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where is the question????
|dw:1358405158149:dw| find force at '1'.(gravitational)

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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'll calculate F4 for you|dw:1358406459709:dw| \[\begin{align} F_{4}&=\frac{Gm.m}{r^{2}}\\ &=\frac{Gm^{2}}{2a^{2}}\\&=\frac{K}{2}\end{align}\] using the same formula, find F3 and F2 you'll end up with K for both of them. taking the vectorial sum of f2 and f3, \[\Large\begin{align} F_{23}&=\sqrt{F_1^2+F_2^2}\\&=\sqrt{2}K\end{align}\] since F2 and F3 are equal in magnitude, F23 is at a 45 degree angle to either of those two forces. i.e.|dw:1358407097861:dw| now, since F4 and F23 are in the same direction, you can just add them up . therefore, \(\Large F_{total}=F_{23}+F_{4}\)
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 \((\sqrt{2}a)^{2}\)

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