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 drawar Feb26-12 03:40 AM

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
$$F_{BC} - mg = ma$$
hence $$F_{BC} = 7.0 N$$

$$F_{AB} - F_{CB} - mg = ma$$
but $$F_{CB} = F_{BC}$$
thus $$F_{AB} = 14.0 N$$
There are no keys available so I really need a clarification to my answer. Thanks!

 LawrenceC Feb26-12 06:12 AM

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 Feb26-12 06:44 AM

No, the acceleration isn't correct.
NET FORCE = ma
You missed out the weight of the system.

 drawar Feb26-12 06:47 AM

Quote:
 Quote by LawrenceC (Post 3784914) 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 Feb26-12 06:49 AM

Quote:
 Quote by drawar (Post 3784954) 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 Feb26-12 06:56 AM

Quote:
 Quote by th4450 (Post 3784956) Its own weight
Then the required pull is (10)(0.6)+(25)(0.4)=16.0 N, right?

 th4450 Feb26-12 07:02 AM

Quote:
 Quote by drawar (Post 3784965) 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 Feb26-12 07:10 AM

Quote:
 Quote by th4450 (Post 3784968) 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 Feb26-12 07:20 AM

Quote:
 Quote by drawar (Post 3784972) 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 Feb26-12 07:32 AM

Quote:
 Quote by LawrenceC (Post 3784914) 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 Feb26-12 07:35 AM

Quote:
 Quote by th4450 (Post 3784980) 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 Feb26-12 07:42 AM

Quote:
 Quote by drawar (Post 3784972) F=(10)(0.6)+(15)(0.4)=12.0 N. Does it seem fine now?
Quote:
 Quote by drawar (Post 3784994) 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 Feb26-12 07:54 AM

Nah, it's not 15 N, it's the acceleration of (B+C)

 th4450 Feb26-12 07:59 AM

Quote:
 Quote by drawar (Post 3785010) 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:

 drawar Feb26-12 09:35 AM