One chain-link pulls another

  • Thread starter drawar
  • Start date
In summary: Do I need to include this in calculation? I am considering the two lower links (B and C) accelerating under the 3 forces.
  • #1
drawar
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0

Homework Statement


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

Homework Equations


F=ma


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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  • #2
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.
 
  • #3
No, the acceleration isn't correct.
NET FORCE = ma
You missed out the weight of the system.
 
  • #4
LawrenceC said:
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?
 
  • #5
drawar said:
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
 
  • #6
th4450 said:
Its own weight

Then the required pull is (10)(0.6)+(25)(0.4)=16.0 N, right?
 
  • #7
drawar said:
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.
 
  • #8
th4450 said:
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?
 
  • #9
drawar said:
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
 
  • #10
LawrenceC said:
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.
 
  • #11
th4450 said:
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)?
 
  • #12
drawar said:
F=(10)(0.6)+(15)(0.4)=12.0 N. Does it seem fine now?

drawar said:
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.
 
  • #13
Nah, it's not 15 N, it's the acceleration of (B+C)
 
  • #14
drawar said:
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. :wink:
 
  • #15
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
 

1. What is the concept behind "One chain-link pulls another"?

The concept behind "One chain-link pulls another" is that when a force is applied to one end of a chain, it will transmit that force to the other end of the chain. This is due to the intermolecular forces between the molecules within the chain, creating a tension that allows for the transmission of the force.

2. How does "One chain-link pulls another" relate to Newton's third law of motion?

"One chain-link pulls another" is an example of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In this case, the force applied to one chain-link is the action, and the force transmitted to the other chain-link is the equal and opposite reaction.

3. What factors affect the strength of the force transmitted in "One chain-link pulls another"?

The strength of the force transmitted in "One chain-link pulls another" is affected by several factors, including the material and thickness of the chain, the angle at which the force is applied, and any additional external forces acting on the chain.

4. Can "One chain-link pulls another" be used to lift heavy objects?

Yes, "One chain-link pulls another" can be used to lift heavy objects, as long as the chain is strong enough and the force is applied at the correct angle. This is the principle behind using a pulley system, where a chain or rope is wrapped around a wheel and used to lift heavy objects.

5. What other real-life examples can be found of "One chain-link pulls another"?

"One chain-link pulls another" can be found in many real-life examples, such as bicycle chains transferring pedal force to the wheels, anchor chains pulling boats in the water, and even in the human body where muscles pull on bones to create movement.

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