## Help with pool ball pyramid problem

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

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!

## Help with pool ball pyramid problem

Those look correct.
As for theta... thats 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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