ch14-p023 - Assuming the pressure at the surface can be...

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23. We can integrate the pressure (which varies linearly with depth according to Eq. 14-7) over the area of the wall to find out the net force on it, and the result turns out fairly intuitive (because of that linear dependence): the force is the “average” water pressure multiplied by the area of the wall (or at least the part of the wall that is exposed to the water), where “average” pressure is taken to mean 1 2 (pressure at surface + pressure at bottom).
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Unformatted text preview: Assuming the pressure at the surface can be taken to be zero (in the gauge pressure sense explained in section 14-4), then this means the force on the wall is 1 2 gh multiplied by the appropriate area. In this problem the area is hw (where w is the 8.00 m width), so the force is 1 2 gh 2 w , and the change in force (as h is changed) is 1 2 gw ( h f 2 h i 2 ) = 1 2 (998 kg/m 3 )(9.80 m/s 2 )(8.00 m)(4.00 2 2.00 2 )m 2 = 4.69 10 5 N....
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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