homework 29 – YOO, HEE – Due: Apr 10 2008, 4:00 am
1
Question 1, chap 14, sect 1.
part 1 of 4
10 points
A uniform brick of length 20 cm is placed
over the edge of a horizontal surface with
the maximum overhang
x
possible without
falling.
g
x
20 cm
Find
x
for a single block.
Correct answer: 10 cm (tolerance
±
1 %).
Explanation:
Let :
L
= 20 cm
For
n
bricks the center of mass is
x
m
≡
n
summationdisplay
i
=1
x
i
m
i
n
summationdisplay
i
=1
m
i
=
1
n
n
summationdisplay
i
=1
x
i
,
where
x
i
is the center of mass position of the
i
th
brick and
m
i
is the mass of the
i
th
brick.
Since
x
1
L
=
1
2
, as measured from the maxi
mum overhang, the center of mass of a single
brick is in its middle or
x
m
L
vextendsingle
vextendsingle
vextendsingle
n
=1
=
1
2
1
=
1
2
of the brick’s length from its maximum over
hang.
x
m
=
1
2
L
=
1
2
(20 cm)
=
10 cm
.
g
x
20 cm
Question 2, chap 14, sect 1.
part 2 of 4
10 points
Two
identical
uniform
bricks
of
length
20 cm are stacked over the edge of a hori
zontal surface with the maximum overhang
x
possible without falling.
g
x
20 cm
Find
x
for two blocks.
Correct answer: 15 cm (tolerance
±
1 %).
Explanation:
Since
m
i
=
m
(all bricks have the same
mass),
n
summationdisplay
i
=1
m
=
n m
.
The bricks will just
balance when the center of mass is over the
fulcrum;
i.e.
, the edge of the horizontal sur
face.
Measurement will be made from the
left edge of the top brick with the maximum
overhang.
To calculate the center of mass
x
m
L
=
1
n L
n
summationdisplay
i
=1
x
i
; when an additional brick
is positioned under the stack of
n

1 bricks,
the additional brick’s left edge is placed at
the balance point of the previous stack of
n
bricks.
The center of mass of the additional brick
is
1
2
of a brick’s length plus the maximum
overhang of the previous stack of
n

1 bricks.
The top brick can extend
1
2
of a brick’s
length from the maximum overhang.
Since
x
2
L
=
x
m
L
vextendsingle
vextendsingle
vextendsingle
n
=1
+
1
2
=
1
2
+
1
2
= 1 as measured
from the maximum overhang, the center of
mass of the two bricks is in their middle or
x
m
L
vextendsingle
vextendsingle
vextendsingle
n
=2
=
1
2
+ 1
2
=
3
4
of a single brick’s length from the maximum
overhang:
x
m
=
3
4
L
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homework 29 – YOO, HEE – Due: Apr 10 2008, 4:00 am
2
=
3
4
(20 cm)
=
15 cm
.
g
x
20 cm
Question 3, chap 14, sect 1.
part 3 of 4
10 points
Three identical uniform bricks of length
20 cm are stacked over the edge of a hori
zontal surface with the maximum overhang
x
possible without falling.
g
x
20 cm
Find
x
for three blocks.
Correct answer: 18
.
3333
cm (tolerance
±
1
%).
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 Spring '08
 Turner
 Physics, Center Of Mass, Mass, Work, Orders of magnitude, maximum overhang

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