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p0536

# p0536 - z and using the z component of Euler’s equation...

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5.36. CHAPTER 5, PROBLEM 36 493 5.36 Chapter 5, Problem 36 Figure 5.14: Accelerating car. It is easiest to work in a coordinate system, ( x, z ) , rotated by an angle α as shown. In this coordinate system, the gravitational vector is g = i g sin α k g cos α where i and k are unit vectors in the x and z directions, respectively. In this coordinate system, the three components of the Euler equation are ρ g sin α = p x + ρ g sin α 0= p y 0= p z ρ g cos α Observe that the car and the fluid within accelerate only in the x direction, i.e., parallel to the inclined plane. Hence, only the x component of g appears on the left-hand side of Euler’s equation. By contrast, both components of the body force act regardless of the motion of the cart, wherefore both components of g appear on the right-hand side. Clearly, since p/ x = p/ y =0 , the pressure is a function only of z , i.e., p ( x, y, z )= f ( z ) where f ( z ) is a function to be determined. Differentiating with respect to

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Unformatted text preview: z and using the z component of Euler’s equation, we find ∂ p ∂ z = f I ( z ) = − ρ g cos α = ⇒ f ( z ) = C − ρ gz cos α 494 CHAPTER 5. CONSERVATION OF MASS AND MOMENTUM where C is a constant. Thus, the pressure throughout the fluid in the car is p ( x, y, z ) = C − ρ gz cos α On the free surface, p = p a . Also, at x = 0 , inspection of Figure 5.13 shows that z = h , wherefore C = p a + ρ gh cos α Therefore, the pressure at any point in the fluid is p ( x, y, z ) = p a + ρ g ( h − z ) cos α On the free surface, p = p a . Thus, the equation of the free surface is p a = p a + ρ g ( h − z ) cos α = ⇒ z = h This means the free surface is parallel to the inclined plane, so that β = α...
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p0536 - z and using the z component of Euler’s equation...

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