Why Does Adding Mass to the Ends of a Chain Affect Its Slipping Time?

[no_brainer]

1. A chain of mass m is on a frictionlesss table, half of the chain is hanging over the table's edge. We let the chain go and after time T1, the chains slips off the table. We repeat this experiment but this time we add two identical objects(each of mass M) at each end of the chain. After time T2 the chains slips off the table.

How would I show that T1 <T2?

The thing I don't get is how did we let the chain go in the first place..was it with a force that we exert on it or what? Also I think that the chain was in equilibrium that's why its not going anywhere..since there is no friction. Why would T2 be less..I think that it should be equal because since there is no friction..

What do you think?

 

[NateTG]

Can you figure out the forces involved.

You might want to look at the situation when only 1/4 of the chain is left on the table.

 

[Necrosis]

well, from what I see, (and know, I'm just a begginer), you have gravity acting on half of the chain, the part hanging over the table, and with no friction, there is no opposing force keeping it from sliding off the table.

so this is kinda what it looks like;


000000
0
0
0
0
You have gravity pulling down creating a force, from there, you can take it, I just woke up ;)