Help with pool ball pyramid problem

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The discussion revolves around a physics problem involving three identical spheres in contact on a horizontal plane with a fourth sphere resting on them. Participants are trying to derive the minimum coefficient of friction required for equilibrium, specifically that it should be at least (rad(3)-rad(2)) for the spheres and (rad(3)-rad(2))/4 between the spheres and the plane. There is confusion regarding the necessity of friction between the spheres and how to calculate the forces acting on them. Participants are attempting to sum forces and derive equations related to the normal force and friction, but struggle with the geometry involved in determining the correct angles. The conversation highlights the complexity of the problem and the challenges in visualizing the forces in a three-dimensional setup.
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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
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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