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.

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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.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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what is the form for rotational intertia
you should have just drawn it :P
I though you'd prefer to draw it yourself :D

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Other answers:

|dw:1435161429650:dw|
I can't add the lengths or the drawing will be nasty.
woah okay, this is confusing drawing
where is the horizontal wall?
Notice that : Fg = weight of the ball Fb = force of pulling in the robe m = friction coefficient r = reaction of the wall
typo not horizontal but vertical.
okay lemme try drawing it and see if i got this right
|dw:1435161896312:dw|
okay so our normal force is Fb* sin theta
did you say the resistance force are both from the frictional + moment of intertia
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|
yes u also have to account for the resistive interial force, trying to remmber how to apply that again
both of these will aid in lowering FB
|dw:1435162483641:dw|
do u know it?
You are getting it slowly.
just need the basic concept here like how do u find the resistive inertial force for a simpler question
|dw:1435162643921:dw|
mr + FBCosH = Fg , FBSinH = r
The moment at any point = 0
okk thats what i want, are u sure its just mr?
u have to add frictional force too there dont u
m = friction coefficient , r = reaction. Just those
maximum friction force = mr . So as to ease things I add mr directly.
reaction force is sintehta Fb
FBSinH = r
we need this part too though|dw:1435162931881:dw|
|dw:1435162940154:dw|
im trying to remember, please clear the concept for me
Intertial force , Maybe up my grade level. Just ignore it.
what does moment of interia give us exactly??
isnt it a torque about that point
maybe we can just completely ignored it?? there will be no rotation?
Bro, if there is inertial force. The problem 'd mention it.
oh okayy alright then lets work now lol
|dw:1435163176004:dw|
|dw:1435163282223:dw|
gotta write an expression for the horizontal distance from point C first as a function of the radius of the circle
No need , just get the torque around (A).
oh i know, umm since its root3 r, we can figure out this angle
|dw:1435163481985:dw|
Did you notice that AC is unknown too ?
yeah
true, ugh just feels like root3 r is giving enough restrictions
Listen to me , Torque around (A) !
okayy howw
Fb, mr will be omitted
i dont really see how there can be any torque about A , should the tourque be about B
R * r sqrt(3) = FG * r so FG = sqrt(3)R R = FG/sqrt(3)
Solved.
nice, i still dont see how u did it
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)

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