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Friction / Newton's Laws

  1. Oct 1, 2007 #1
    Two blocks, one M = 88kg, and a smaller one m = 16kg which is held to the bigger block by a coefficient of static friction between the two of us = 0.38, but the surface between M is frictionless. What is the minimum horizontal force required to hold m against M?

    the force of gravity acting on the smaller block is 16kg * 9.81m/s2
    Fg = 156.96N
    Ff = usFn
    Ff = 16*9.81m/s2*.38
    Ff = 59.6448

    The difference in forces is 97.3152 downwards
    I'm not sure to to proceed with a horizontal force to keep the m held to M

    Thanks
     
  2. jcsd
  3. Oct 1, 2007 #2

    Doc Al

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    I'm going to assume that you need to find the force with which you need to push against M in order to keep m from falling.
    So far, so good.
    Not sure what you're doing here. What vertical forces act on the small mass? For vertical equilibrium, what must the vertical forces add to?

    Use that fact to figure out what the normal force must be between the two masses. Then apply Newton's 2nd law, first to the small mass, then to both masses together.
     
  4. Oct 1, 2007 #3
    I'm going to assume that you need to find the force with which you need to push against M in order to keep m from falling.

    Yes

    On the small masses the verticle forces would be the force of gravity and the force of friction
    For equilibrium they must be equal
    So can I assume that Fg = Ff ?
    If so

    Fg = 156.96N
    Ff = Fg
    Ff = usFn
    156.96N = .38*Fn
    Fn = 413.05N
     
  5. Oct 1, 2007 #4

    Doc Al

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    Good! Keep going. Analyze the horizontal forces acting on the small mass. What must be its acceleration? Deduce everything that you can.
     
  6. Oct 1, 2007 #5
    Fg = 156.96N
    Ff = Fg
    Ff = usFn
    156.96N = .38*Fn
    Fn = 413.05N

    The only horizontal force on the small mass would be the applied force since there is no friction on the surface, correct?

    I'm not sure how to calculate the acceleration, is it related to the verticle forces?
     
  7. Oct 1, 2007 #6

    Doc Al

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    Right.

    Just apply Newton's 2nd law in the horizontal direction. (The vertical forces on "m" cancel, remember?)
     
  8. Oct 1, 2007 #7
    Fg = 156.96N
    Ff = Fg
    Ff = usFn
    156.96N = .38*Fn
    Fn = 413.05N

    So in the horizontal direction i need to use
    Fnet = ma
    I know mass: 16kg
    Fnet = 16kg*a
    I need to figure out Fnet and a
    I wouldn't use the acceleration of gravity since it's in the horizontal direction
    The only other force I can think of would be the normal force, but that's also in the vertical direction and wouldn't give me what I need to find

    Do I need to use the mass of the larger block to solve for anything?
    The normal force of the larger block?
    Fn = 88kg*9.81m/s2
     
  9. Oct 1, 2007 #8

    Doc Al

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    Reread the earlier posts and you'll realize that you already know the net force on m.
    The acceleration due to gravity is not relevant, except in calculating the weight of m.
    The normal force is normal (perpendicular) to the surface between M and m. What direction is that?

    Eventually.
    If you mean the force that m exerts on M, that's the same normal force you've already calculated--consider Newton's 3rd law. (But you don't need it.)
     
  10. Oct 1, 2007 #9
    Ah, I see my mistake, I forgot how normal force worked

    Fg = 156.96N
    Ff = Fg
    Ff = usFn
    156.96N = .38*Fn
    Fn = 413.05N

    So to keep it up would the applied force need to be = to the normal force of m?

    Fn = 413.05N = Fa

    But then there would be nothing left for me to calculate

    Or do I use that to find the acceleration?
    Fn = 413.05N = Fnet
    Fnet = 413.05N = 16kg*a
    a = 25.816m/s^2

    and then find fa using the combined mass?

    Fnet = Fa - Fn = mtotal*a - 413.05N
    Fa = (16kg+88kg)(25.816m/s^2) - 413.05N
    Fa = 2684.8N - 413.05N
    Fa = 2271.775

    or do I not need to consider normal force for the last step leaving me with Fa = 2684.8N ?
     
  11. Oct 1, 2007 #10

    Doc Al

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    You changed your post while I was responding. :wink:

    Absolutely.

    If you look at both masses together, then the applied force is the only horizontal force you need. The normal forces are internal forces and they cancel out.

    If you look at the big mass alone, then you would have to consider the normal force:
    Fa - Fn = Ma
    Thus Fa = Ma + Fn

    Either way you'd get the same answer. Redo your final calculation correctly (you made a few errors).

    Here's a tip: Whenever possible, stick with symbols until the last step, then plug in numbers. Less chance for error.

    For example:
    You found the normal force:
    [tex]F_n = mg/\mu[/tex]

    Plugging that into F=ma gives you:
    [tex]mg/\mu = ma[/tex]

    Thus:
    [tex]a = g/\mu[/tex]

    (Much less calculator work this way.)

    Edit: I messed this up, as pointed out by theunloved below. Apologies to Destrio! :redface:
     
    Last edited: Oct 6, 2011
  12. Oct 1, 2007 #11
    Let's try this all the way through

    Fg = ma = 16kg*9.81m/s^2
    Fg = 156.96N
    Ff = Fg
    Ff = usFn
    156.96N = .38*Fn
    Fn = 413.05N
    Fn = 413.05N = Fnet
    Fnet = 413.05N = 16kg*a
    a = 25.816m/s^2

    Fnet = Fa = mtotal*a
    Fa = (88kg+16kg)*(25.816m/s^2)
    Fa = 2684.8N

    Or
    Fnet = Fa - Fn = Ma
    Fa = Ma + Fn
    Fa = 88kg*(25.816m/s^2) + 413.05
    Fa = 2684.8N

    I'm still getting the same final answer (.8 when i keep numbers in my calculator, .9 if I dont)

    Thanks
     
  13. Oct 1, 2007 #12

    Doc Al

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    Now you've got it. :smile:
     
  14. Oct 1, 2007 #13
    Excellent

    Thanks your your help once again :)
     
  15. Oct 5, 2011 #14
    I'm sorry to bring it up again, but I'm having the same problem and reading this post helps me figure out how to do it; however, I think that what you're saying is not right Doc Al.
    I draw a FBD of m and M, and here are all the forces that I've found:
    Taking downwards as positive, x+ from left to right

    For m
    mg - fs = 0 (1) ( it's not slipping down, so a must be equal 0 )
    F - Fc (F contact) = ma (2)

    For M
    Mg - N = 0
    Fc = Ma (3)

    Basically, I just did the same with Destrio for the first part
    (1) -----> mg = fs = 156.8N
    and we also have that fs = us * Fc
    ----> Fc = fs / us = 156.8 / 0.38 = 412.6 N

    From here, we can see that there are two forces acting on m, not one like you stated Doc. But there is only ONE force acting on M. So, am going to use the equation (3) to find acceleration.
    From (3) we have a = Fc / M = 412.6 / 88 = 4.68 m/s^2
    Now I have acceleration, I can plug it in (2) to find F

    F = Fc + ma = 412.6 + 16(4.68) = 487.48 N

    I'm pretty sure that I did it right, but since am a noob when it comes to physics, I hope that someone else could point out my mistakes. Thanks a lot.
     
  16. Oct 6, 2011 #15

    Doc Al

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    You are absolutely correct. I mixed up my 'm' and 'M' in my previous post. Good catch!
     
  17. Oct 6, 2011 #16

    Doc Al

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    Just for the record, I'm going to point out my mistakes in my post #10 above.

    Correct.

    Correct.
    Oops! That should be a small m, of course:
    Fa - Fn = ma

    Or, considering the big mass alone:
    Fn = Ma

    Looks who's talking! :redface:

    That's true.

    OK.

    Oops... once again I mixed up my m's. It should be:
    [tex]mg/\mu = Ma[/tex]

    This should be:
    [tex]a = \frac{mg}{\mu M}[/tex]

    Thanks again to theunloved!
     
  18. Oct 6, 2011 #17
    I should be the one to say thanks to all your hard work Doc, not the other way around ;)
     
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