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Forces and Friction on a Block and a Beam

  1. Jun 30, 2015 #1
    1. The problem statement, all variables and given/known data
    A block of mass M = 1 kg is placed between a vertical wall and the end of A
    beam mass m = 10 kg . If μ S = 0.6 A , determine the minimum value of the angle θ min
    θ for which the block remains in equilibrium and the coefficient of friction
    Static B μ S corresponding to prevent the beam from sliding B.

    Picture of the problem in attachment.

    2. Relevant equations


    3. The attempt at a solution
    Block:
    ∑Fy = +fsablock -Wblock = 0
    ∑Fx = +Nblock = 0

    Beam:
    ∑Fx = -fsabeam + NA =0
    ∑Fy = -Wbeam +NB + fsabeam =0

    is this correct? and how can I possibly find the min θ?

    thank you.
     

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    Last edited: Jun 30, 2015
  2. jcsd
  3. Jun 30, 2015 #2

    haruspex

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    You seem to be using x as the vertical axis and y as the horizontal, which is a bit confusing.
    In general, a static frictional force is not ##N\mu_s##. That is its maximum value.
    The minimum angle is the one at which the system is only just stable, i.e. the angle at which (some or all of) the static frictional forces are at their maximum values. So merely by assuming those forces to be ##N\mu_s## you will find the minimum angle.
    However, there is a catch here. Consider how the system will move when it does slip. Only those interfaces that will necessarily slide can be assumed to have been at their maximal frictional forces the instant before slipping. Which ones are they?

    There are two, equal and opposite, normal forces acting on the block. One of these is equal and opposite to a force on the plank.
     
    Last edited: Jun 30, 2015
  4. Jun 30, 2015 #3
    Yeah sorry, I had them correct on paper for the axies x/y on the block, I edited the thread for that part.

    ∑Fy = +fsablock -Wblock = 0
    ∑Fx = +Nblock -Nplank = 0

    Now that should be correct.

    But I still dont understand how to use the angle in the situation because in my ∑Fx/∑Fy on both the block and the beam, I don't use any angle.
     
  5. Jun 30, 2015 #4
    "Consider how the system will move when it does slip. Only those interfaces that will necessarily slide can be assumed to have been at their maximal frictional forces the instant before slipping. Which ones are they?"
    EDIT: When it does slip, the block will slide on with the plank and the wall. and the plank will slide on the floor and the block. So the 3 frictional forces on the drawing ?
     
    Last edited: Jun 30, 2015
  6. Jun 30, 2015 #5

    SammyS

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    probbb111-png.85393.png
    It's helpful - to me anyway - to show a reasonable sized image.

    You need to consider torque to get the angle θ involved. Torque about either point B or about point A, whichever makes solving less complicated.
     
  7. Jun 30, 2015 #6
    But I dont even have any distance in meter, how am I supposed to consider torque?
     
  8. Jun 30, 2015 #7

    SammyS

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    Maybe use length, L ? Otherwise, pick some length. Either way, it will cancel out.
     
  9. Jun 30, 2015 #8
    ∑MA = +NB*(tan(θ)*L) -fsbplank * L -Wplank * (tan(θ)*L)/2 = 0

    L is the vertical length A to B.
     
    Last edited: Jun 30, 2015
  10. Jun 30, 2015 #9
    One thing I am wondering.

    How is the traction force between the plank and the block? is it Vertical? or same direction as the plank with the angle?
    if that is the case, I probably dont need to use torque.
     
  11. Jun 30, 2015 #10

    haruspex

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    Certainly the block will slide down the wall.
    Is it possible that the base of the plank will not slip? Imagine point B as a hinge.
    Is it possible that the plank and block will move together? Imagine point A as a hinge?
     
  12. Jun 30, 2015 #11

    haruspex

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    There are two frictional forces on the block. Don't (yet) assume they're the same.
     
  13. Jun 30, 2015 #12
    Yeah well if I imagine both point as a hinge, block and plank may move together and base of plank wont slip.

    But I don't understand because if that is the case then there is no friction in those point ?
     
  14. Jun 30, 2015 #13
    Oh yeah true, do I make both in the same direction or opposite ? I believe its in the same direction.
     
  15. Jun 30, 2015 #14

    haruspex

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    No, not both at once - one or the other.
    Imagining B as a hinge, you can see that the block could slip down while the plank stays still. This means the horizontal force at B might be less than the maximum possible there, i.e. ##F_B<\mu_{sB}N_B##.
    Imagining (instead) A as a hinge, the block and plank could both slip down, with the points of contact between block and plank staying together. This means the vertical frictional force between block and plank might be less than the maximum value there, ##F_A<\mu_{sA}N_A##.
    But only one of these will be true. So there are two cases to consider.
    In both cases, the block will slide against the wall, so just before sliding ##F_{wall}=\mu_{sA}N_A##.
    Yes, same direction.
     
  16. Jun 30, 2015 #15
    Hey sorry, I understand it as a text, but how do I test the two cases ?

    Block:
    ∑Fy = +Nblock*0.6 + Nblock1*0.6 - 9.81block = 0
    ∑Fx = +Nblock - Nblock1 = 0
    Nblock = 8.175N
    Nblock1 = 8.175N

    Plank:
    That woul mean NA = 8.175?

    ∑Fx = -NA*0.6+ NA =0
    -NA*0.6+ NA =0
    ∑Fy = -98.10plank +NB + NA*06 =0
    ∑MA = +NB*(tan(θ)*L) -NB*us * L -98.10 * (tan(θ)*L)/2 = 0

    Thank you very much
     
    Last edited: Jun 30, 2015
  17. Jun 30, 2015 #16

    SammyS

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    Looks like you have some errors regarding the Plank, at least you do for Σ Fx .

    Isn't the normal force at A, NA, the same as Nblock ?
     
  18. Jun 30, 2015 #17
    Block:
    ∑Fy = +Nblock*0.6 + Nblock1*0.6 - 9.81block = 0
    ∑Fx = +Nblock - Nblock1 = 0
    Nblock = 8.175N
    Nblock1 = 8.175N

    Plank:
    Would that mean NA is = 8.175N?
    ∑Fx = -NB*usb+ 8.175 =0
    ∑Fy = -98.10plank +NB + 8.175*06 =0
    When I solve these 2:
    usb= 0.087719
    NB = 93.195N

    ∑MA = +93.195*(tan(θ)*L) -93.195*usb * L -98.10 * (tan(θ)*L)/2 = 0

    L = vertical length A to B.
    I cant figure out how to solve Torque equation, keeps saying false in my calculator.
    Is my torque equation right ?

    Thank you very much
     
  19. Jul 1, 2015 #18

    haruspex

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    It's not a good idea to leap into numerics straight away. Keep it all symbolic until the end.
    ##N_A\mu_{sA} + F_A - Mg=0##
    If we are considering the case where the plank would stay still, ##N_A\mu_{sA} = F_A##, ##2N_A\mu_{sA} = Mg##.
    ##F_B=N_A##
    If we are considering the case where the plank slips down, ##F_B=N_B\mu_{sB}##. But to avoid confusion we need to do one case at a time, and we haven't finished the static plank case yet.
    Check your signs. Which way does the frictional force at A act on the plank?
     
  20. Jul 1, 2015 #19
    ∑Fy = -98.10plank +NB - 8.175*06 =0

    Okay,
    but I don't understand why we have to consider if one slips etc if in the problem it says I have to find usb and the min angle when everything is in equilibrium? it doesn't say anything about when it starts moving.
     
  21. Jul 1, 2015 #20

    haruspex

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    I should read the question more carefully. I was taking both coefficients as given.
    OK, so we also want ##\mu_{sB}## at its minimum.
    We need to find the forces etc. when it is just about to slip. But in order to do that we have to think about how it will move when it does slip. Those contacting surfaces that will slide in relation to each other must be the ones that are at limit of static friction before slipping.
    So, we can assume that just before it slips the wall/block interface is at limit of friction, and the plank/floor interface is at limit of friction. (If the second were not true we could make ##\mu_{sB}## less.)
    For now, we still shouldn't assume that the wall/plank interface is at its limit. We can reconsider that later.
    So we have ##N_A\mu_{sA} + F_A - Mg=0##
    ##F_B=N_B\mu_{sB}##

    Now have another go at the equations for Fy for the plank and the moments on the plank.
     
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