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Homework Help: Finding the force of kinetic friction without a coefficient

  1. Dec 18, 2016 #1
    1. The problem statement, all variables and given/known data
    In a lab assignment we created a Fletcher's trolley system and were asked to calculate the normal force, the force of friction, and then create a graph of both to find the coefficient of kinetic friction.

    So far, the variables I have are:

    Hanging mass: 0.100 kg
    Sliding mass: 0.1553 kg
    Distance traveled by the sliding mass: 0.398 m
    Time taken for the sliding mass to reach the ground: 0.89 s
    Acceleration of the system: 1.01 m/s^2
    Normal force: 1.52 N
    Force of friction : Unknown

    2. Relevant equations
    The acceleration of the system was found using the equation a=(2(d-vit))/t^2 which i simplified to a=2d/t^2 since vi was 0 m/s.

    The normal force of the sliding object was found using Fn = mg

    The force of friction would be calculated using Ffk = Fn*μk but I don't have μk.

    3. The attempt at a solution
    As stated above, I tried using Ffk = Fn*μk but i didn't have all the variables.
  2. jcsd
  3. Dec 18, 2016 #2
    Did you draw free body diagrams for each of the 2 blocks and then write equations to relate sum of the forces to acceleration for each block?
  4. Dec 18, 2016 #3
    Okay so I tried doing that and essentially what I got was that Ffk = Mg - ma where Mg is the force of the hanging object and ma is the force of both masses of the system multiplied by the system's acceleration because Fnet = Fapp -Ffk.
  5. Dec 18, 2016 #4
    I'm not used to working with systems. I would normally write separate equations for each block and then solve the two equations with the two unknowns. But I think your equation below makes sense.
    So I think you have everything you need to solve that equation, provided you rewrite Ffk in terms of the normal force. You know the masses and acceleration.

    I did not know what Fapp was.

    Edit: P.S. Welcome to Physics Forums.
  6. Dec 18, 2016 #5
    Fapp was a formula for applied force which I used to derive my new formula for Ffk. Other than that, I now have everything I need to calculate the force of friction. Thank you!
  7. Dec 18, 2016 #6
    Let me know what you get for a final answer. I always like to check my work.
  8. Dec 18, 2016 #7
    So I subbed in the values resulting in the equation becoming Ffk = (0.100 kg(9.81m/s^2))-((0.100kg+0.1553kg)(1.01m/s)) which resulted in the answer being 0.723 N rounded to three significant digits.
  9. Dec 18, 2016 #8
    Okay, that looks right. For some reason, I was thinking that the goal of the problem was to find the coefficient of friction. And as I now re-read the problem statement, finding the coefficient of friction is indeed the final question. I just don't know why you would need to create a graph to do that.
  10. Dec 18, 2016 #9
    It's just a procedure my professor told us to use even though we could find the coefficient using μk = Ffk/Fn after Ffk was determined. Either way, thanks again for the help.
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