## AravindG 4 years ago qn on gravitation figure attached

1. AravindG

|dw:1328026421121:dw|

2. AravindG

wel i hav three masses m making an equilateral triangle

3. JamesJ

So the sides aren't solids themselves, they're just there to show the triangle; the mass is at the points, got it.

4. AravindG

and a mass 2m lies in the centroid

5. anonymous

wat shud we find grav.potential at centre?

6. AravindG

wel 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??

7. AravindG

and pls tel me hw to take the sign i am confused in that part

8. JamesJ

The 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.

9. AravindG

oh wait 10 min pls i will be back mom's calling

10. JamesJ

Let 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.

11. anonymous

|dw:1328026727603:dw||dw:1328026672304:dw|

12. JamesJ

13. AravindG

for FAB it is 2Gm^2 j

14. AravindG

FCG is 2 GM^2 (cos 30 -sin 30)

15. AravindG

SRRY

16. AravindG

FGA IS 1ST ONE

17. JamesJ

Notice 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 $$F_b = (F \cos(4\pi/3), F \sin(4\pi/3) ) = ...$$.

18. AravindG

2ND ONE FGC

19. JamesJ

$F_c = (F \cos(-\pi/3), F\sin(-\pi/3)) = ....$ 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.

20. AravindG

AND FGB i got 2GM^@ (-cos 30 - sin 30) adding i got 0

21. JamesJ

ok. I'm not following your notation, but ok.

22. AravindG

james i hav another doubt

23. AravindG

u see in my text it is written

24. JamesJ

sorry, what text?

25. AravindG

|dw:1328029900180:dw|

26. JamesJ

right.

27. AravindG

i dont get the 2 nd equality

28. JamesJ

In the first equation, the vector r at the end is r-hat, $$\hat{r}$$ the unit vector. In the second equation, the vector r at the end should NOT be r-hat, but simply r, $$r$$....

29. JamesJ

...and that's because $\hat{r} = \frac{r}{|r|}$

30. JamesJ

Hence $\frac{GMm}{|r|^2}\hat{r} = \frac{GMm}{|r|^3} r$

31. AravindG

oh then it must be a printing mistake i was very confused with that

32. JamesJ

fair enough too

33. AravindG

wel can we write

34. AravindG

|dw:1328030176734:dw|

35. JamesJ

No. Definitely not, because if r is a vector, then 1/r is meaningless.

36. JamesJ

$\hat{r} = \frac{r}{|r|}$

37. AravindG

oh!!

38. AravindG

thx

39. AravindG

lemme cntinu stdying contact u wen a doubt arises

40. AravindG

bye