Still another way of stating pascals law is that the

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Unformatted text preview: till another way of stating Pascal's law is that the pressure is the same for any two points at the same elevation in a continuous body of fluid. The following figure illustrates this point. 2 Hydrostatics.nb Stress in Hydrostatic Systems The mathematical statement of Cauchy's First Law is a statement for the balance of linear momentum of a fluid body Ρv V t m Ρb V t m t t m n (1) A t We showed in a previous set of notes that the stress vector t n at a point in a fluid can be expressed in terms of the stress tensor T: t nT n (2) where the stress tensor T is defined as a sum of dyadic products T Tij ei ej For a fluid at rest Eqn. (1) becomes Ρb V 0 m t t m n A (3) t And since a fluid at rest cannot support any shear forces, the stress vector becomes t pn n (4) where p is the hydrostatic pressure. To show that the pressure is the same in every direction we consider a small tetrahedron shown in Figure 2 D C A B Our tetrahedron has 3 mutually orthogonal faces (ADC, ABD, ABC) with areas x, y, z , and a slant face (BDC) with area . The outward directed unit normals to these 4 faces are -i, -j, -k, and n. We apply (4) to our tetrahedron to get Ρg V 0 m t t m n A (5) t Since the volume of our tetrahedron is small we express the volume integral in terms of average values: Hydrostatics.nb Ρg V Ρg V (6) t m Let the stress acting on the face with outward directed normal -i be t with normals -j and -k be t t m A n t t and t j t i where we have used the fact that t related to the area of the slant face in jn kn z j t t y j py x , and in a similar way on faces A k t n A z t x i px y A j y t i . Thus the surface integral in (6) can be written as k A i x x 3 y k pz k t z z (7) n n pn p n . The surface areas of the 3 mutually orthogonal faces rae by simple projections nx ny nz (8) Thus our balance of hydrostatic forces becomes Ρg 0 V n pn Now we divide through by 0 n pn i nx px j ny py and take the limit V/ i nx px j ny py k nz pz (9) 0 to get k nz pz (10) Recall that n i nx j ny k nz (11) Substituting for n in (11) we get 0 i nx pn px j ny pn py k nz pn pz (12) If we take the scalar product of (13) with i , j , k respectively we get pn px 0, pn py 0, pn pz 0 (13) Hence the stress vector on any arbitrary surface in a fluid at rest is given by t pn n (14) Note that the stress vector t n acts in a direction opposite to that defined by the unit normal...
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