A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

AravindG

  • 4 years ago

qn on gravitation figure attached

  • This Question is Closed
  1. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1328026421121:dw|

  2. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    wel i hav three masses m making an equilateral triangle

  3. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and a mass 2m lies in the centroid

  5. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wat shud we find grav.potential at centre?

  6. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  8. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  10. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

  13. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    for FAB it is 2Gm^2 j

  14. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  15. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    SRRY

  16. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    FGA IS 1ST ONE

  17. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    2ND ONE FGC

  19. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[ F_c = (F \cos(-\pi/3), F\sin(-\pi/3)) = .... \] 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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  21. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  22. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    james i hav another doubt

  23. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    u see in my text it is written

  24. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    sorry, what text?

  25. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1328029900180:dw|

  26. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    right.

  27. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i dont get the 2 nd equality

  28. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  30. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  31. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  32. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    fair enough too

  33. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    wel can we write

  34. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1328030176734:dw|

  35. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  36. JamesJ
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[ \hat{r} = \frac{r}{|r|} \]

  37. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh!!

  38. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thx

  39. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    lemme cntinu stdying contact u wen a doubt arises

  40. AravindG
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    bye

  41. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.