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AravindG
 4 years ago
qn on gravitation figure attached
AravindG
 4 years ago
qn on gravitation figure attached

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AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1wel i hav three masses m making an equilateral triangle

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2So the sides aren't solids themselves, they're just there to show the triangle; the mass is at the points, got it.

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1and a mass 2m lies in the centroid

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wat shud we find grav.potential at centre?

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1wel 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
 4 years ago
Best ResponseYou've already chosen the best response.1and pls tel me hw to take the sign i am confused in that part

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2The 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
 4 years ago
Best ResponseYou've already chosen the best response.1oh wait 10 min pls i will be back mom's calling

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Let 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328026727603:dwdw:1328026672304:dw

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1FCG is 2 GM^2 (cos 30 sin 30)

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Notice 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 \( F_b = (F \cos(4\pi/3), F \sin(4\pi/3) ) = ... \).

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2\[ F_c = (F \cos(\pi/3), F\sin(\pi/3)) = .... \] 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
 4 years ago
Best ResponseYou've already chosen the best response.1AND FGB i got 2GM^@ (cos 30  sin 30) adding i got 0

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2ok. I'm not following your notation, but ok.

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1james i hav another doubt

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1u see in my text it is written

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1i dont get the 2 nd equality

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2In the first equation, the vector r at the end is rhat, \( \hat{r} \) the unit vector. In the second equation, the vector r at the end should NOT be rhat, but simply r, \( r \)....

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2...and that's because \[ \hat{r} = \frac{r}{r} \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Hence \[ \frac{GMm}{r^2}\hat{r} = \frac{GMm}{r^3} r \]

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1oh then it must be a printing mistake i was very confused with that

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2No. Definitely not, because if r is a vector, then 1/r is meaningless.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2\[ \hat{r} = \frac{r}{r} \]

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1lemme cntinu stdying contact u wen a doubt arises
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