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Unformatted text preview: z = h at x = 0 when the coffee is at the brink of spilling. Hence, C = p a + ρ gh and ax + gz = gh 486 CHAPTER 5. CONSERVATION OF MASS AND MOMENTUM Then, at the front of the cup where x = d , gh min = gh − ad = ⇒ h min = h − ad g Now, we are given the fact that h o = 1 2 ( h min + h ) Substituting for h min from above, h o = 1 2 X h − ad g + h ~ = ⇒ h o = h − ad 2 g So, if the acceleration is a = λ g , we have h o = h − λ 2 d But, we are given h o = 0 . 75 h , which means . 75 h = h − λ 2 d = ⇒ λ = 0 . 5 h d Finally, for the given values λ = 0 . 5 w 4 in 3 . 5 in W = 0 . 57...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.
 Spring '06
 Phares

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