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Force required to maintain Equilibrium

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    My problem here is that I keep trying to make substitutions for the normal forces and cannot come up with the number for P. I saw 4.92 and don't understand how there is a range of force for which equilibrium can be maintained. The force I came up with for P is way too big, the answer is 0.955 lb
     

    Attached Files:

  2. jcsd
  3. Sep 23, 2013 #2
    You are assuming that the forces of friction are ## F = \mu N ##. That is not correct for static friction. In statics, the force of friction can be anywhere in the range ##[0, \ \mu N]##.
     
  4. Sep 23, 2013 #3

    SteamKing

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    Just to save you a little calculation, it's perfectly OK to calculate moments in units of inch-pounds rather than converting all distances to feet.
     
  5. Sep 23, 2013 #4
    So in this problem, where am I going wrong as far as why the number isn't coming out right? I believe my moment sign convention is consistent, and I am doing the necessary calculations to solve for N_1 and N_2, yet it doesn't come out right. Anyone know where it went wrong?

    Voko, if the domain can be between [0,μN], then why doesn't the minimum force P=0?
     
  6. Sep 23, 2013 #5
    One problem is that the last term in your moment equation involves ##N_1##. How that could happen for moments about B is a mystery.

    Another problem, as I already said, is that you assume the forces of friction are at their maximum.
     
  7. Sep 23, 2013 #6
    Good catch with the N_1, I see the mistake. I am not sure though what you mean about the assumption that the forces of friction are at their maximum. How do I mathematically express the minimum force if I know the maximum force is F=μN?
     
  8. Sep 23, 2013 #7
    You could introduce two independent friction forces, and inequalities for their range. That will give you inequalities for P.
     
  9. Sep 23, 2013 #8
    What do you mean by that? I don't understand it
     
    Last edited: Sep 23, 2013
  10. Sep 23, 2013 #9
    I do not understand how the force of friction at A can be directly replaced with some expression involving the normal force at B - before the rest of the equations are written and analysed at the very least.

    Anyway, as I said earlier, you should treat the forces of friction independently of the normal forces, so the whole issue is moot.
     
  11. Sep 23, 2013 #10
    How do I treat them independently of the normal force? Are you saying the normal force at A doesn't cause a moment at point B? The force of friction is directly proportional to the normal force by the factor of the coefficient of friction. How else can I express the force of friction other than F=μN?

    If the range is [0,μN], why isn't the minimum force just zero?
     
    Last edited: Sep 23, 2013
  12. Sep 23, 2013 #11
    Just use symbols ##F_A## and ##F_B## for the forces of friction. Do not equate them with ##\mu N##. Instead, use inequalities ## 0 \le F \le \mu N ##.
     
  13. Sep 23, 2013 #12
    Hmm...I just thought of something, are you meaning to say that I should just consider the forces of friction at points A and B as zero, thus only finding the moments from the factors of W and N? Or are you saying since F_a is zero, then N is zero as well??
     
    Last edited: Sep 23, 2013
  14. Sep 23, 2013 #13
    I do not mean any of that. Re-read my previous message.
     
  15. Sep 23, 2013 #14
    I'm looking at it. Using the inequalities, I get that. However, what does that means as far as the minimum force of P required? Does the minimum force of P occur when the frictions are maximums? That should mean that since the friction is a maximum, the force P should not have to exert as much to keep the pole from staying on the wall, since the friction would be doing most of the work.
     
  16. Sep 24, 2013 #15
    Still not sure how to incorporate the inequalities into my strategy for solving the problem. I have 4 equations and 5 unknowns (N_1, N_2, F_a, F_b, P), so this thing is not solvable with what I have. Is taking a moment around A and B to get 4 equations (2 from the horizontal and vertical forces, and 2 equations from the moments at point A and B) valid, or is there a redundancy and the moments at A and B are not independent?
     

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  17. Sep 24, 2013 #16
    I said earlier that the forces of friction are in the range ##[0, \ \mu N]##. I now think that was wrong, because that assumes a particular direction. I think the correct range is ##[-\mu N, \ \mu N]##.

    As for your question, you have 5 unknowns and 3 equations. That means three unknowns can be expressed as functions of two unknowns. In particular, everything, including ##P##, can be expressed as a function of the friction forces. Then, using the inequalities on the friction forces, you can find the range of ##P##.
     
  18. Sep 24, 2013 #17
    Problem has been solved, thank you.
     
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