T
0.154newton
=
or
T
m
Al
g
⋅
BF

=
The tension in the supporting cord plus the buoyant force must equal the
weight of the block if it is to be in equilibrium (see figure above). That is
T + BF = mg,
from which
BF
0.091newton
=
or
BF
V
ρ
w
⋅
g
⋅
=
The block displaces a volume
V
of water when submerged, so the
buoyant
force
on it is
(b)
V
9.26cm
3
=
or
V
m
Al
ρ
Al
=
Because
ρ
= m/V
, we have
(a)
Solution
ρ
w
1000
kg
m
3
⋅
=
ρ
Al
2700
kg
m
3
⋅
=
Densities:
m
Al
25 gm
⋅
=
Mass of aluminum:
System
Parameters
The mass of a block of
aluminum is
m
Al
= 25 gm.
(a) What is its volume? (b)
What will be the tension in a
string that suspends the
block, when the block is
totally submerged in water?
The density of aluminum is
ρ
Al
=
2700 kg/m
3
.
Schaum's
Problem 13.14
Archimedes'
Principle
13.2
Chapter 13
Fluids at Rest
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View Full Documentm
w
73 gm
⋅
=
Density of water:
ρ
w
1000
kg
m
3
⋅
=
Solution
The figure above shows the situation when the object is in water. From the
figure,
BF + T = mg
, so
BF
m
alloy
g
⋅
m
w
g
⋅

=
or
BF
0.13newton
=
.
But
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 Spring '10
 dr.brosky
 Buoyant Force, BF BF malloy⋅

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