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saisukruth
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answer
Krishnadas
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where is the question????
Ruchi.
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|dw:1358405158149:dw| find force at '1'.(gravitational)
Ruchi.
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@Krishnadas
Krishnadas
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F=GM1M2/r^2....and use vector addition
Ruchi.
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plz solve it i hav forgotten
Ruchi.
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@krishnadas
Ruchi.
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can any one hlp me out with this question plz....:((((
Ruchi.
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@hartnn
Ruchi.
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hey plz reply that among u can anyone can solve it or not?
rajathsbhat
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it's easy.|dw:1358406095611:dw|
find out the three forces and then add them vectorially.
rajathsbhat
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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, \(\Large F_{total}=F_{23}+F_{4}\)
rajathsbhat
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I'm so sorry. Huge mistake in the direction of forces.|dw:1358407602553:dw|
Ruchi.
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i'm nt getting u @rajathsbhat
rajathsbhat
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which part?
Ruchi.
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|dw:1358408936334:dw| may i right
Ruchi.
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k/2?i hav nt understood
rajathsbhat
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i have just used k to make things look clean
\[\Large K=\frac{GM^{2}}{a^{2}}\]
Ruchi.
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f^23?
rajathsbhat
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F23 is the resultant force of F2 and F3.
Ruchi.
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f total=k/2+root 2k?
rajathsbhat
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yup
Ruchi.
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root 2a?
rajathsbhat
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what?
Ruchi.
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?\[\sqrt{2a}\]
rajathsbhat
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no \(\sqrt{2}*a\)
Ruchi.
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y?
rajathsbhat
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the pythagoras theorem:
that side=\(\sqrt{a^{2}+a^{2}}=\sqrt{2a^{2}}=\sqrt{2}a\)
Ruchi.
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gm^2/2a?
rajathsbhat
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no \[\large \frac{Gm^2}{2a^{2}}\]
rajathsbhat
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the denominator is \((\sqrt{2}a)^{2}\)