Help with pool ball pyramid problem

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Homework Help Overview

The problem involves three identical spheres resting on a horizontal plane, with a fourth sphere positioned above them, touching all three. The task is to demonstrate the minimum coefficients of friction required for equilibrium among the spheres and between the spheres and the plane.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss summing forces and the role of friction in maintaining equilibrium. Questions arise about the necessity of friction between the spheres and the calculation of forces acting on the spheres, particularly the weight distribution of the top sphere on the bottom spheres.

Discussion Status

Some participants have provided equations related to the forces acting on the spheres and are exploring geometric relationships to find the angle involved. There is recognition of the complexity of the problem, with varying degrees of understanding and attempts to clarify the relationships between the forces.

Contextual Notes

Participants express uncertainty regarding the angles involved in the geometry of the arrangement and the implications of additional forces, particularly the friction between the spheres. There is a mention of conflicting angle values found online that do not align with the expected results.

Blade707
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Three identical spheres lie in contact with one another on a horizontal plane. A fourth
sphere rests on them, touching all three. Show that, in equilibrium, the coefficient of
friction between the spheres is at least (rad(3)-rad(2)) and that the coefficient between each sphere and the plane is at least (rad(3)-rad(2))/4

The only thing I can think to use here is summing forces. But woah, I am so stuck. Do I have the right idea even? Any help would be appreciated it!

Thanks!
 
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Thats pretty intense!
I'm having trouble conceptualizing why there has to be any friction between the spheres... but i think i can help with the second part:
The weight of the top sphere is the only one disrupting equilibrium right? i.e. no top sphere, no motion (even without friction).
How much top sphere (TS) weight is on each bottom sphere (BS)?
How much is parallel to the plane they're resting on?
How much friction is required to balance that - keeping them all stationary?
 
Uh I'm having troubles grasping this. Ok so using what you said before about the second part, I summed forces on one ball.

(1) In the x: (mg/3)cos(theta)=N(cf) mg/3 is the weight of the top ball on the bottom sphere, N is the normal force at floor, and (cf) is my coefficient of friction.

(2) In the y: N=mgsin(theta)+mg

Putting (2) into (1):
(mg/3)cos(theta)=(cf)(mgsin(theta)+mg)
reduced:
cf=cos(theta)/(3sin(theta)+3)

Is this the right equation to work with? If I knew what angle to put in anyway!
 
Those look correct.
As for theta... that's just a little geometry.
The pyramid will be a tetrahedron, I'm sure you can look up or derive the angle somehow... I'm terrible with 3d geo.
 
Yeah I'm awful at it as well I found some angles online like 55 degrees and 71 degrees but neither of them get me to the correct answer given in the question. could it be because I need another force for the friction between the top ball and bottom ball? If so I just get more lost because in introduces a new cf
 

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