Chain Friction Partition Problem

Question

A chain with a uniform linear mass density (mass per unit length) is placed on a surface with a static coefficient of friction µ. What is the maximum fraction of the chain's total length that can hang over the edge of the surface without the chain slipping off? Tried to split the chain into different parts to look at. Maybe that will help?

Answer 1

Hi Luke!

 

Splitting the chain into a "hanging" and "on the table" section will help, I think.  In the ideal consideration, this falls into the same category as the question you posted about the two masses, one on a table and one hanging, connected by a string slung over a pulley.  So I will appeal to that question in the setup; refer to that one for the concept I am using here.

 

The force that would pull the chain in the "down to the floor/toward the edge of the table" direction (which I will call positive) is the force of gravity on the hanging part of the chain.

 

The force acting against that, and thus pointing in my negative direction, is the force of static friction (assuming the chain to be stationary).

 

So, Newton's 2nd law along the axis of motion looks like:

 

F_net = m_Hg - F_s-table = ma = 0 (for the chain being stationary)

 

where

 

m_H = mass of the chain that is hanging off the table

F_s-table = force of static friction between chain and table

 

To have as much of the chain hanging off the table as possible, we assume we have the maximum static friction force trying to hold it on the table.  As long as friction on this length of chain is acting simply -- which we always assume in this case -- we have:

 

m_Hg - μ_sN = 0

 

where

 

N = normal force of the table on the section of chain on the tabletop

μ_s = coefficient of static friction

 

The normal force of the table on the section of chain on top of it equals the weight of that section of chain:

 

N = m_Tg

 

where 

 

m_T = mass of the chain on the table

 

Thus, we have for our Newton's 2nd law expression:

 

m_Hg - μ_sm_Tg = 0

 

Now, with a constant mass per unit length (let's call that λ), we can recast the masses above in terms of lengths of the chain they represent.

 

m = λL  (meaning a given length of chain L has mass m)

 

Thus:

 

m_H = λL_H

m_T = λL_T

 

And our Newton's 2nd law expression becomes:

 

λL_Hg - μsλL_Tg = 0

 

The λ and g cancel out, which leaves:

 

L_H - μ_sL_T

 

Thus, L_H/L_T = μ_s  (meaning more length has to be on the table than hanging, which we might expect)

 

The fraction of the total length (which is just L_T + L_H) hanging off the table can be written as:

 

H = L_H/(L_T+L_H) = μsL_T/(L_T+μ_sL_T) = μ_s/(1+μ_s)

 

This meets the "minimum success criterion," in that it is less than 1, which a fraction of the total length should be.

 

I hope this helps!  If you have any questions about it, or see any problems, just let me know.