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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.
 Fall '08
 SPRUNGER
 Physics

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