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Static Friction Problem

  1. Mar 16, 2004 #1
    I'm struggling on a presentation problem in my physics class. The problem reads: A crate containing machine parts sits unrestrained on the back of a flatbed truck traveling along a straight road at a speed of 80 km/h. The driver applies a constant braking force and comes to stop in a distance of 22 m. What is the minimum coefficient of friction between the crate and the truck bed if the crate is not to slide forward?

    Ok since the crate doesn't slide forward I'm thinking this would be static friction correct?? Other than that I'm clueless, I don't necessarily want the problem solved for me, just some formulas, ideas, or just how to setup this problem up. Thanks a lot everyone!!
  2. jcsd
  3. Mar 16, 2004 #2

    Chi Meson

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    The static friciton is the only force that can cause the crate to accelerate (negatively) as the truck comes to a stop. Since static friction has a maximum quantity, then the magnitude fo acceleration has a maximum quantity. You will need to combine Newton's 2nd Law formula with the frictional force formula.
  4. Mar 16, 2004 #3


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    Yes, since the crate doesn't move at all it is static friction.
    The next thing you need to know is what force is stopping the truck-the static friction force must be that in order to stop the crate also (so that it doesn't continue forward). You know that the truck decelerated from 80 km/h to 0 in 80 m (= 0.08 km) at a constant acceleration. You should be able to figure out what acceleration does that.

    Once you know the acceleration, you will need to calculate the force on the crate- you won't actually be able to do that because force = mass times acceleration and you don't know the mass of the crate! Fortunately the problem asks for "coefficient of friction", which is the number you multiply by the weight of the crate (9.8 times the mass). Set up that equation and the mass cancels out.
  5. Mar 16, 2004 #4


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    friction = force from acceleration
    uN = ma
    umg = ma
    ug = a
    u = a/g
  6. Mar 17, 2004 #5
    Couple of more questions.

    Chi Meson- what is the frictional force formula?

    HallsofIvy- I understand how to find the acceleration, I believe I got -145.5. Is this correct? However I don't understand the second paragraph of your response, could you please help me understand it better.

    ShawnD- what does u stand for in your equations.

    Thanks everyone for all your help!
  7. Mar 17, 2004 #6


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    u is coefficient of friction.
  8. Mar 17, 2004 #7
    It's probably described as [tex]\mu_k[/tex] (for kinetic friction) and [tex]\mu_s[/tex] (for static friction) in your books.
  9. Mar 18, 2004 #8
    Alright that makes sense, but I'm still struggling on solving this problem. I know I'm going to need the equation coefficient of friction = force of friction/force of N. But i don't know how to solve it. Thanks!
  10. Mar 18, 2004 #9
    First of all you need to find the constant deceleration that brings the truck to a stop. You can do find this using this equation:

    [tex]V_f^2 = V_0^2 + 2ax[/tex]

    That is also the deceleration of the crate, since it does not move in relation to the truck itself. But there has to be a force that will cause this deceleration of the crate, and that is the friction. This means that:

    [tex]\Sigma F = {f_s}_{max} = ma[/tex]

    You shold know that:

    [tex]{f_s}_{max} = N\mu_s = mg\mu_s[/tex]

    So subtitute that in the formula above and you would get:

    [tex]mg\mu_s = ma[/tex]
    [tex]g\mu_s = a[/tex]
    [tex]\mu_s = \frac{a}{g}[/tex]

    And that is the minimal coefficient of (static) friction that will stop the create from sliding. Because if the coefficient was any smaller than that, [tex]{f_s}_{max}[/tex] would not be large enough to enforce on the create the same deceleration of the truck, which means they will move in relation to one another.
  11. Mar 18, 2004 #10
    Well said Chen! Nicely put.
  12. Mar 18, 2004 #11
    Alright I followed the steps and got coefficient of friction =
    -145.5/-9.81 = 14.83 Is this correct? Thanks Chen, the help was excellent. If this is correct could someone also help me with a way to prove this problem that would be awesome! Thanks again!
  13. Mar 18, 2004 #12


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    I would suspect that it is not correct. Can you show how you got -145.5 as the acceleration? I have no idea how you got that....
  14. Mar 19, 2004 #13
    You forgot to turn 80kmh into it's m/s equivalent. :smile: The coefficient should be something around 1.15.

    Also what do you mean by "proving this problem"?
  15. Mar 19, 2004 #14
    Thanks Chen, I did forget to do that. What I mean by proving the problem is I need to use a different equation or take a different path of solving to prove that this answer is correct. It's part of the presentation requirements. Again everyone has been a lot of help and I understand this a great deal better!
  16. Mar 19, 2004 #15
    Have you studied about work and energy?

    You could say that the friction force does work that brings the crate's kinetic energy from its original value to zero:

    [tex]W_{f_s} = f_sx\cos \theta = \Delta E_k = E_{kf} - E_{k0}[/tex]

    Expand [tex]f_s[/tex] and [tex]E_{k0}[/tex], and remember that the angle between the friction and the displacement is 180 degrees. You should be able to figure the rest out. :smile: If you can't I will explain it further.
    Last edited: Mar 19, 2004
  17. Mar 19, 2004 #16
    Chen, we actually can't. We just started on work and energy as it's right after Forces. My teacher has requested that we don't use any of the work and energy stuff. However he did suggest to go back and solve for one of the given values. In my case that'd be velocity or distance or possibly acceleration.
  18. Mar 19, 2004 #17
    I'm not sure how else you can solve the problem... you could rewrite some of the equations, or pretend to be somewhere else (an inertial observer), but it would be just that - a rewrite.
  19. Mar 19, 2004 #18
    That's what I'll end up doing. Since i'm limited in what I can do, i'm pretty much just going backwords. As long as I attempt this I think it'll be ok. Thanks for all your help Chen.
  20. Mar 19, 2004 #19
    You could take the equations:

    [tex]V_f^2 = V_0^2 + 2ax[/tex]
    [tex]{f_s}_{max} = mg\mu_s = ma[/tex]

    And play with them a little... for example you can divide the first one by 2, multiply it by the crate's mass and substitute the acceleration times mass:

    [tex]\frac{1}{2}mV_f^2 - \frac{1}{2}mV_0^2 = {f_s}_{max}x[/tex]

  21. Mar 19, 2004 #20
    It's ridiculous to disallow the use of some method. Physics doesn't care if you make an energy argument or a momentum argument or whatever kind of argument you want, as long as it's correct.

    What's he going to do, tell you it's wrong?

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