Coefficient of static friction of chain

[coolcat_ka]

Hi! I'm studying for an exam on Friday, and I'm stuck on this problem:

A uniform chain of length 8.00m initially lies stretched out on a horizontal table.
A. Assuming the coefficient of static friction between chain and table is 0.600, show that the chain will begin to slide off the table if at least 3.00m of it hangs over the edge of the table.
B. Determine the speed of the chain as its last link leaves the table, given that the coefficient of kinetic friction between the chain and the table is 0.400

Possible Equations to use?
m=z(y), y=dm/dx
dm/dt=dz/dt(y)

Any help would be appreciated!

 

[tiny-tim]

Welcome to PF!

Hi coolcat_ka! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[coolcat_ka]

ok, so for part a:

m=ro(x), where x=length.
m1=5rx, m2=3r0

F(gravity)-F(friction)=0

so m2g-mv(m1g)=0

(9.8)(3)r0-(9.8)(5.)(0.6)r0-0
3-(5(0.6))=0

And I believe that is how I can prove that the chain starts to slide, but I wanted to makesure I had done that correctly.

But I am stuck on part b - I don't even know where to start.

 

[tiny-tim]

Hi coolcat_ka!

m=ro(x), where x=length.
m1=5rx, m2=3r0

F(gravity)-F(friction)=0

so m2g-mv(m1g)=0

(9.8)(3)r0-(9.8)(5.)(0.6)r0-0
3-(5(0.6))=0

Goodness, that's a mess!

(if you don't have a rho on your keyboard, then use something you do have … like k … it isn't compulsory to use rho for density )

i] you could have divided your equation by g, so that you didn't have to put 9.8 in at all

ii] it would be better practice if you used a general amount, y, instead of 3, and then showed that y = 3.

For part b), use the same basic method (but with kinetic instead of static friction) … Newton's second law … weight of y less friction of (8-y) = mass times acceleration.

 

[coolcat_ka]

Excellent, thank you!