In Eq. 7.14, it is assumed that the density of the animal is greater than thedensity of the fluid in which it is submerged. If the situation is reversed,the immersed animal tends to rise to the surface, and it must expendenergy to keep itself below the surface. How is Eq. 7.14 modified forthis case? similar to the hovering flight we discussed in Chapter 6, but our approach tothe problem will be different.Because a fraction f of the animal is submerged, the animal is buoyed upby a force FB given byFB = 8f Vpu(7.9)where py is the density of water. The force Fg is simply the weight of thedisplaced water.The net downward force FR on the animal is the difference between itsweight g Vp and the buoyant force; that is,Fp = gVp- gVfpw= gV(p- fPw)(7.10)To keep itself floating, the animal must produce an upward force equal toFp. This force can be produced by pushing the limbs downward against thewater. This motion accelerates the water downward and results in the upwardreaction force that supports the animal.If the area of the moving limbs is A and the final velocity of the acceleratedwater is v, the mass of water accelerated per unit time in the treading motionis given by (see Exercise 7-1)m = Avpu(7.11)Because the water is initially stationary, the amount of momentum imparted tothe water each second is mv. (Remember that here m is the mass acceleratedper second.)Momentum given to the water per second = mv88Chapter 7 FluidsThisthe rate of change of momentum of the water. The force producingthis change in the momentum is applied to the water by the moving limbs.The upward reaction force FR, which supports the weight of the swimmer,is equal in magnitude to Fp and is given byFR = Fp = gV(p - fPw) = mv(7.12)Substituting Eq. 7.11 for m, we obtainPw Av? = gV (p- SPu)orgV (p- fpw)(7.13)ApuThe work done by the treading limbs goes into the kinetic energy of the accel-erated water. The kinetic energy given to the water each second is half theproduct of the mass accelerated each second and the squared final velocity ofwater. This kinetic energy imparted to thegenerated by the limbs; that is.each second is the powerKE/sec = Power generated by the limbs,P=Substituting equations for m and v, we obtain (see Exercise 7-1)w(1-4)P=(7.14)ApwHere W is the weight of the animal (W = gVp).It is shown in Exercise 7-2 that a 50-kg woman expends about 7.8 W tokeep her nose above water. Note that, in our calculation, we have neglectedthe kinetic energy of the moving limbs. In Eq. 7.14 it is assumed that thedensity of the animal is greater than the density of water. The reverse case isexamined in Exercise 7-3.7.6 Buoyancy of FishThe bodies of some fish contain porous bones or air-filled swim bladders that

The density of the animal ρ is less than the density of the water ρw ρ<ρw The body naturally…

In Eq. 7.14, it is assumed that the density of the animal is greater than the density of the fluid in which it is submerged. If the situation is reversed, the immersed animal tends to rise to the surface, and it must expend energy to keep itself below the surface. How is Eq. 7.14 modified for this case?

In Eq. 7.14, it is assumed that the density of the animal is greater than the density of the fluid in which it is submerged. If the situation is reversed, the immersed animal tends to rise to the surface, and it must expend energy to keep itself below the surface. How is Eq. 7.14 modified for this case?

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter8: Centroids And Distributed Loads

Section: Chapter Questions

Problem 8.113P: Calculate the resultant force caused by the water acting on the parabolic arch dam. The water levels...

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