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One chain-link pulls another

by drawar
Tags: chainlink, pulls
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drawar
#1
Feb26-12, 03:40 AM
P: 132
1. The problem statement, all variables and given/known data
A student tries to raise a chain consisting of three identical links. Each link has a mass of 200 g. The three-piece chain is connected to light string and then suspended vertically, with the student holding the upper end of the string and pulling upward. Because of the student's pull, an upward force 15.0 N is applied to the chain by the string. Find the force exerted by the top link on the middle link.
A) 3.0 N
B) 6.0 N
C) 8.0 N
D) 10.0 N
E) None of the above

2. Relevant equations
F=ma


3. The attempt at a solution
Due to the force, the chain accelerates upwards at a=15/0.6=25 m/s^2.
Let A, B, C be the top, middle and bottom link respectively.
Apply Newton's second law twice to C and B, we have
[tex]F_{BC} - mg = ma[/tex]
hence [tex]F_{BC} = 7.0 N[/tex]

[tex]F_{AB} - F_{CB} - mg = ma[/tex]
but [tex]F_{CB} = F_{BC} [/tex]
thus [tex]F_{AB} = 14.0 N[/tex]
There are no keys available so I really need a clarification to my answer. Thanks!
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LawrenceC
#2
Feb26-12, 06:12 AM
P: 1,195
You have computed the acceleration correctly. But after that you went awry. At the bottom of the top link you have 0.4 kg suspended and accelerating upward. Draw a FBD of the bottom of the top link and look at the forces. There are three. One is the upward pull that you seek.
th4450
#3
Feb26-12, 06:44 AM
P: 38
No, the acceleration isn't correct.
NET FORCE = ma
You missed out the weight of the system.

drawar
#4
Feb26-12, 06:47 AM
P: 132
One chain-link pulls another

Quote Quote by LawrenceC View Post
You have computed the acceleration correctly. But after that you went awry. At the bottom of the top link you have 0.4 kg suspended and accelerating upward. Draw a FBD of the bottom of the top link and look at the forces. There are three. One is the upward pull that you seek.
But I can only figure out 2 forces. One is the upward force exerted by the top link and the other is the weight of the two lower links. Am I missing something?
th4450
#5
Feb26-12, 06:49 AM
P: 38
Quote Quote by drawar View Post
But I can only figure out 2 forces. One is the upward force exerted by the top link and the other is the weight of the two lower links. Am I missing something?
Its own weight
drawar
#6
Feb26-12, 06:56 AM
P: 132
Quote Quote by th4450 View Post
Its own weight
Then the required pull is (10)(0.6)+(25)(0.4)=16.0 N, right?
th4450
#7
Feb26-12, 07:02 AM
P: 38
Quote Quote by drawar View Post
Then the required pull is (10)(0.6)+(25)(0.4)=16.0 N, right?
The acceleration of the system isn't 25m/s^2 because you missed out the weight of the system which is 6N.
drawar
#8
Feb26-12, 07:10 AM
P: 132
Quote Quote by th4450 View Post
The acceleration of the system isn't 25m/s^2 because you missed out the weight of the system which is 6N.
Ah I see, it should have been 15 m/s^2, then F=(10)(0.6)+(15)(0.4)=12.0 N. Does it seem fine now?
th4450
#9
Feb26-12, 07:20 AM
P: 38
Quote Quote by drawar View Post
Ah I see, it should have been 15 m/s^2, then F=(10)(0.6)+(15)(0.4)=12.0 N. Does it seem fine now?
The weight of link A (2N) is still missing
LawrenceC
#10
Feb26-12, 07:32 AM
P: 1,195
Quote Quote by LawrenceC View Post
You have computed the acceleration correctly. But after that you went awry. At the bottom of the top link you have 0.4 kg suspended and accelerating upward. Draw a FBD of the bottom of the top link and look at the forces. There are three. One is the upward pull that you seek.
Whoops, it was a little too early in the morning when I typed this. My first sentence was incorrect.
drawar
#11
Feb26-12, 07:35 AM
P: 132
Quote Quote by th4450 View Post
The weight of link A (2N) is still missing
Do I need to include this in calculation? I am considering the two lower links (B and C) accelerating under the 3 forces.
I'm getting confused. Is the force acting on (B+C) by A equals the force acting on B by A? What if the chain now consists of m identical links and we are to find the force acting on the nth link by the (n-1)th link (n<=m)?
th4450
#12
Feb26-12, 07:42 AM
P: 38
Quote Quote by drawar View Post
F=(10)(0.6)+(15)(0.4)=12.0 N. Does it seem fine now?
Quote Quote by drawar View Post
I am considering the two lower links (B and C) accelerating under the 3 forces.
If you are considering B and C, then it should not be 15 N there.
drawar
#13
Feb26-12, 07:54 AM
P: 132
Nah, it's not 15 N, it's the acceleration of (B+C)
th4450
#14
Feb26-12, 07:59 AM
P: 38
Quote Quote by drawar View Post
Nah, it's not 15 N, it's the acceleration of (B+C)
Oh I see, sorry.
The net force acting on B+C is the vector sum of the tension in the mid string (answer required) and the weight of B+C, and this net force = ma = (0.2+0.2)(15)
You should get the correct answer by now.
drawar
#15
Feb26-12, 09:35 AM
P: 132
So if we consider the motion of (B+C), then there are only 2 forces acting on it,
One is the upward force exerted by the top link and the other is the weight of the two lower links
which result in an acceleration of 15 m/s^2.
Thank you for clearing it up for me :D


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