Find center of mass in equilateral triangle if it's side a and masses is 4
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Let the corner of the triangle
{
(
A
,
100
)
}
be at the origin
O
(
0
,
0
)
Then,
the corner
{
(
B
,
150
)
}
wiil be at
(
0.5
,
0
)
and the corner
{
(
C
,
200
)
}
will be at
(
0.25
,
0.25
√
3
)
Let
G
(
x
G
,
y
G
)
be the center of mass of the particles
Then,
x
G
=
a
x
a
+
b
x
b
+
c
x
c
a
+
b
+
c
=
100
⋅
0
+
150
⋅
0.5
+
200
⋅
0.25
100
+
150
+
200
=
125
450
=
0.28
y
G
=
a
y
a
+
b
y
b
+
c
y
c
a
+
b
+
c
=
100
⋅
0
+
150
⋅
0
+
200
⋅
0.25
√
3
100
+
150
+
200
=
51.7
450
=
0.12
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