Three equal masses of m kg each are fixed at the vertices of an
equilateral triangle ABC of side a. What is the force acting on each mass?
Answers
Answer:
no idea
Explanation:
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ANSWER
From Figure, the gravitational force on mass 2m at G due to mass at A is
F
1
=G
1
2
m×2m
=2Gm
2
along GA
Gravitational force on mass 2m at G due to mass at B is
F
2
=G
1
2
m×2m
=2Gm
2
along BG
Gravitational force on mass 2m at G due to mass at C is F
3
=G
1
2
m×2m
=2Gm
2
along GC
Draw DE parallel to BC passing through point G. Then ∠EGC=30
o
=∠DGB.
Resolving
F
2
and
F
3
into two rectangular components, we have
F
2
cos30
o
along GD and F
2
sin30
o
along GH
F
3
cos30
o
along GE and F
2
sin30
o
along GH
Here, F
2
cos30
o
and F
3
sin30
o
are equal in magnitude and acting in opposite directions, and cancel out each other. The resultant force on mass 2m at G is
F
1
−(F
2
sin30
o
+F
3
sin30
o
)
=2Gm
2
−(2Gm
2
×
2
1
+2Gm
2
×
2
1
)
=0
b. When mass at A is 2m, then gravitational force on mass 2m at G due to mass 2m at A is
F
1
′
=G
1
2
2m×2m
=4Gm
2
along GA
The resultant force on mass 2m at G due to masses at A, B and C is
F
1
=(F
2
sin30
o
+F
3
sin30
o
)
=4Gm
2
−(2Gm
2
×
2
1
+2Gm
2
×
2
1
)
=2Gm
2
along GA.