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.

- TrojanPoem

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- dan815

what is the form for rotational intertia

- dan815

you should have just drawn it :P

- TrojanPoem

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

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## More answers

- TrojanPoem

|dw:1435161429650:dw|

- TrojanPoem

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

- dan815

woah okay, this is confusing drawing

- dan815

where is the horizontal wall?

- TrojanPoem

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

- TrojanPoem

typo not horizontal but vertical.

- dan815

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

- dan815

|dw:1435161896312:dw|

- dan815

okay so our normal force is Fb* sin theta

- dan815

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

- 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|

- dan815

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

- dan815

both of these will aid in lowering FB

- dan815

|dw:1435162483641:dw|

- dan815

do u know it?

- TrojanPoem

You are getting it slowly.

- dan815

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

- TrojanPoem

|dw:1435162643921:dw|

- TrojanPoem

mr + FBCosH = Fg , FBSinH = r

- TrojanPoem

The moment at any point = 0

- dan815

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

- dan815

u have to add frictional force too there dont u

- TrojanPoem

m = friction coefficient , r = reaction. Just those

- TrojanPoem

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

- dan815

reaction force is sintehta Fb

- TrojanPoem

FBSinH = r

- dan815

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

- dan815

|dw:1435162940154:dw|

- dan815

im trying to remember, please clear the concept for me

- TrojanPoem

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

- dan815

what does moment of interia give us exactly??

- dan815

isnt it a torque about that point

- dan815

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

- TrojanPoem

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

- dan815

oh okayy alright then lets work now lol

- dan815

|dw:1435163176004:dw|

- dan815

|dw:1435163282223:dw|

- dan815

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

- TrojanPoem

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

- dan815

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

- dan815

|dw:1435163481985:dw|

- TrojanPoem

Did you notice that AC is unknown too ?

- dan815

yeah

- dan815

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

- TrojanPoem

Listen to me , Torque around (A) !

- dan815

okayy howw

- TrojanPoem

Fb, mr will be omitted

- dan815

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

- TrojanPoem

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

- TrojanPoem

Solved.

- dan815

nice, i still dont see how u did it

- 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)

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