1 Million CN Question!

[Fortress]

13. [1pt]
A flexible chain of length 1 m, and mass 2 kg is held from one end so that it hangs vertically, the lower end just touching the surface of a table. The upper end is suddenly released so that the chain falls onto the table and coils up in a small heap, each link coming to rest as soon as it strikes the table. What is the force (in newtons) exerted by the table on the chain when 0.9 m of chain have landed on the table?

1 million plus ubër-props for the solution and explanation to this problem. You have until October 9th, 4 pm or until I figure it out.

[Mr_Cynic]

I'm almost sure I'm wrong, but I think 20 Newtons.  I'd really need an illustration of a problem that oddly worded.

[—-]

How the hell do you figure that out.

[Delta1212]

um...

18N?

0.9 * 2 = 1.8 * 10 = 18

[Ganon5]

Is this your homework fortress?

[Grand Poobah Marx]

um...

18N?

0.9 * 2 = 1.8 * 10 = 18

Newtons are measure in kg, not pounds, you'd have to convert.

[Delta1212]

2 kg

[IronSoldier820]

I hated Physics. Mainly because my teacher was a Football Coach who didn't know a single bit of Physics either.

[Delta1212]

I hated Physics. Mainly because my teacher was a Football Coach who didn't know a single bit of Physics either.

Mine was an oblivious Ukranian woman.

[Vladimir]

7.56 N

F=MA
2kg*4.2M/SEC^2

Probably wrong...

[Fortress]

I'm almost sure I'm wrong, but I think 20 Newtons.  I'd really need an illustration of a problem that oddly worded.

um...

18N?

0.9 * 2 = 1.8 * 10 = 18

7.56 N

F=MA
2kg*4.2M/SEC^2

Probably wrong...

I'm pretty sure that these are not correct. I thought it was simple, too, but the hint that my professor gave us was this:

Hint: This is a hard problem, but there are several different approaches. Use F(total)=d(MV)/dt, and remember that the total force on the chain includes the force of gravity on it. Here, both M and V are functions of time. Alternatively use the center of mass position and total mass which is not a function of time, and F = d(MV_cm)/dt.

I'm not quite sure what exactly that means... hence the prize. I also agree that the question is a strange one.

[Fortress]

Is this your homework fortress?

lol perhaps... >_<

I hated Physics. Mainly because my teacher was a Football Coach who didn't know a single bit of Physics either.

Mine was an oblivious Ukranian woman.

For some reason I have two professors, one German and one Russian. Neither of them teach other than just my class... o.O

[Vladimir]

Managed to dig up this from http://en.allexperts.com/q/Physics-1358/force-chain.htm, hope it helps.

"The problem is that you are only considering the weight of the chain. The table also has to exert an upward force to stop the chain from falling! That can be determined by using the "impulse" form of Newton's 2nd Law F=m*Dv/Dt.
In this case F=v*Dm/Dt Dm is the mass of one little piece of the chain. SInce the chanin has a mass of 2 kg and a length of 1m the mass per unit length of the chain is lambda=mass/length=2kg/m. If you multiply this mass per unit length by the length of a piece of the length of the chain Dl you will get the mass of the little piece.
Dm=lambda*Dl. Newton's 2nd Law then becomes
F=v*Dm/Dt=v*lambda*Dl/Dt where Dl/Dt is the velocity of the chain therefore F=v*lambda*v=v^2*lambda.
Before finishing the problem you need to determine the time t when the chain contacts the table. Do=0, Df=-0.6m, Vo=0m/s, a=-9.8m/s^2, Vf=? and t=?
Using the displacement equation determine the time
Df=1/2*a*t^2+Vo*t+Do
-0.6=1/2*(-9.8)*t^2+0+0 solve for t.
Then use the velocity equation to determine the velocity of the chain link at the time it strikes the table.
Vf=a*t+Vo
Vf=-9.8*t+0 solve for Vf. The magnitude of this velocity is the speed of the chain when it collides with the table.
Use this velocity to calculate the force required to stop the falling chain.  
Finally, the forc exerted by the table will be equal to the sum of the force needed to support the weight of the chain plus the force needed to stop the link of the falling chain."

[Fortress]

Amazing. I don't believe it. That question on the website is the same exact question posted on my homework, except a difference in the chain length hitting the table. That can only mean some kid like me did the exact same thing I did except on a different site. Hahaha...

A process which gives me a final result of ~52.95N. Vladimir, I thank you for using whatever search engine that came up with that site and then thus posted the general explanation here.

We have a $1 million CN winner!

Also, since you're apparently new, I'll do the EA a favor and also contribute joining funds when you're validated. Expect funds soon.

[schneereich]

lol i'm gonna start posting my homework on RIA too:

whats the fourth derivative of sqrt(4-x^3) ?

[xristosx]

lol i'm gonna start posting my homework on RIA too:

whats the fourth derivative of sqrt(4-x^3) ?

im assuming either you just made some number up or you are doing maths for NASA

well i did do it with my calculator took me 2 mins

so im just going to write it in 2 sections far out this is big

NOTE: the --------- is the fraction

9x^2(x^3-40)                    
----------------------------
4(x^3-4)^2*sqrt(4-x^3)

..45x^2(x^6-80*x^3-128)  
+-----------------------------
................16(4-x^3)^(7/2)

aint technology great ^^