Three-point masses of equal masses m are placed at the vertices of an equilateral triangle. They are kept in
equilibrium by unspecified forces. Find the net gravitational force on any one mass due to other two masses.
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G Draw a perpendicular AD to the side BC.
∴AD=ABsin60=
2
3
l
Distance AO of the centroid O from A is
3
2
AD.
∴AO=
3
2
(
2
3
l)=
3
l
By symmetry, AO=BO=CO=
3
l
Force on mass 2m at O due to mass m at A is
F
OA
=
(l/
3
)
2
Gm(2m)
=
l
2
6Gm
2
along OA
Force on mass 2m at O due to mass m at B is
F
OB
=
(l/
3
)
2
Gm(2m)
=
l
2
6Gm
2
along OB
Force on mass 2m at O due to mass m at C is
F
OC
=
(l/
3
)
2
Gm(2m)
=
l
2
6Gm
2
along OC
Draw a line PQ parallel to BC passing through O. Then ∠BOP=30=∠COQ
Resolving
F
ˉ
OB
and
F
ˉ
OC
into two components.
Components acting along OP and OQ are equal in magnitude and opposite in direction. So, they will cancel out while the components acting along OD will add up.
∴ The resultant force on the mass 2m at O is
F
R
=F
OA
−(F
OB
sin30+F
OC
sin30)
=
l
2
6Gm
2
−(
l
2
6Gm
2
×
2
1
+
l
2
6Gm
2
×
2
1
)=0
solution
∴AD=ABsin60=
2
3
l
Distance AO of the centroid O from A is
3
2
AD.
∴AO=
3
2
(
2
3
l)=
3
l
By symmetry, AO=BO=CO=
3
l
Force on mass 2m at O due to mass m at A is
F
OA
=
(l/
3
)
2
Gm(2m)
=
l
2
6Gm
2
along OA
Force on mass 2m at O due to mass m at B is
F
OB
=
(l/
3
)
2
Gm(2m)
=
l
2
6Gm
2
along OB
Force on mass 2m at O due to mass m at C is
F
OC
=
(l/
3
)
2
Gm(2m)
=
l
2
6Gm
2
along OC
Draw a line PQ parallel to BC passing through O. Then ∠BOP=30=∠COQ
Resolving
F
ˉ
OB
and
F
ˉ
OC
into two components.
Components acting along OP and OQ are equal in magnitude and opposite in direction. So, they will cancel out while the components acting along OD will add up.
∴ The resultant force on the mass 2m at O is
F
R
=F
OA
−(F
OB
sin30+F
OC
sin30)
=
l
2
6Gm
2
−(
l
2
6Gm
2
×
2
1
+
l
2
6Gm
2
×
2
1
)=0
solution
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