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Calculating acceleration from the coefficient of kinetic friction

  1. Feb 6, 2019 #1
    1. The problem statement, all variables and given/known data
    At a construction side, a small crane is raising two boxes of nails on a plank to the roof. One box has already been opened and is half full, while the other box is new. The boxes, including the nails, weight 10 kg and 20 kg, respectively, and are the same size.

    Diagram:
    BLCZgpB.png

    a) If the coefficient of static friction is 0.4, draw an FBD for each box of nails and use it to calculate the angle at which each box beings to slide.
    b) If the coefficient of kinetic friction is 0.3, how fast will the boxes accelerate along the plank, once they start to slide?

    2. Relevant equations


    3. The attempt at a solution

    a)

    Box 1 = 10 kg, box 2 = 20 kg. Both FBDs are the same except for the mass of the box. For my FBD (if it's correct) should I include mgcosθ and mgsinθ?
    hr6DMXh.png

    Using the FBD:

    \begin{array}{l}
    μF_N=F_g \\
    μF_N=mg\sinθ \\
    (0.4)(mg\cosθ)=mg\sinθ \\
    0.4=\frac {\sin}{\cos} \\
    θ=tan^{-1}(0.4) \\
    θ=21.8°
    \end{array}

    This is the answer for both boxes because mass and gravity cancel out in the equation, meaning that the mass of the box does not effect the angle at which each box begins to slide but the coefficient of friction does.

    b) I don't know where to start for this question. I thought of using F=ma but I need to use the angle at which they begin to slide somewhere so F=ma wont work.
     
    Last edited: Feb 6, 2019
  2. jcsd
  3. Feb 6, 2019 #2
    Your diagram and your equations are not consistent. But your equation is OK. (By the way, put a backslash \ before "sin", "cos" etc and the function name will appear in a non-italic font).
    1. ##F_g## according to your equation is the component of gravity parallel to the ramp. It's the force that would cause the object to move, with friction opposing it. So the direction on the diagram is incorrect.

    2. ##F_n##, according to your equation, is the other component of gravity, the one normal to the surface. So it should be normal to the surface.

    3. And since the two forces in that equation are not parallel / anti-parallel, they shouldn't look that way in your diagram.

    b) F = ma is correct. That would be the net force, the force which pulls it down the ramp (which depends on the angle in the way you showed in part a), minus the frictional force (which also depends on the angle, in the way you showed in part a).
     
  4. Feb 6, 2019 #3
    Yeah I am really confused. I edited my equations right as you posted this because something looked wrong. Also I am sort of confused about why mg sin and mg cos are actually doing in this problem and why they are where they are.

    Is this FBD better?

    hv7HlsU.png
     
    Last edited: Feb 6, 2019
  5. Feb 6, 2019 #4
    Yes, if ##F_g## means ##mg##. That is the weight vector, and it points vertically downward. Although as I noted that's not the way you used ##F_g## when you wrote your equation.

    Look at the triangle you have drawn, that has ##F_g## as hypotenuse and two dashed lines as legs. Those are the components of ##F_g## which are perpendicular and parallel to the ramp.

    The angle between the normal dashed line and ##F_g## is the same ##\theta## as between the ramp and the horizontal. It might take some trig to convince yourself of that, but it's true. One argument that might help you think about it is to think what happens as the ramp angle goes to 0. When the ramp is horizontal, that component is the whole weight. The angle between ##mg## and the normal component is 0.

    So the normal component is the adjacent of ##\theta##, meaning the normal component is ##F_g \cos \theta##. That's why ##F_N## is ##mg \cos \theta##. And the other component, the one parallel to the ramp, the one that pulls the object down the ramp, is the opposite of angle ##\theta## and so the component is ##F_g \sin \theta ##.

    Again, think about what happens when ##\theta## goes to 0 and the sine goes to 0. When the ramp is horizontal, there's no gravity acting in the parallel direction. So something that goes to 0 is the right trig function to use here.

    The equation you write is that static friction ##\mu F_N## is equal and opposite to the component of gravity pulling down the ramp, ##mg \sin \theta##. And ##F_N## as we said is ##mg \cos \theta##.
     
  6. Feb 6, 2019 #5
    Wow that makes perfect sense. Much easier to understand than how my notes explain it. Thanks so much!

    For part B you said I can use ##F=ma## . I know it can be rearranged to ##\displaystyle a=\frac {F_{net}} {m}## . I don't really understand how I am supposed to use it for this question. I know that the boxes start to slide at 21.8 degrees, and that the coefficient of kinetic friction is 0.3.
     
  7. Feb 6, 2019 #6

    haruspex

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    Just before it started to slide, everything was in balance. The static friction was equal to the downslope component of the gravitational force.
    As it starts to slide, what forces change and by how much? What does that tell you about the net force parallel to the slope now?
     
  8. Feb 7, 2019 #7
    I'm not sure. I don't really understand this. Maybe the net force is equal to the normal force minus the force of kinetic friction? I'm lost. I think that if I find the net force I can use F=ma to find acceleration.
     
  9. Feb 7, 2019 #8

    haruspex

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    Please try again to answer this question.
    Let's go through the list. We have
    • Gravitational force
    • Normal force
    • Frictional force
    Which of those might change when it goes from stationary to sliding?
    There are three forces. The net force comes from adding all three as vectors.

    It may help to go back to the equations you used to solve the first part but make them more general.
    You wrote ##\mu F_N=mg\sin(\theta)##.
    We can generalise that by allowing for acceleration and not assuming maximum friction: ##ma=\Sigma F=mg\sin(\theta)-F_{friction}##.
    For the first part, you could plug in a=0 and ##F_{friction}=\mu_sF_N##. In the second part we need different substitutions.
     
  10. Feb 8, 2019 #9
    Normal force and frictional force. Gravitational force would not change.

    I'm still confused.. sorry. What does this part mean? ##ma=\Sigma F=mg\sin(\theta)-F_{friction}##, I haven't seen learned (##\Sigma##) this yet. Does it mean ##ma=mg\sin(\theta)-F_{friction}##? If so, then you said to let ##a=0## and ##F_{friction}=\mu_sF_N##? If this is the correct way to write this, does that mean that mass cancels out again just like in part a?
     
  11. Feb 8, 2019 #10
    ##\Sigma## is a capital "sigma". It means "sum." ##ma = \Sigma F## means "##ma## is the sum of all the forces."
     
  12. Feb 8, 2019 #11
    Thank you.
     
  13. Feb 8, 2019 #12

    jbriggs444

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    Would the normal force change very much in a transition from just-before slipping to just-after slipping?
     
  14. Feb 8, 2019 #13
    I guess it wouldn't. Would friction be the only force that changes then?
     
  15. Feb 8, 2019 #14

    jbriggs444

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    Yes, That is what @haruspex was trying to get you to realize.
     
  16. Feb 8, 2019 #15
    Okay. If only friction changes how can I use F=ma to solve for acceleration? I'm lost.
     
  17. Feb 8, 2019 #16

    jbriggs444

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    Before it broke loose, it was stationary. Acceleration was what?
    After it broke loose, friction changed. By how much did it change?
    As a result of that change, how much should acceleration change?
    So how much acceleration will result?

    That is an outline of the strategy I would use.
     
  18. Feb 8, 2019 #17
    Before it broke loose the acceleration was ##a=0.0m/s^2##
    After it broke loose, the friction changed from ##\mu_s=0.4## to ##\mu_k=0.3##. I guess this would result in a change of 0.1 in friction? I haven't had to do that yet.
    How can I find how the acceleration will change from the change in friction?
     
  19. Feb 8, 2019 #18

    jbriggs444

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    Yep.
    The relevant coefficient of friction changed by 0.1. If we just knew the normal force, we could translate that into a change in frictional force. Had we calculated the normal force yet? If not, we can write a force balance equation in the normal direction to figure it out.
     
  20. Feb 8, 2019 #19
    Alright. This would be the equation to calculate the normal force. I would have to do this for each box because of the different masses right? 10kg and 20kg.

    Hopefully this is correct. I just have to plug everything in.

    \begin{array}{l}
    F_{net}=F_N-F_g \\
    F_{net}=F_N-mg \\
    0 N=F_N-mg \cos \theta \\
    F_N=mg\cos \theta
    \end{array}
     
  21. Feb 8, 2019 #20

    jbriggs444

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    Good. Now with that in hand, can you take the next steps?
     
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