Static problem with three cylinders

[sangol]

Homework Statement

Static problem

Three identical cylinders of mass m are arranged in a triangle as shown:

http://img440.imageshack.us/img440/2040/cylindersjo4.jpg

The friction coefficient between cylinders is "u1" and between cylinders and ground "u2". What condition must satisfy "u1" and "u2" to remain the system in equilibrium ?

N1 is normal force between left cylinder and top cylinder (Fr1 friction force in the same point (u1*N1))
N2 is normal force between right cylinder and top cylinder (Fr2 friction force in the same point (u1*N2))
N3 is normal force between left cylinder and right cylinder (Fr3 friction force in the same point (u1*N3))
N4 is normal force between ground and left cylinder (Fr4 friction force in the same point (u2*N4))
N5 is normal force between ground and right cylinder (Fr5 friction force in the same point (u2*N5))

Homework Equations

solve({(1/2)*N1*sqrt(3)+(1/2)*N2*sqrt(3)+(1/2)*Fr2-(1/2)*Fr1-m*g = 0, (1/2)*Fr1*sqrt(3)+(1/2)*N1+(1/2)*Fr2*sqrt(3)-(1/2)*N2 = 0, N4+Fr3+(1/2)*Fr1-(1/2)*N1*sqrt(3)-m*g = 0, Fr4-N3-(1/2)*Fr1*sqrt(3)-(1/2)*N1 = 0, N5-m*g-Fr3-(1/2)*N2*sqrt(3)-(1/2)*Fr2 = 0, N3-Fr5+(1/2)*N2-(1/2)*Fr2*sqrt(3) = 0,Fr1+Fr2 = 0, Fr4+Fr3+Fr1 = 0, Fr2+Fr3-Fr5 = 0, Fr1 = u1*N1, Fr2 = u1*N2, Fr3 = u1*N3, Fr4 = u2*N4, Fr5 = u2*N5}, {u1, u2, Fr1, Fr2, Fr3, Fr4, Fr5, N1, N2, N3, N4, N5})

The Attempt at a Solution

{Fr2 = 0, Fr3 = 0, Fr1 = 0, Fr4 = 0, u1 = 0, N5 = (3/2)*m*g, N4 = (3/2)*m*g, N3 = -(1/6)*m*g*sqrt(3), N1 = (1/3)*m*g*sqrt(3), N2 = (1/3)*m*g*sqrt(3), u2 = 0, Fr5 = 0}



Possible interpretation of the result:

The solution is correct (i suppose) to frictionless case, so with friction:

Fr4 + N3 = 0 -> u2 * N4 = - N3 -> u2 * (3/2)*m*g = (1/6)*m*g*sqrt(3) -> u2 = sqrt(3)/9 ~ 0.19 ?

u2 = 0.19 and u1 = 0 is sufficient to equilibrium !?

 

[sangol]

Ok i solved it !

If you put u1, u2, u3, u4, u5 as friction coefficients (instead of u1 and u2) in equations:

Homework Equations

:[/B]

solve({(1/2)*N1*sqrt(3)+(1/2)*N2*sqrt(3)+(1/2)*Fr2-(1/2)*Fr1-m*g = 0, (1/2)*Fr1*sqrt(3)+(1/2)*N1+(1/2)*Fr2*sqrt(3)-(1/2)*N2 = 0, Fr1+Fr2 = 0, N4+Fr3+(1/2)*Fr1-(1/2)*N1*sqrt(3)-m*g = 0, Fr4-N3-(1/2)*Fr1*sqrt(3)-(1/2)*N1 = 0, Fr4+Fr3+Fr1 = 0, N5-m*g-Fr3-(1/2)*N2*sqrt(3)-(1/2)*Fr2 = 0, N3-Fr5+(1/2)*N2-(1/2)*Fr2*sqrt(3) = 0, Fr2+Fr3-Fr5 = 0, Fr1 = u1*N1, Fr2 = u2*N2, Fr3 = u3*N3, Fr4 = u4*N4, Fr5 = u5*N5},{u1, u2, u3, u4, u5, Fr1, Fr2, Fr3, Fr4,Fr5, N1, N2, N3, N4, N5})

3. Solutions:

N1 = (1/2)*m*g
N2 = (1/2)*m*g
N3 = 0
N4 = (3/2)*m*g
N5 = (3/2)*m*g

Fr1 = 1/2*(sqrt(3)-2)*m*g
Fr2 = -1/2*(sqrt(3)-2)*m*g
Fr3 = 0
Fr4 = -1/2*(sqrt(3)-2)*m*g
Fr5 = -1/2*(sqrt(3)-2)*m*g

u1 = -2+sqrt(3) ~ -0.268 (negative because Fr1 have opposite direction)
u2 = 2-sqrt(3) ~ 0.268
u3 anyone
u4 = -(1/3)*sqrt(3)+2/3 ~ 0.089
u5 = -(1/3)*sqrt(3)+2/3 ~ 0.089

So, the solutions are:

Friction coefficient between cylinders: 0.268
Friction coefficient between cylinders and ground: 0.089

I though that friction was more important between ground and cylinders that between cylinders for stability but the result show opposite to my logic.

If someone finds any error in operations, please reply the thread.

Greetings!