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Slippery safe problem

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data
    There must be a thousand variations on this but my step daughter brought this to me after i volunteered to help her with her Physics I class.

    Burglar wants safe--decides to ice the floor thinking this will help him push it out the door. Coefficient of friction same for buglar and safe, both static and kinetic. Applies a fixed force at 30 degrees off vertical. I've change the constants provided, as I just want to make sure I understand principles.
    Mass(burglar)=60Kg Mass Safe=600Kg Force applied=1200N MUs=0.2 MUk=.15

    Does safe move, burglar or both, if so at what rate(s) of acceleration? If not what force is required at this angle to move the safe?


    2. Relevant equations

    F=ma, Frictional F=Normal*(mu)



    3. The attempt at a solution

    My thinking was that the force applied somewhat vertically helped in his efforts--it lightened the safe, reducing its Frictional force, while the reactive component helped dig in his heels (cramp-ons were not allowed). So I first computed the frictional forces.

    If I'm thinking about this correctly,
    frictional force for the burglar is his weight plus the vertical component of the reactive force, times Mu,

    Making g=10m/s^2

    that would be (60*10 + cos(30)*1200))*0.2=327.8

    for the safe ((600*10 minus cos(30)*1200))*0.2=992.2

    Since the horizontal component of his force is sin(30)*1200=600

    it would not be enough to overcome the static friction of the safe, yet is larger than his frictional force, so he would accelerate backwards while the safe wouldn't budge. And that would just be the two differences in the forces (after adjusting the frictional force with the kinetic MU)/mass.

    In order to get the safe to move:

    6000*0.2/(.2*cos(30)+sin(30))=1782 N
    At which point his frictional force will have increased to 482, which is still less than the reactive horizontal force, so he will still go backwards. And that the only way for this not to happen is to make the push more vertical still. This just doesn't seem intuitive.

    Am I all wet here? :confused:
    JS
     
    Last edited: Jan 27, 2007
  2. jcsd
  3. Jan 27, 2007 #2

    cristo

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    Before I try and plough through the maths, can I just ask one thing: is the burgular pushing or pulling the safe? Your first few lines imply pulling (since you say this will balance out the weight of the safe) but in the last line you talk about the burgular pushing the safe. Which one is correct?
     
  4. Jan 27, 2007 #3
    Pushing the safe, sorry if that was unclear.
     
  5. Jan 27, 2007 #4

    cristo

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    In that case, I think you reaction force for the safe is incorrect. To see this, draw a diagram of just the safe, and the forces acting upon the safe. You have the weight 6000 acting vertically downwards, and the force of the burglar pushing will be acting down on the safe at an angle of 30 to the vertical. Thus, the vertical component of this force will be added to the weight of the safe when calculating the normal reaction force.
     
  6. Jan 27, 2007 #5
    The push is directed upward on the safe, tho. So I ges I am confused, I thought this would reduce the Normal force on the safe and increase it on the burglar. In other words if the burgler was under the safe and pushing upwards the safe would "weigh" less and there would be less friction.
     
  7. Jan 27, 2007 #6

    cristo

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    Ahh, ok. I was imagining a situation of the burglar pushing downwards at an angle of 30 to the safe (i.e. the safe was much smaller than him) , but if he is pushing upwards, then you are correct in saying that the vertical component of this force will make the normal contact force of the safe on the ground less.

    Ok, now I have the situation right in my head, I'll have a read through your maths, and get back to you.
     
  8. Jan 27, 2007 #7
    Thanks, I should have drawn a diagram, but wasn't sure how as my scanner is dead.
     
  9. Jan 27, 2007 #8

    cristo

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    Well, I've had a look through, and quite frankly, I'm a little confused myself! I'll blame this on the fact that it's quite late! So, since I've been totally useless, I'll leave a note in the homework helpers forum, so hopefully someone who can help will drop by! Sorry I couldn't be of more help!
     
    Last edited: Jan 27, 2007
  10. Jan 28, 2007 #9
    just trying to keep this on the radar.
    John s
     
  11. Jan 28, 2007 #10

    AlephZero

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    That's right. There is no limit to the force pushing vertically up, because there is no the side force and no friction force, so he won't slip.

    For no slipping, the resultant of the burglars weight and the pushing force must be in within the friction angle of tan-1(0.2) or about 11.3 degrees to the vertical. Draw a diagram of the two vectors and the resultant, and it's obvious that as the push force increases it must be nearer the vertical, to keep the resultant in the "no-slip" part of the diagram.
     
  12. Jan 28, 2007 #11
    Sure, the arctan that comes up in inclines all the time. Same issues here. Thanks that was insightful. Is the rest of the solution OK?
     
  13. Jan 28, 2007 #12

    Gokul43201

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    Haven't read through real carefully but it looks okay. Why do you think requiring a more vertical push is unintuitive? After all, that is the easy way to increase the friction on the burglar, while decreasing the friction on the safe.
     
  14. Jan 28, 2007 #13
    AZ,
    I figured since i was tutoring, making sure my reasoning was solid, a good use of the forum. These kind of problems have always been tougher for me than others, and she flunked the course last go. I guess i want to make sure she's not digging a deeper hole with me than with a paid tutor. I didn't want to violate any rule to be sure, it was my first use of the forum, and hope I have paid back by helping along a couple others. As to the burglar, if he could muster the force, he is practically lifting it and the ice only gets in his way. He should have brought a portable winch.
    John
     
  15. Jan 28, 2007 #14

    Gokul43201

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    This is true.

    Also, if this helps any, I'm going to rewrite the same calculation you did above but with some additional pointers.

    For the burglar to stay put (rather than slip backwards), you want:
    [tex]\mu_s (mg + Fcos \theta) > Fsin \theta [/tex]

    If you increase F at fixed [itex]\theta[/itex] so that F is really large compared to mg (physically unrealistic, but for the sake of argument), then, at best, you get the condition
    [tex]\mu _sFcos \theta > F sin \theta ~ \implies~ tan \theta < \mu _s [/tex].
    So even if you indefinitely increase F, you get nowhere unless the angle with the vertical is small (specifically, smaller than the angle of friction, [itex]tan ^{-1}( \mu _s) [/itex]).

    For the safe to slide, you need:
    [tex]\mu_s (Mg - Fcos \theta) < Fsin \theta [/tex]

    Clearly, this condition is satisfied for any push angle, so long as F is big enough (hardly an easy task) that the left hand side be made as close to zero as desired.

    This tells you that the constraint isn't so much from making the safe move, as it is from keeping the burglar stationary.

    There's a third outside-the-strongbox solution (not applicable at this level). Why does the burglar have to be stationary? There's nothing saying the burglar shouldn't recoil, if in the process he gets the safe to move forward.
     
    Last edited: Jan 28, 2007
  16. Jan 29, 2007 #15
    AZ,

    Thanks again. I see that all your posts have the same links, so I misinterpreted. And on reflection, agree that the best approach would be to apply all the push horizontally just as one intuitively does when pushing out a car stuck in the snow, an event that has been happening here quite a bit as of late, along with an occasional recoil.
     
  17. Jan 29, 2007 #16

    cristo

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    Just to confirm; the links to the rules are in Gokul's signature, and are indeed not implying you violated the forum rules. In fact, helping in other threads is just the spirit appreciated here at PF! Good luck with the tutoring!
     
  18. Jan 29, 2007 #17
    Right I realized that later re confusing Gokul's with AZ, thanks--I used to get a lot of satisfaction out of tutoring, if not much $$, and she's a good kid, just doesn't have a real solid foundation in trig and calculus, which for engineering physics at a place like school of mines is a real impediment we will have to overcome. But I think we'll do it, she is plenty motivated and working on problems not due for another week--thats one skill I never had.
    John
     
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