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where is the question????

|dw:1358405158149:dw| find force at '1'.(gravitational)

F=GM1M2/r^2....and use vector addition

plz solve it i hav forgotten

can any one hlp me out with this question plz....:((((

hey plz reply that among u can anyone can solve it or not?

it's easy.|dw:1358406095611:dw|
find out the three forces and then add them vectorially.

I'm so sorry. Huge mistake in the direction of forces.|dw:1358407602553:dw|

i'm nt getting u @rajathsbhat

which part?

|dw:1358408936334:dw| may i right

k/2?i hav nt understood

i have just used k to make things look clean
\[\Large K=\frac{GM^{2}}{a^{2}}\]

f^23?

F23 is the resultant force of F2 and F3.

f total=k/2+root 2k?

yup

root 2a?

what?

?\[\sqrt{2a}\]

no \(\sqrt{2}*a\)

y?

the pythagoras theorem:
that side=\(\sqrt{a^{2}+a^{2}}=\sqrt{2a^{2}}=\sqrt{2}a\)

gm^2/2a?

no \[\large \frac{Gm^2}{2a^{2}}\]

the denominator is \((\sqrt{2}a)^{2}\)