Solve Newtons Laws Problem: Find Acceleration & Force

[iluvphys]

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

Homework Statement

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.

Homework Equations

F = m*a

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 doesn't depend on the variables m and a.
Is there another way to describe this? Please help.

 

[alphysicist]

Hi iluvphys,

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

Homework Statement

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.

Homework Equations

F = m*a

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 doesn't depend on the variables m and a.
Is there another way to describe this? Please help.

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?

 

[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

 

[alphysicist]

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

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?

 

[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-

 

[alphysicist]

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-

My guess is you just have an algebra error somewhere.

The force you are looking for is:

[tex]
f = 2mg + 2 ma
[/tex]

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

[tex]
f = 2 mg + 2 m \left(\frac{F-3mg}{3m}\right)
[/tex]

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.

 

[iluvphys]

My goodness I solved it. I couldn't do it before because I was bringing the equations to the left hand side as well that's why i got my odd solution.
Now thanks to your advice i get 2/3 F.
So the force applied on the middle link by the upper one is 2/3 F. Am i correct?

 

[alphysicist]

My goodness I solved it. I couldn't do it before because I was bringing the equations to the left hand side as well that's why i got my odd solution.
Now thanks to your advice i get 2/3 F.
So the force applied on the middle link by the upper one is 2/3 F. Am i correct?

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).

 

[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!

 

[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.