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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?

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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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?
similar to the hovering flight we discussed in Chapter 6, but our approach to the problem will be different. Because a fraction f of the animal is submerged, the animal is buoyed up by a force FB given by FB = 8f Vpu (7.9) where py is the density of water. The force Fg is simply the weight of the displaced water. The net downward force FR on the animal is the difference between its weight 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 to Fp. This force can be produced by pushing the limbs downward against the water. This motion accelerates the water downward and results in the upward reaction force that supports the animal. If the area of the moving limbs is A and the final velocity of the accelerated water is v, the mass of water accelerated per unit time in the treading motion is given by (see Exercise 7-1) m = Avpu (7.11) Because the water is initially stationary, the amount of momentum imparted to the water each second is mv. (Remember that here m is the mass accelerated per second.) Momentum given to the water per second = mv 88 Chapter 7 Fluids This the rate of change of momentum of the water. The force producing this 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 by FR = Fp = gV(p - fPw) = mv (7.12) Substituting Eq. 7.11 for m, we obtain Pw Av? = gV (p- SPu) or gV (p- fpw) (7.13) Apu The 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 the product of the mass accelerated each second and the squared final velocity of water. This kinetic energy imparted to the generated by the limbs; that is. each second is the power KE/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) Apw Here 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 to keep her nose above water. Note that, in our calculation, we have neglected the kinetic energy of the moving limbs. In Eq. 7.14 it is assumed that the density of the animal is greater than the density of water. The reverse case is examined in Exercise 7-3. 7.6 Buoyancy of Fish The bodies of some fish contain porous bones or air-filled swim bladders that
Transcribed Image Text:similar to the hovering flight we discussed in Chapter 6, but our approach to the problem will be different. Because a fraction f of the animal is submerged, the animal is buoyed up by a force FB given by FB = 8f Vpu (7.9) where py is the density of water. The force Fg is simply the weight of the displaced water. The net downward force FR on the animal is the difference between its weight 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 to Fp. This force can be produced by pushing the limbs downward against the water. This motion accelerates the water downward and results in the upward reaction force that supports the animal. If the area of the moving limbs is A and the final velocity of the accelerated water is v, the mass of water accelerated per unit time in the treading motion is given by (see Exercise 7-1) m = Avpu (7.11) Because the water is initially stationary, the amount of momentum imparted to the water each second is mv. (Remember that here m is the mass accelerated per second.) Momentum given to the water per second = mv 88 Chapter 7 Fluids This the rate of change of momentum of the water. The force producing this 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 by FR = Fp = gV(p - fPw) = mv (7.12) Substituting Eq. 7.11 for m, we obtain Pw Av? = gV (p- SPu) or gV (p- fpw) (7.13) Apu The 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 the product of the mass accelerated each second and the squared final velocity of water. This kinetic energy imparted to the generated by the limbs; that is. each second is the power KE/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) Apw Here 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 to keep her nose above water. Note that, in our calculation, we have neglected the kinetic energy of the moving limbs. In Eq. 7.14 it is assumed that the density of the animal is greater than the density of water. The reverse case is examined in Exercise 7-3. 7.6 Buoyancy of Fish The bodies of some fish contain porous bones or air-filled swim bladders that
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