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Pressure is the change in force per unit of area

  1. Oct 3, 2004 #1
    i read somewhere that pressure is the change in force per unit of area and can be represented as the derivative
    P(A) = dF/dA
    but in order for it to be the derivative shouldn't it be
    P(A) = dF/dA
    lim dA -> 0
     
  2. jcsd
  3. Oct 3, 2004 #2
    no in order for it to be a d/dx the orginal formula would be lim A -> 0 F/A thus A becomes dA...etc
     
  4. Oct 3, 2004 #3

    arildno

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    The expression P(A)=dF/dA is rather misleading, since basically, through the limiting procedure, we find the pressure at a point, not on some area.
     
  5. Oct 3, 2004 #4

    T@P

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    er p(a) = dF/dA is not a limit... its really another notation for the change in force per unuit of area. its just like any dx/dy (if you've done it before).
     
  6. Oct 3, 2004 #5
    i don't quite understand this :grumpy:
    a derivative of a function is defined as

    dy/dx
    lim dx ->0

    for example if the function is y=x^2 then the derivative is
    (x1^2-x2^2)/(x1-x2)
    lim x2 -> x1
    or
    y'=2x1
    if (x1,x1^2) and (x2,x2^2) are points on the function graph

    is this correct? or am i confusing something (i probably am)
     
    Last edited: Oct 3, 2004
  7. Oct 3, 2004 #6

    arildno

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    Given a point "a", a sequence of surfaces including "a", [tex]A_{n}[/tex], (which decreases to a as n goes to infinity) with associated forces acting upon them, [tex]F_{n}[/tex],
    we define the pressure at a as:
    [tex]p(a)=\lim_{n\to\infty}\frac{F_{n}}{A_{n}}=\frac{dF}{dA}[/tex]

    This should really be done with stresses and oriented surfaces, but it shows that the argument in the pressure function is a point, not some area (which p(A) might misleadingly be represented as).

    (It goes without saying that p is to be the limit for every chosen surface sequence which includes "a")
     
    Last edited: Oct 3, 2004
  8. Oct 3, 2004 #7
    this is a bit new to me i only studied basic analysis
    i always thought dx is defiend as the difference between two function variables

    (for example if the function is f(x) = x^2 then dx = x1 - x2 and dy = x1^2 - x2^2)
     
  9. Oct 3, 2004 #8

    arildno

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    Very good, Anzas!
    However, since an area is a two-dimensional structure (not only a one-dimensional length), the maths get a bit trickier..
    FIRST:
    If you have a curve and wants to find the slope (tangent value) at some point, you may do this by
    a)finding the average slopes given by secant values to the curve (where one of the points defining the secant is the one you want to find the slope at)
    AND THEN
    b) "Shrinking" the distance between the secant points to zero, i.e, computing the limit known as the derivative (that is the tangent slope)
    Agreed?
     
  10. Oct 3, 2004 #9

    arildno

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    SECOND:
    Compare this "average" and "shrink" technique to how we define pressure with the help of forces&areas upon which these forces act.
     
  11. Oct 4, 2004 #10
    yes the "average" and "shrink" method is actually describing in the words the limit lim dx -> 0
    i was thinking is it not possible simply making a function of force which has a variable of an area like this
    f(a)
    then the derivative function f(a1) - f(a2) / a1 - a2
    lim a2 -> a1
    or
    df(a)/da
    lim da -> 0

    f'(a)=p(a)
    is the pressure function
     
    Last edited: Oct 4, 2004
  12. Oct 4, 2004 #11
    anyone? :grumpy:
     
  13. Oct 4, 2004 #12

    arildno

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    1.Let's think of a piece A0 of a plane, shall we? (keeping it simple)
    2. Now, let us say that there acts a net force F0 normal on that plane.
    3. The AVERAGE pressure on A0 can then be written as F0/|A0|, where |A0| is the area of A0.
    Agreed thus far?
     
  14. Oct 4, 2004 #13

    arildno

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    Just a note:
    Calling that AVERAGE pressure P0, We get P0|A0|=F0
    (direction along the normal)
    4. Now, of course, the local distribution of force on A0, might vary from point to point.
    5. Let's therefore look at a piece A1 lying inside A0.
    Some force F1 works on A1, we may define an AVERAGE pressure P1=F1/|A1|
    6. Now, what happens when we continue this process?
    We then end up at some point "a" (lying inside all the An's), which can be given an associated pressure P(a), the limiting value of the ratios Fn/|An| as n goes to infinity.
     
  15. Oct 5, 2004 #14
    i see now, thanks for your help :smile:
     
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