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Homework Help: Negative Values of the Frictional Force?

  1. Mar 20, 2017 #1
    1. The problem statement, all variables and given/known data

    6110bc6d-047f-4ce3-8462-1248e3134b33.jpe

    Ignore B & C for now

    The block in the figure reaches a velocity of 40 m/sec in 100 m, starting from rest. Compute the coefficient of kinetic friction between the block and the ground.

    W = 100N

    F = 40 N

    2. Relevant equations

    ∑Fy = 0 so Normal Force = Weight
    ∑Fx = Force - Inertia - Frictional Force - Weight = 0

    3. The attempt at a solution

    0.5 (V^2 -Vo^2) = as

    0.5 (40^2 - 0) = a(100m)

    a = 8 m/s^2

    So I got this equation:

    40 N = 100 N ( cos90) + 100N (8m/s^2 / 9.806 ) + 100 N ( μ)

    I ended up getting a negative coefficient of friction which is

    -0.415827. Is there something wrong with my equation?
     
  2. jcsd
  3. Mar 20, 2017 #2
    You cannot have a negative coefficient of friction.
    Also, write out your equation in variable form.
     
  4. Mar 20, 2017 #3
    Think about the directions of the forces, how they add or subtract.
     
  5. Mar 20, 2017 #4
    0.5 (V2 -Vo2) = 2 * a * d

    ∑Fy = 0

    W = 100N
    N - W = 0
    so W =100 N


    ∑Fx = 0

    F = 40 N

    F - Ff - W ( a / g ) - W = 0
     
  6. Mar 20, 2017 #5

    The inertia naturally resists the motion so it's negative, frictional force too. Weight is multiplied by the cosine of 90 so it's technically zero. Force appears to be going to the right so it's positive. I'm still having trouble why it's a negative value when computed
     
  7. Mar 20, 2017 #6
    If Ff is the frictional force, what is W?
     
  8. Mar 20, 2017 #7
    Weight, so are you saying Weight does not exist on a horizontal movement ?
     
  9. Mar 20, 2017 #8
    But the acceleration must have the same direction and the same sign as the net force.
     
  10. Mar 20, 2017 #9
    It has no effect that isn't already accounted for: weight by definition acts only vertically, and in this case is balanced by the normal force. You have the coresponding mass in the term that includes g.
     
  11. Mar 20, 2017 #10

    kuruman

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    I see a couple of problems with this equation.
    1. You say that the net force (sum of all the forces) is zero. That is true only if the acceleration is zero. This is not true here because the block starts from zero velocity and reaches 40 m/s some time later. Therefore the acceleration cannot possibly be zero.
    2. You show 4 forces acting in the horizontal direction. I can only see 2 that make sense, "Force" (pushing the block) and "Frictional force" (from the ground contact). Where does "Inertia" come from? What about "Weight"? Is that in the horizontal direction?
     
  12. Mar 20, 2017 #11
    I got the same answer. Maybe we are both crazy.
     
  13. Mar 20, 2017 #12

    Ok , so this gave me an idea

    ∑Fx = 0

    F = 40 N

    F - Ff + W ( a / g ) = 0

    So if I compute again

    40 - Ff + 100 ( 8m/s2 / 9.806)

    I would get a Ff of 121.5827 N

    divide that by 100 and I get a coefficient of 1.22

    Seems pretty large for a coefficient of friction
     
  14. Mar 20, 2017 #13
    This is wrong.

    Read what @kuruman said.
     
  15. Mar 20, 2017 #14
    If a force of 40 N is applied to a mass of 10 kg, you get an acceleration of 4 m/s^2. So how can you get an acceleration of 8 m/s^2 unless you have another force acting in the same direction. Thus, a negative μ.

    Maybe it's too earl for me and I'm not fully awake yet.
    Edit: See ^. I can't even type "early".
     
  16. Mar 20, 2017 #15
    Interestingly enough, I also got that same negative answer. I'm not sure what could've gone wrong here unless the data from the original question was copied down incorrectly, since a negative coefficient of friction is physically impossible.
     
  17. Mar 20, 2017 #16
    Either written down wrong, or a poorly-thought-out problem.
     
  18. Mar 20, 2017 #17
    Yeah both are very possible
     
  19. Mar 20, 2017 #18
    @AilingLore21 Can you check if all of the values in the problem are written down correctly.
     
  20. Mar 20, 2017 #19
    I think you have the wrong sign for the acceleration; but I agree something is strange here. The block was accelerated at almost g, which as far as I can see is incompatible with a 40 N force and a 100/g kg mass. Have you checked the given data, including units?
     
  21. Mar 20, 2017 #20
    I somewhat figured out that
    They are apparently. I'm used to solving the inclined versions of these. I wonder why I'm having this much trouble on a horizontal one
     
  22. Mar 20, 2017 #21
    Actually, your first answer of μ = -0.416 looks like it was correct. But I could not understand your method. It seemed like there were some problems.
    For example, if you could explain how this equation came about.
     
  23. Mar 20, 2017 #22

    kuruman

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    Look, if the contact were frictionless, the acceleration would be
    $$a=\frac{F}{W/g}=\frac{40}{100/9.8} = 3.92~m/s^2$$
    How can the acceleration with friction be about twice a much? If the numbers are copied correctly, they are not thought out correctly.
     
  24. Mar 20, 2017 #23
    Actually, the more I look at your equation, the more it makes sense. The only thing that is really questionable in that equation is the 100cos90 term. Where did that come from?

    One other tip: It is best to leave out units in these equations. Just include them in the final result.

    If I throw out the 100cos90 term in your equation, I am left with:
    40 = (100)(8/9.8) + 100μ, which is derived from:
    ΣF = ma
    F - f = ma (where F is the 40 N applied force and f is the friction force)
    40 - μ(100) = (100/9.8)(8)
    Solving gives μ = -0.416.

    So I think you were pretty much on the right track, except for that 100cos90 term.
     
  25. Mar 20, 2017 #24

    kuruman

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    You can safely do that, ##\cos 90^o =0##.
     
  26. Mar 20, 2017 #25
    For sure. I just think it shows a lack of understanding that it was there in the first place.
     
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