1. Oct 1, 2008

### jturn

1. The problem statement, all variables and given/known data
A chain consisting of five links, each of mass 0.105 kg, is lifted vertically with a constant acceleration of a = 3.5 m/s2. Find the magnitude of the force that link 3 exerts on link 2.

2. Relevant equations
F = ma

3. The attempt at a solution
I did:
F = ma
FU - (0.105 * 3)(9.8) = (0.105) * 3.5
FU = 4.19 N

where FU is the force pulling up, (0.105*3) is the mass of the link's 5, 4, and 3 weighing down on link 2 (mg=weight) and then the mass*accel on the other side.

This is wrong. why? and how do i get the right answer?

2. Oct 1, 2008

### jturn

does no one know how to do it? ...if it helps. i found the force required to move the top link at that acceleration to be 6.98 N....at it was the right answer...

3. Oct 2, 2008

### PhanthomJay

It looks like your calculation and answer is correct, although you left out a 3 in an apparent typo omission (see above in red). Unless, the links are numbered 1 thru 5 bottom to top, rather than 1 thru 5 top to bottom??