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## anonymous 3 years ago Question?

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1. anonymous

answer

2. anonymous

where is the question????

3. anonymous

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

4. anonymous

@Krishnadas

5. anonymous

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

6. anonymous

plz solve it i hav forgotten

7. anonymous

@krishnadas

8. anonymous

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

9. anonymous

@hartnn

10. anonymous

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

11. anonymous

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

12. anonymous

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, $$\Large F_{total}=F_{23}+F_{4}$$

13. anonymous

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

14. anonymous

i'm nt getting u @rajathsbhat

15. anonymous

which part?

16. anonymous

|dw:1358408936334:dw| may i right

17. anonymous

k/2?i hav nt understood

18. anonymous

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

19. anonymous

f^23?

20. anonymous

F23 is the resultant force of F2 and F3.

21. anonymous

f total=k/2+root 2k?

22. anonymous

yup

23. anonymous

root 2a?

24. anonymous

what?

25. anonymous

?$\sqrt{2a}$

26. anonymous

no $$\sqrt{2}*a$$

27. anonymous

y?

28. anonymous

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

29. anonymous

gm^2/2a?

30. anonymous

no $\large \frac{Gm^2}{2a^{2}}$

31. anonymous

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

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