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One chainlink pulls another 
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#1
Feb2612, 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 threepiece 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! 


#2
Feb2612, 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.



#3
Feb2612, 06:44 AM

P: 38

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


#4
Feb2612, 06:47 AM

P: 132

One chainlink pulls another



#5
Feb2612, 06:49 AM

P: 38




#6
Feb2612, 06:56 AM

P: 132




#7
Feb2612, 07:02 AM

P: 38




#8
Feb2612, 07:10 AM

P: 132




#9
Feb2612, 07:20 AM

P: 38




#10
Feb2612, 07:32 AM

P: 1,195




#11
Feb2612, 07:35 AM

P: 132

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 (n1)th link (n<=m)? 


#12
Feb2612, 07:42 AM

P: 38




#13
Feb2612, 07:54 AM

P: 132

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



#14
Feb2612, 07:59 AM

P: 38

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. 


#15
Feb2612, 09:35 AM

P: 132

So if we consider the motion of (B+C), then there are only 2 forces acting on it,
Thank you for clearing it up for me :D 


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