AravindG
  • AravindG
qn on gravitation figure attached
Physics
schrodinger
  • schrodinger
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AravindG
  • AravindG
|dw:1328026421121:dw|
AravindG
  • AravindG
wel i hav three masses m making an equilateral triangle
JamesJ
  • JamesJ
So the sides aren't solids themselves, they're just there to show the triangle; the mass is at the points, got it.

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AravindG
  • AravindG
and a mass 2m lies in the centroid
anonymous
  • anonymous
wat shud we find grav.potential at centre?
AravindG
  • AravindG
wel i am asked what is the force acting on the 2m mass . I know its zero but can u help me get the answer vectorially??
AravindG
  • AravindG
and pls tel me hw to take the sign i am confused in that part
JamesJ
  • JamesJ
The force from each of the masses at A,B,C has the same magnitude. And each force is 120 degrees apart. They must add up to zero.
AravindG
  • AravindG
oh wait 10 min pls i will be back mom's calling
JamesJ
  • JamesJ
Let Fa, Fb and Fc denote the three forces. First we note that |Fa| = |Fb| = |Fc|. Call that quantity F, a scalar. Now, write down each for in components using a coordinate system where the force Fa is in the y direction. Then \[ F_a = (0,F) \] Now write down the vectors Fb and Fc in components.
anonymous
  • anonymous
|dw:1328026727603:dw||dw:1328026672304:dw|
JamesJ
  • JamesJ
AravindG
  • AravindG
for FAB it is 2Gm^2 j
AravindG
  • AravindG
FCG is 2 GM^2 (cos 30 -sin 30)
AravindG
  • AravindG
SRRY
AravindG
  • AravindG
FGA IS 1ST ONE
JamesJ
  • JamesJ
Notice that we don't need to use explicit numbers. It's enough that \[ F = |F_a| = |F_b| = |F_c| \] Now, \( F_a = (0, F) \). Then \( F_b = (F \cos(4\pi/3), F \sin(4\pi/3) ) = ... \).
AravindG
  • AravindG
2ND ONE FGC
JamesJ
  • JamesJ
\[ F_c = (F \cos(-\pi/3), F\sin(-\pi/3)) = .... \] Simplify the expressions for F_b and F_c to find them explicitly. Then add up the vectors \[ F_a + F_b + F_c \] and show it's zero.
AravindG
  • AravindG
AND FGB i got 2GM^@ (-cos 30 - sin 30) adding i got 0
JamesJ
  • JamesJ
ok. I'm not following your notation, but ok.
AravindG
  • AravindG
james i hav another doubt
AravindG
  • AravindG
u see in my text it is written
JamesJ
  • JamesJ
sorry, what text?
AravindG
  • AravindG
|dw:1328029900180:dw|
JamesJ
  • JamesJ
right.
AravindG
  • AravindG
i dont get the 2 nd equality
JamesJ
  • JamesJ
In the first equation, the vector r at the end is r-hat, \( \hat{r} \) the unit vector. In the second equation, the vector r at the end should NOT be r-hat, but simply r, \( r \)....
JamesJ
  • JamesJ
...and that's because \[ \hat{r} = \frac{r}{|r|} \]
JamesJ
  • JamesJ
Hence \[ \frac{GMm}{|r|^2}\hat{r} = \frac{GMm}{|r|^3} r \]
AravindG
  • AravindG
oh then it must be a printing mistake i was very confused with that
JamesJ
  • JamesJ
fair enough too
AravindG
  • AravindG
wel can we write
AravindG
  • AravindG
|dw:1328030176734:dw|
JamesJ
  • JamesJ
No. Definitely not, because if r is a vector, then 1/r is meaningless.
JamesJ
  • JamesJ
\[ \hat{r} = \frac{r}{|r|} \]
AravindG
  • AravindG
oh!!
AravindG
  • AravindG
thx
AravindG
  • AravindG
lemme cntinu stdying contact u wen a doubt arises
AravindG
  • AravindG
bye

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