FLUID MECHANICS PROBLEMS I SOLUTIONS

FLUID MECHANICS PROBLEMS I SOLUTIONS - FLUID MECHANICS...

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FLUID MECHANICS PROBLEMS I SOLUTIONS 1) a) The buoyant force provides the upwards thrust that essentially reduces the weight of the object. Hence, ( 29 F N F N B B = - = 3 0 1 4 1 6 . . . . b) ( 29 ( 29 ( 29 ( 29 F V g N k g m V m s V m B w a t e r w a t e r w a t e r w a t e r = = = × - ρ 1 6 1 0 9 8 1 6 3 1 0 3 3 2 4 3 . . . . c) Since the plastic toy is under water, then the volume of the toy is equivalent to the volume of displaced fluid. Thus, V V m w a t e r p l a s t i c = = × - 1 6 3 1 0 4 3 . . Neglecting the buoyant force due to air, the mass of the plastic toy is m N m s m k g p l a s t i c p l a s t i c = = × - - 3 9 8 3 0 6 1 0 2 1 . . . Hence, the density of the plastic toy is ρ p l a s t i c p l a s t i c p l a s t i c m V = = × × = - - ρ ρ p l a s t i c p l a s t i c k g m k g m 3 0 6 1 0 1 6 3 1 0 1 8 7 7 1 4 3 . . 3 . 2) a) V m V k g k g m V m s t o n e s t o n e s t o n e s t o n e s t o n e = = = × - ρ 2 4 2 5 0 0 9 6 1 0 3 3 3 . . b) ( 29 ( 29 F k g m s F N g g = = - 2 4 9 8 2 3 5 2 2 . . . c) ( 29 ( 29 ( 29 ( 29 F V g F k g m m m s F N B w a t e r w a t e r B B = = × = - ρ 1 0 0 0 9 6 1 0 9 8 9 4 1 3 3 2 3 . . . . d) In order to lift the stone, the boy must supply the difference between the weight of the stone and the buoyant force. Hence, the boy must supply ( 29 F N F N a p p l i e d a p p l i e d = - = 2 3 5 2 9 4 1 1 4 1 1 . . . . 3) The volume of the block of wood is ( 29 ( 29 ( 29 V c m c m c m V c m w o o d w o o d = = 1 2 1 5 1 8 3 2 4 0 3 . Since the buoyant force supports the weight of the block of wood, then we must first find the buoyant force. If 67.5% of the volume is submerged, then the volume of water is ( 29 ( 29 ( 29 ( 29 V V V c m V c m w a t e r
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