for the two of them. You can now get the coefficient of friction.
     
  6. Mar 1, 2009 #5
    yea thanks that is the basic step for the formula with that i was able to get :

    m*a = F - Ff
    F(75) - Ff = m*a(.5)
    F(81) - Ff = m*a(.75)

    so from there: (using Ff)

    m*a(.5) + F(75) = F(81) + m*a(.75)

    that really dosn't seem to fit a lot of the formulas make sence to me but i just feel like im missing something!
     
  7. Mar 1, 2009 #6

    rock.freak667

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    F(75)=75,F(81)=81, a(0.5)=0.5, a(.75)=0.75

    Subtracting 1 from 2

    F(81)-F(75)=ma(0.75)-m(0.5)
    => 81-75=0.75m-0.5m

    Solve for m
     
  8. Mar 1, 2009 #7
    yay dumb algebra on my part! the coefficient of friction is no problem.....surprisingly =)

    thanks to u who are spending their time helpin so much, personaly i get most of it i just need a few points in the right direction!
     
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