Free-body diagrams and Newton's laws with a suspended chain

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Korisnik
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Homework Statement


The chain comprising three rings (each of mass ##0.25kg##) is suspended from a massless rope, and a pulling force ##\left(F=9N\right)## is exerted upwards on the rope. Picture: http://i.imgur.com/xeaiBsc.jpg?1.

I need to find the values of all the unknowns.

Homework Equations


$$a:\ F_y=F_R +w_a=m_aa_a\\
b:\ F_y=F_R+w_b+F_{c,b} =m_ba_b\\
c:\ F_y=F_{b,c}+w_c+F_{d,c} =m_ca_c\\d:\ F_y=F_{c,d}+w_d =m_da_d$$

The Attempt at a Solution


I'm not sure how to solve the problem without assuming that the acceleration of each ring is equivalent to the acceleration of the whole chain ##\left(a_a=a_b=a_c=a_d\right)##. However, I don't know the law/fact that hypothesis is a consequence of.
 
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Korisnik said:
I'm not sure how to solve the problem without assuming that the acceleration of each ring is equivalent to the acceleration of the whole chain ##\left(a_a=a_b=a_c=a_d\right)##. However, I don't know the law/fact that hypothesis is a consequence of.
Suppose you assign variables to the heights of the rings. Assuming they are rigid, what is the relationship between those variables? What do you get if you differentiate those relationships?
 
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haruspex said:
Suppose you assign variables to the heights of the rings. Assuming they are rigid, what is the relationship between those variables? What do you get if you differentiate those relationships?
Hmm, I think I see what you're trying to say: let ##h_i## be height of body ##i## as a function of time, and ##\Delta h## a constant: then ##h_b=h_c+\Delta h##. Differentiating the equation $$\begin{align}\frac{\mathrm d{h_b}}{\mathrm d{t}}&=\frac{\mathrm d}{\mathrm d{t}}(h_c+\Delta h)\\ &\Rightarrow v_b=v_c \\&\Rightarrow a_b=a_c.\end{align}$$Did I do it correctly?
 
Korisnik said:
Hmm, I think I see what you're trying to say: let ##h_i## be height of body ##i## as a function of time, and ##\Delta h## a constant: then ##h_b=h_c+\Delta h##. Differentiating the equation $$\begin{align}\frac{\mathrm d{h_b}}{\mathrm d{t}}&=\frac{\mathrm d}{\mathrm d{t}}(h_c+\Delta h)\\ &\Rightarrow v_b=v_c \\&\Rightarrow a_b=a_c.\end{align}$$Did I do it correctly?
Yes.
 
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