Chap9Add - Chapter 9 Additional Notes! P1 A h Want to look...

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Chapter 9 Additional Notes P 1 A mg P 1 A h Want to look at when in equilibrium.
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Since we want equilibrium we can write F = 0 P 2 A P 1 A mg = 0 However, from the definition of density we can write the mass as m = ρ V We can then write the sum of the forces as P 2 A P 1 A − ρ Vg = 0
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However, we can write the volume as V = Ah We can now write sum of the forces as P 2 A P 1 A − ρ Ahg = 0 So we can write P 2 = P 1 + ρ hg
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Car Lift Example: a) Since the system is closed, the pressure at each spot must be the same. So we can then write P 1 = P 2 F 1 A 1 = F 2 A 2 Solving for force 1 gives F 1 = A 1 A 2 F 2
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Now using that the area is that of a circle we can write F 1 = r 1 r 2 2 F 2 = 1478 N b) Using the definition of pressure, we can then write F 1 A 1 = 1478 N π r 1 2 = 188157 Pa
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Floating Example: Since the object is floating, the net force acting on it must be zero. So we can write F = 0 B mg = 0 Or B = mg ρ W V W g = mg Where we used the definition of B.
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So using the density we can write the mass of the object as ρ W V W g = mg W V W g = O V O g Now we can write the Volume of the water as V = Ah Hence we have W Ah = O V O
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Solving for the height h gives h = ρ O W V O A = 0.063 m
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This note was uploaded on 01/02/2010 for the course PHY 101 taught by Professor Pralle during the Fall '08 term at SUNY Buffalo.

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Chap9Add - Chapter 9 Additional Notes! P1 A h Want to look...

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