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Unformatted text preview: 5.40. CHAPTER 5, PROBLEM 40 633 5.40 Chapter 5, Problem 40 Problem: A truck carries a tank of water that is open at the top with length , width w and depth h. If the truck is half full and = 5h, what is the maximum acceleration, a, that can be sustained without spilling any water? Assume constant acceleration, and note that the pressure is constant and equal to its atmospheric value at the free surface. . ..................... ...................... ... . ... . ................... .................. w..................................................................................................................................................................................................................................................................................................................... ´ ... ´ ´ . ................................................................................................................................................................ .....´................................................................................................................................................................................................ .. .................. .. .... ... .... ........................................................................................................................... .................. . .............. ........... .´...................................................................................................´................................. . .... ................................................................................................................................................................................................................. ....... ............................... ............................................................ ...... .... ..... . ...... ...... .................... ...................´............................................................................................................................................ .. .....´..................... .......... ......................... . .......... .................... ...................... ...... .......................... ......................................................... .... ..... ..... .. ´.............................. ...................... . ..................... . . . .... .................´................................................................................................................................. ................. . ............ ............................... ............... .................... ............. . .................................................................. ...................................................... ....................... ..... ..... ........ . .. .... .´...................................................................................................................´...................................................s .................................. . . ................. . ................ ............. . .............................. . . .......... ............... . ............................................................................................................................................................................c .................................... .... ................. .... ......... .... .............. . s. ........... .............................................................................................................................................................................. . .. . . ............... ..... . . c. . ... ..... ................................................ ... .......................................................................................................................................................................................................................................................................... . . . . ..... .. .. . . ........ .......... . . . .. . ........... . ......................... . . ...... .. ................................................................................................................................................................................................................... . .. .. . ...... ... ............................................... .. ... ............. ...... ......................................... . . ....... .. . ...................................................................................................................v...v......... . . . ... .... ..... ....... ........... . . .. .................................................................... .......... ................ . ...f...f... .......................................................................•...............................•.........•......... f v f v. • . .... .... ... . . . .. .. . . . . . . h . . . . . .. . .. . . . . . . . . . .. . . . ho . .. .. . . . . . . . Not moving x=a ¨ ´ ´ ´ aaa ´ ´ aa´ aa a a aa c s aa c s aa zy ff vv f v f v •• • • x Accelerating ......... ......................................................................................................... ......................................................................................................................................... . .. ....... ....... ..... .................................................................................................................................................................... ................ ............. ... ......... .... .. . ................................................. ...... ... .. ... ... ................. .. ........................ ...................... . ..... ........................................................................................................................................................................... ...................................................................................................................................................................................................... .. . .. . ................................................... .................................................................................................................................................................. ... . ......................................... . .... . .................... ......... .... ............................................................................................ ... ................................... ... . .................................................................................................................................................................... .................... . .......................... . .. . .. ..................... . ........... . ....................................... ................... ......... . ............................................................................................................................................................................................ ................. . ......................... . ................... .............. . ............ . ............................................. . .. ..... ............................................................................. ...................... ............. ....... .. . . ... ............................................................................................................................................................ . .............................................................. .. ............................... . ...................... .. . . ................ ............ . .......... ....... ..................................... ............ . ................. ......... ... ........................................................................................................................................................................................ . ......... . . . . . ................. . . . . .. . . . . ........................................................................... .... . .............................................................. . ........................ ...... ................................................ ............................................................................................................................................................................................................................................. ...... ........... . .... . . . . . . .. . . . .. . .. ... .. ................ . ............ . ..... ...................................................................................................................... ...................................................................................................................................... .. ...... ....................................................................................................................................... . .. . .. . . .. . . ... ................................................................................................................................... . .. . . . ....... ..... . . . .. .......................... ... ......... .. .. ............. ... . . ................................................................................................ .......... . ...... . . Solution: When the truck accelerates, the three components of the Euler equation are ∂p ∂x ∂p 0=− ∂y ∂p 0 = − − ρg ∂z ρa = − Clearly, since ∂ p/∂ y = 0, the pressure is at most a function of x and z . Integrating first over x, p(x, z ) = −ρax + f (z ) where f (z ) is a function of integration. Then, differentiating with respect to z and using the z component of Euler’s equation, we find ∂p = f (z ) = −ρg ∂z =⇒ f (z ) = C − ρgz where C is a constant. Thus, the pressure throughout the fluid in the truck is p(x, z ) = C − ρax − ρgz On the free surface, p = pa . So, the equation of the free surface is C − ρax − ρgz = pa 634 CHAPTER 5. MASS AND MOMENTUM PRINCIPLES Now, if we let x = 0 at the back of the truck, we will have z = h at x = 0 when the fluid is at the brink of spilling (see figure below). Hence, C = pa + ρgh and ax + gz = gh . . . . .. . . . . . . . . . . . . . . . . . h . . . . . . . . . . . . . . . . . .. .. . . ..................................................................................................................................................................................... .......................... ..... . .............................................................................................................................................................................................................................................................................................................................................. . .. .. ....... ....................... ..... ...... ..... ..... ....... 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.................................................................................................................................................................................................. x=0 . . .. .. . . . . . H . . . . . .. .. . . x= To evaluate H , note that =⇒ a + gH = gh H =h− a g To conserve mass, the volume of liquid must remain constant. Before any motion occurs, the volume is V = who During acceleration, the fluid in the tank has the shape of a rectangular parallelepiped for which the volume is 1 V = w(H + h) 2 Thus, we must have who = 1 w(H + h) 2 =⇒ 1 ho = (H + h) 2 So, substituting for H from above, ho = But, we are given ho = 1 h and 2 1 a a h− +h =h− 2 g 2g = 5h. Substituting these values gives 1 5ah h=h− 2 2g Thus, the acceleration is =⇒ 1 a= g 5 1 5a = 2 2g ...
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