alexander (jra2623) – oldhomework 20 – Turner – (92510)
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001
(part 1 of 2) 10.0 points
A carpenter’s square of uniform density has
the shape of an L, as shown in the figure.
Assume:
A (
x, y
) coordinate frame with
the origin at the lower left corner of the car
penter’s square. The
x
axis is horizontal and
to the right. The
y
axis is vertically upward.
Given:
In the figure,
B
= 12 cm,
C
=
3
.
7 cm,
D
= 5 cm,
E
= 20 cm.
Because the square is uniform in thickness
and has a small thickness, we can assume that
the weight of each segment of the square is
proportional to its area.
D
E
B
C
What is the
x
coordinate of the center of
gravity?
Correct answer: 3
.
73431 cm.
Explanation:
Basic Concepts:
Center of mass:
vectorr
c
=
∑
i
m
i
vectorr
i
∑
i
m
i
.
Solution:
Take the origin to be at the outside
corner of the
L
. Divide the carpenter’s square
into two rectangles (simple shapes) and find
the area and the center of mass of each piece.
Each piece will behave as if it were a point
particle with mass equal to the mass of the
piece and located at the piece’s center of mass.
Thus, the center of mass of the whole square
will be
vectorr
c
=
m
top
vectorr
c
−
top
+
m
bot
vectorr
c
−
bot
m
top
+
m
bot
,
where
m
top
and
m
bot
are the masses of the two
pieces and
r
c
−
top
and
r
c
−
bot
are the centers of
mass of the two pieces.
The mass of each
piece is proportional to its area, so if we let
σ
represent the uniform mass per area of the
pieces,
m
top
=
σ A
top
and
m
bot
=
σ A
bot
.
For objects where the mass is uniformly
distributed, (
i.e.
, the density is uniform for
thick objects or the mass per area is uni
form for thin objects) the center of mass is
at the same position as the object’s geomet
ric center.
This means that the (
x
cm
, y
cm
)
coordinates of the left/top (5 cm
×
20 cm)
rectangle are
parenleftbigg
D
2
,
E
2
parenrightbigg
, or (2
.
5 cm
,
10 cm),
and the (
x
cm
, y
cm
) coordinates of the bottom
(3
.
7 cm
×
7 cm) rectangle are
parenleftbigg
B
+
D
2
,
C
2
parenrightbigg
,
or (8
.
5 cm ,
1
.
85 cm), since
B

D
2
+
D
=
B
+
D
2
. Thus,
A
top
=
D
×
E
= (5 cm)
×
(20 cm)
= 100 cm
2
A
bottom
=
C
×
W
= (3
.
7 cm)
×
(7 cm)
= 25
.
9 cm
2
x
cm
=
σ
bracketleftbigg
A
top
D
2
+
A
bottom
B
+
D
2
bracketrightbigg
σ
(
A
top
+
A
bottom
)
=
(100 cm
2
) (2
.
5 cm)
(100 cm
2
) + (25
.
9 cm
2
)
+
(25
.
9 cm
2
) (8
.
5 cm)
(100 cm
2
) + (25
.
9 cm
2
)
=
470
.
15 cm
3
125
.
9 cm
2
= 3
.
73431 cm
.
002
(part 2 of 2) 10.0 points
What is the
y
coordinate of the center of grav
ity?
Correct answer: 8
.
32339 cm.
Explanation:
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alexander (jra2623) – oldhomework 20 – Turner – (92510)
2
y
cm
=
σ
bracketleftbigg
A
top
E
2
+
A
bottom
C
2
bracketrightbigg
σ
(
A
top
+
A
bottom
)
=
(100 cm
2
) (10 cm)
(125
.
9 cm
2
)
+
(25
.
9 cm
2
) (1
.
85 cm)
(125
.
9 cm
2
)
=
(1047
.
91 cm
3
)
(125
.
9 cm
2
)
= 8
.
32339 cm
.
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 Summer '08
 Kaplunovsky
 Mass, Work, Correct Answer, kg, Alexander

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