Question
what is dx ,dy and ds in this and how depth is 1 in this
PASCAL'S LAW According to Pascal's Pascal's law states that in a statie fluid, the iatensity of pressureaw, the pressure at a point in afluid at rest is of the orientation of the at a point is same in all directions. Proot Comsider a small wedge shaped ee in a aidat omel independent rest as shown in the Fig. 3.1. The forces acting on the element are (i) Weight of the element acting vertically downward. (ii) Pressure forces acting normal to the surface. g vertically downward.ufacebout the P, HYDROSTATIC PRESSURE P pressure acting on face AB p, pressure acting on face AC. P- pressure acting on face BC w- specifie weight of the liquid. Force on face AB p,x area of face AB p,x (dy x 1) p,dy Force on face Force on face BC-p,x area of face BC,-P,x (ds x 1) pds Weight of fluid element-specific weight volume ACspyx area of face AC-p, x (drx1) w= w x (area of triangular element x depth) (dx x dy x 1) 2 x As the fluid element is in equilibrium, EF, Algebraie sum of all the forces acting on x-direction 0 i.e., prdy-p,ds sin =0 i.e. pdy pads sine P‘ ds sin e is the horizontal component of force on face Ba dysd8 sin ] (3.3) PrP F Algebraic sum of all the forces along y-direction 0
Explanation / Answer
In this proof, we need to take an element with wedge shape. The sides of the elemental wedge is dx, dy and ds. And we are considering unit depth or simply thickness of the wedge. We know, force= pressure× area. Here , area = length of side × thickness (1 unit).
