- #1
Rasalhague
- 1,387
- 2
If pressure at some point on a surface is defined, as here ( http://en.wikipedia.org/wiki/Pressure ), as
[tex]\frac{\mathrm{d}||F_n||}{\mathrm{d}A}[/tex]
(apparently meaning the rate of change wrt area of the magnitude of force normal to the specified surface), why is pressure at a point not, in general, 0? How does the force at a point in space depend on the size of an area, given that no area appears in the definition of force
[tex]\textbf{F} = \frac{\mathrm{d} \textbf{p}}{\mathrm{d} t}.[/tex]
Likewise, a stress vector (traction vector) for a given surface is defined here ( http://en.wikipedia.org/wiki/Stress_(physics) ) as
[tex]\frac{\mathrm{d} \textbf{F}}{\mathrm{d}A},[/tex]
but, given the force acting at a point, how is the force at that point changed at all by changing the value of [itex]A[/itex] in the equation?
Alternatively, if the magnitude of the force perpendicular to the an area is denoted by [itex]\Delta F[/itex] (Fishbane et al.: Physics, 2nd ed., p. 436), and pressure p defined thus:
[tex]p = \lim_{\Delta A \rightarrow 0} \frac{\Delta F}{\Delta A} = \frac{\mathrm{d} F}{\mathrm{d} A},[/tex]
why isn't pressure always infinite at every point, given that for any choice of point, as one choses a smaller area to divide it by, that choice should have no affect on the magnitude of a vector belonging to a force field, nor on its normal component, since what is changing is only the area we're choosing to divide the magnitude of the force by, and not the orientation of the surface.
[tex]\frac{\mathrm{d}||F_n||}{\mathrm{d}A}[/tex]
(apparently meaning the rate of change wrt area of the magnitude of force normal to the specified surface), why is pressure at a point not, in general, 0? How does the force at a point in space depend on the size of an area, given that no area appears in the definition of force
[tex]\textbf{F} = \frac{\mathrm{d} \textbf{p}}{\mathrm{d} t}.[/tex]
Likewise, a stress vector (traction vector) for a given surface is defined here ( http://en.wikipedia.org/wiki/Stress_(physics) ) as
[tex]\frac{\mathrm{d} \textbf{F}}{\mathrm{d}A},[/tex]
but, given the force acting at a point, how is the force at that point changed at all by changing the value of [itex]A[/itex] in the equation?
Alternatively, if the magnitude of the force perpendicular to the an area is denoted by [itex]\Delta F[/itex] (Fishbane et al.: Physics, 2nd ed., p. 436), and pressure p defined thus:
[tex]p = \lim_{\Delta A \rightarrow 0} \frac{\Delta F}{\Delta A} = \frac{\mathrm{d} F}{\mathrm{d} A},[/tex]
why isn't pressure always infinite at every point, given that for any choice of point, as one choses a smaller area to divide it by, that choice should have no affect on the magnitude of a vector belonging to a force field, nor on its normal component, since what is changing is only the area we're choosing to divide the magnitude of the force by, and not the orientation of the surface.