Graphics Reference
In-Depth Information
1
C
2
A
¢
1
D
4
2
2
1
C
¢
A
B
Fig. 11.20.
The masses assigned to
A, B
and
C
to determine
D
.
The total mass is 7 (1 + 2 + 4), therefore
D
=
4
7
A
+
2
7
B
+
1
7
C
The point
E
is on the intersection of lines
BB
and
AA
. Therefore, we begin
by placing a mass of 1 at
A
. Then, for line
CA
to balance at
B
amassof2
must be placed at
C
. Similarly, for line
BC
to balance at
A
amassof4must
be placed at
B
. This configuration is shown in Figure 11.21.
The total mass is still 7, therefore
E
=
1
7
A
+
4
7
B
+
2
7
C
From the symmetry of the triangle we can state that
F
=
2
7
A
+
1
7
B
+
4
7
C
2
C
1
2
B
¢
A
¢
2
E
1
1
4
A
B
Fig. 11.21.
The masses assigned to
A, B
and
C
to determine
E
.
