Pressure in Accelerated Rigid-Body Motions

We now repeat the derivation of Eq. 2.1 for the case in which an entire mass of fluid is in some kind of accelerated rigid-body motion. Again we use the small, cubical element of fluid shown in Fig. 2.1 and consider it to be part of a larger mass of fluid. In Sec. 2.1 we showed that if the fluid was not being accelerated, then the sum of the forces on it must be zero. If the fluid was being accelerated, then the sum of the forces acting on it, in the direction of the acceleration, must equal the mass times the acceleration. For the cubical element of fluid being accelerated in the vertical direction, we rewrite Eq. 2.1 as 

Dividing by Ax Ay Az and taking the limit as Az approaches zero, we find 

Example 2.18. An open tank contains water 5 m deep. It is sitting on an elevator. Calculate the gauge pressure at the bottom of the tank (a) when the elevator is standing still, (b) when the elevator is accelerating upward at the rate of 5 m/s and (c) when the elevator is accelerating downward at the rate of 5 m/s.  

Example 2.19. A rectangular tank of orange juice on ja cart is moving in the x direction with a steady acceleration of 1 ft/s. See Fig. 2.21. What angle does its free surface make with the horizontal j 

However, we may also calculate the gauge pressure at C by using Eq. 2.36 for the horizontal direction, in which case we have a = x, cos = 0, and 

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