# Newtons Laws Problem

1. Oct 21, 2008

### iluvphys

Dear all, I hope you could help me out a little.

1. The problem statement, all variables and given/known data
A person tries to raise a chain of 3 identical links; where each of the links has mass m.
The chain is connected to a string is suspended vertically with the person holding the upper end of the string and pulling upward.
Because of the person´s pull an upward force of F is applied to the chain by the string.

a) Find the acceleration of the chain.
b) Find the force exerted by the top link on the middle link.

2. Relevant equations
F = m*a

3. The attempt at a solution

I was able to get the acceleration which is (F-3mg)/3m!

For part B i was able to get the answer: 2(mg+ma). Which i believe is correct. But the program i am using tells me the answer is wrong and doesnt depend on the variables m and a.

2. Oct 21, 2008

### alphysicist

Hi iluvphys,

I think they want you to write the force on the middle link (from the upper link) in terms of the applied force F. What would that be?

3. Oct 22, 2008

### iluvphys

Hello alphysicist, thank you for your response.
I figured since the first equation describes the chain which oncisist of the links i must break it down like you said, so:

m2*a2 = F/3 - m1*g

Is that correct? Thanks again

4. Oct 22, 2008

### alphysicist

I don't believe this equation is correct here.

In my last post I was saying that you already have found an expression for the force you are looking for (between block 1 and 2):

f = 2mg + 2 ma

and you already have an expression for the acceleration (that has the person's force F):

a = (F-3mg)/3m

Can you put those together and get rid of m and a?

5. Oct 22, 2008

### iluvphys

I put those two equations together by replacing a with F-3mg/3m.
But what i get is mf=0.
So i am doing something wrong!?!
How else can i put these equations together?
Thanks for your help by the way-

6. Oct 22, 2008

### alphysicist

My guess is you just have an algebra error somewhere.

The force you are looking for is:

$$f = 2mg + 2 ma$$

and you said you plugged in your value for acceleration a which would turn this into:

$$f = 2 mg + 2 m \left(\frac{F-3mg}{3m}\right)$$

So now simplify the right hand side. What do you get? I believe you should find that the masses cancel, but F (the external applied force on block 1) is still present. If you still do not get the answer, please post the steps you are taking.

7. Oct 22, 2008

### iluvphys

My goodness I solved it. I couldnt do it before because I was bringing the equations to the left hand side as well thats why i got my odd solution.
So the force applied on the middle link by the upper one is 2/3 F. Am i correct?

8. Oct 22, 2008

### alphysicist

That looks right to me. And does that make sense? If F is required to accelerate three blocks, (2/3)F is required to accelerate two blocks (because the force that m1 puts on m2 is responsible for "pulling" both m2 and m3).

9. Oct 22, 2008

### iluvphys

Thank you very much. I "could" have solved it using my brain since your explanation is so simple and true, I mean not having to depend on equations.
Well, thank you very much!

10. Oct 22, 2008

### alphysicist

Glad to help! And it's important to learn how to use the formalism, because you will certainly encounter complicated problems where your intuition might not be much help.