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182 PHYSICS ⊳ this element of area corresponding to the normal Example 9.1 Thetwothighbones(femurs), 2 forces F a Fband F ac shown in Fig. 9.2 on the each of cross-sectional area10 cm support faces BEFC, ADFC and ADEB denoted by A , A a b the upper part of a human body of mass 40 and A respectively. Then kg. Estimate the average pressure c Fbsinθ = F c F bosθ = F a (by equilibrium) sustained by the femurs. Absinθ = A c A bosθ = A (ay geometry) Thus, Answer Total cross-sectional area of the 2 –4 2 F F F femurs is A = 2 × 10 cm = 20 × 10 m . The b = c = a; Pb= Pc = Pa force acting on them is F = 40 kg wt = 400 N Ab Ac A a (9.4) –2 (taking g = 10 m s ). This force is acting Hence, pressure exerted is same in all vertically down and hence, normally on the directions in a fluid at rest. It again reminds us femurs. Thus, the average pressure is that like other types of stress, pressure is not a F 5 −2 vector quantity. No direction can be assigned Pav =A = 2×10 N m ⊳ to it. The force against any area within (or bounding) a fluid at rest and under pressure is 9.2.1 Pascal’s Law normal to the area, regardless of the orientation of the area. The French scientist Blaise Pascal observed that Now consider a fluid element in the form of a the pressure in a fluid at rest is the same at all horizontal bar of uniform cross-section. The bar points if they are at the same height. This fact is in equilibrium. The horizontal forces exerted may be demonstrated in a simple way. at its two ends must be balanced or the pressure at the two ends should be equal. This proves that for a liquid in equilibrium the pressure is same at all points in a horizontal plane. Suppose the pressure were not equal in different parts of the fluid, then there would be a flow as the fluid will have some net force acting on it. Hence in the absence of flow the pressure in the fluid must be same everywhere in a horizontal plane. 9.2.2 Variation of Pressure with Depth Fig. 9.2 Proof of Pascal’s law. ABC-DEF is an Consider a fluid at rest in a container. In element of the interior of a fluid at restFig. 9.3 point 1 is at height h above a point 2. This element is in the form of a right- angled prism. The element is small so thatThe pressures at points 1 and 2 are P and1P 2 respectively. Consider a cylindrical element of the effect of gravity can be ignored, but fluid having area of base A and height h. As the has been enlarged for the sake of clarity. fluid is at rest the resultant horizontal forces Fig. 9.2 shows an element in the interior of a should be zero and the resultant vertical forces should balance the weight of the element. The fluid at rest. This element ABC-DEF is in the form of a right-angled prism. In principle, this forces acting in the vertical direction are due to prismatic element is very small so that every the fluid pressure at the top (P A) ac1ing downward, at the bottom (P A) acting upward. part of it can be considered at the same depth 2 from the liquid surface and therefore, the effect If mg is weight of the fluid in the cylinder we of the gravity is the same at all these points. have (P − P ) A = mg (9.5) But for clarity we have enlarged this element. 2 1 The forces on this element are those exerted by Now, if ρ is the mass density of the fluid, we the rest of the fluid and they must be normal to have the mass of fluid to be m = ρV= ρhA so the surfaces of the element as discussed above. that Thus, the fluid exerts pressures P , P and P on P2−P 1 ρgh (9.6) a b c Reprint 2026-27