## TrojanPoem one year ago Statics: A metal homogeneous opaque ball with radius r was fastened from a point on its surface with a robe and the other end was fastened on point (A) on a rough vertical wall, the ball stationed in balance and It was about to slip down the wall at point (B). If AB = sqrt(3) r and the friction coefficient between the ball and wall is 1/sqrt(3).Prove that the tan of the angle which the robe makes with the wall is sqrt(3)/2 notice that the the action line of the ball weight affects on it's center.

1. dan815

what is the form for rotational intertia

2. dan815

you should have just drawn it :P

3. TrojanPoem

I though you'd prefer to draw it yourself :D

4. TrojanPoem

|dw:1435161429650:dw|

5. TrojanPoem

I can't add the lengths or the drawing will be nasty.

6. dan815

woah okay, this is confusing drawing

7. dan815

where is the horizontal wall?

8. TrojanPoem

Notice that : Fg = weight of the ball Fb = force of pulling in the robe m = friction coefficient r = reaction of the wall

9. TrojanPoem

typo not horizontal but vertical.

10. dan815

okay lemme try drawing it and see if i got this right

11. dan815

|dw:1435161896312:dw|

12. dan815

okay so our normal force is Fb* sin theta

13. dan815

did you say the resistance force are both from the frictional + moment of intertia

14. TrojanPoem

When the ball tries to slip , friction force will be generated = reaction of the wall x friction coefficient it's similar to this : |dw:1435162343747:dw|

15. dan815

yes u also have to account for the resistive interial force, trying to remmber how to apply that again

16. dan815

both of these will aid in lowering FB

17. dan815

|dw:1435162483641:dw|

18. dan815

do u know it?

19. TrojanPoem

You are getting it slowly.

20. dan815

just need the basic concept here like how do u find the resistive inertial force for a simpler question

21. TrojanPoem

|dw:1435162643921:dw|

22. TrojanPoem

mr + FBCosH = Fg , FBSinH = r

23. TrojanPoem

The moment at any point = 0

24. dan815

okk thats what i want, are u sure its just mr?

25. dan815

u have to add frictional force too there dont u

26. TrojanPoem

m = friction coefficient , r = reaction. Just those

27. TrojanPoem

maximum friction force = mr . So as to ease things I add mr directly.

28. dan815

reaction force is sintehta Fb

29. TrojanPoem

FBSinH = r

30. dan815

we need this part too though|dw:1435162931881:dw|

31. dan815

|dw:1435162940154:dw|

32. dan815

im trying to remember, please clear the concept for me

33. TrojanPoem

Intertial force , Maybe up my grade level. Just ignore it.

34. dan815

what does moment of interia give us exactly??

35. dan815

isnt it a torque about that point

36. dan815

maybe we can just completely ignored it?? there will be no rotation?

37. TrojanPoem

Bro, if there is inertial force. The problem 'd mention it.

38. dan815

oh okayy alright then lets work now lol

39. dan815

|dw:1435163176004:dw|

40. dan815

|dw:1435163282223:dw|

41. dan815

gotta write an expression for the horizontal distance from point C first as a function of the radius of the circle

42. TrojanPoem

No need , just get the torque around (A).

43. dan815

oh i know, umm since its root3 r, we can figure out this angle

44. dan815

|dw:1435163481985:dw|

45. TrojanPoem

Did you notice that AC is unknown too ?

46. dan815

yeah

47. dan815

true, ugh just feels like root3 r is giving enough restrictions

48. TrojanPoem

Listen to me , Torque around (A) !

49. dan815

okayy howw

50. TrojanPoem

Fb, mr will be omitted

51. dan815

i dont really see how there can be any torque about A , should the tourque be about B

52. TrojanPoem

R * r sqrt(3) = FG * r so FG = sqrt(3)R R = FG/sqrt(3)

53. TrojanPoem

Solved.

54. dan815

nice, i still dont see how u did it

55. TrojanPoem

Get torque around (A), sum of forces up = 0 , sum of forces down = 0 , You get 4 equations with 4 unknowns solve and you will get tan ( I am using paper for ease :D)