P15_026

# P15_026 - numerical uncertainties associated with that...

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26. (a) The pressure (including the contribution from the atmosphere) at a depth of h top = L/ 2 (corre- sponding to the top of the block) is p top = p atm + ρgh top =1 . 01 × 10 5 + (1030)(9 . 8)(0 . 300) = 1 . 04 × 10 5 Pa where the unit Pa (Pascal) is equivalent to N/m 2 . The force on the top surface (of area A = L 2 = 0 . 36 m 2 )is F top = p top A =3 . 75 × 10 4 N. (b) The pressure at a depth of h bot =3 L/ 2 (that of the bottom of the block) is p bot = p atm + ρgh bot =1 . 01 × 10 5 + (1030)(9 . 8)(0 . 900) = 1 . 10 × 10 5 Pa where we recall that the unit Pa (Pascal) is equivalent to N/m 2 . The force on the bottom surface is F bot = p bot A =3 . 96 × 10 4 N. (c) Taking the diFerence F bot F top cancels the contribution from the atmosphere (including any
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Unformatted text preview: numerical uncertainties associated with that value) and leads to F bot − F top = ρg ( h bot − h top ) A = ρgL 3 = 2180 N which is to be expected on the basis of Archimedes’ principle. Two other forces act on the block: an upward tension T and a downward pull of gravity mg . To remain stationary, the tension must be T = mg − ( F bot − F top ) = (450)(9 . 8) − 2180 = 2230 N . (d) This has already been noted in the previous part: F b = 2180 N, and T + F b = mg ....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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