# Homework Help: Static problem with three cylinders

1. Jun 3, 2007

### sangol

1. The problem statement, all variables and given/known data

Static problem

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

http://img440.imageshack.us/img440/2040/cylindersjo4.jpg [Broken]

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))

2. Relevant 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})

3. 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}

:uhh:

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 !?

Last edited by a moderator: May 2, 2017
2. Jun 3, 2007

### sangol

Ok i solved it !!

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

2. Relevant 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, 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. :rofl: