(adsbygoogle = window.adsbygoogle || []).push({}); 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!

**Physics Forums - The Fusion of Science and Community**

# One chain-link pulls another

Have something to add?

- Similar discussions for: One chain-link pulls another

Loading...

**Physics Forums - The Fusion of Science and Community**