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39th International Physics Olympiad  Hanoi  Vietnam  2008
Theoretical Problem No. 1 /Solution
1
3
−
==
Solution
1. The structure of the mortar
1.1. Calculating the distance TG
The volume of water in the bucket is
V
.
The length of the
bottom of the bucket is
.
33
1000cm
10 m
00
60
0 74 0 12
60 m
0 5322m
tan
( .
.
tan
)
.
dL
h
=−
=
−
=
(as the initial data are given with two significant digits, we shall keep only two
significant digits in the final answer, but we keep more digits in the intermediate steps).
The height
of the water layer in the bucket is calculated from the formula:
c
21
/
2
b
0
(2
3
/
)
tan60
2
3
cd
V
d
Vb
c
dbc
c
+
−
=+
⇒
=
Inserting numerical values for
,
and
, we find
b
d
0.01228m
c
=
V
.
When the lever lies horizontally, the distance, on the horizontal axis, between the rotation
axis and the center of mass of water N, is
o
TH
60
0 4714m
24
tan
.
dc
a
≈+ +
=
, and
(see the figure below).
TG
(
/
)TH
0.01571m
mM
H
T
N
K
R
S
P
Answer:
.
TG
0.016m
=
α
1
2
1.2. Calculating the values of
and
.
1
When the lever tilts with angle
, water level is at the edge of the bucket. At that
point the water volume is
. Assume
10 m
−
PQ
d
<
. From geometry
,
from which
P
. The assumption
PQ/ 2
Vh
b
=×
PQ
d
<
Q
0.1111m
=
is obviously satisfied
(
).
0.5322m
d
=
QS=
PQ+ 3
tan
/
/(
).
hh
h
1
=
1
To compute the angle
, we note that
From this
we find
.
o
20.6
1
=
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View Full Document39th International Physics Olympiad  Hanoi  Vietnam  2008
Theoretical Problem No. 1 /Solution
2
When the tilt angle is
, the bucket is empty
:
.
o
30
o
30
α
2
=
G
h
T
R
N
Q
P
I
S
β
1.3. Determining the tilt angle
of the lever and the amount of water in the bucket
m
μ
when the total torque
on the lever is equal to zero
(m)
x
=
Denote
PQ
.
The
amount
of
water
in
the
bucket
is
water
9(
k
g
)
2
xhb
mx
ρ
==
.
=0
when the torque coming from the water in the bucket cancels out the torque
coming from the weight of the lever. The cross section of the water in the bucket is the
triangle PQR in the figure. The center of mass N of water is located at 2/3 of the meridian
RI, therefore NTG lies on a straight line. Then:
TN
TG
mg
Mg
×
=×
or
TN
TG
30 0.1571
0.4714
mM
×=
×
=
(1)
Calculating
from
x
then substitute (1) :
TN
2
TN
(
3
)
0.94 0.08 3
0.8014
32
3
3
x
xx
La
h
=+−
+ =
−
−=
−
which implies
(2)
2
TN
9 (0.8014
/3)
3
7.213
x
x
−
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 Spring '11
 NA
 Physics

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