Doc Al said:
Here are a few hints. At any given time, the scale must:
(1) support the weight of the chain segment already on it
(2) stop the momentum of the moving chain hitting it
#2 is where dp/dt comes in. Consider an element of chain mass (dm) hitting the scale in time dt. You'll need to figure out how fast the chain is falling as a function of time (and thus length). Then figure out how much mass hits in time dt (what length of chain hits in that time). Then you can figure out dp/dt.
This is somewhat less accurate than it needs to be, since it ignores the possible effects of induced stress in the chain.
Here's how to do it slightly more accurate:
For each time interval \delta{t}, we consider the partition P^{(\delta{t})} of the chain into 3 parts, each that has constant mass in the said time interval between instants t and t+\delta{t}:
a) C_{U}^{(\delta{t})}(t), the upper part of the chain that remains strictly above the scale in the whole interval.
We assign to this part the velocities v_{U}^{(\delta{t})}(t),v_{U}^{(\delta{t})}(t+\delta{t}) where v_{U}^{(\delta{t})}(t) is the velocity at time t and v_{U}^{(\delta{t})}(t+\delta{t}) is the velocity at time t+\delta{t}.
We require that the velocity function is continuous.
The part has mass m^{(\delta{t})}(t) throughout our interval.
b) C_{L}^{(\delta{t})}(t), the lower part of the chain that was strictly lying on the scale at time t, and still is so ever after.
This is assumed to be at rest throught the time interval; its mass is m_{U}^{(\delta{t})}(t)
c)C_{E}^{(\delta)}(t), the connecting element that at time t has velocity v_{U}^{(\delta{t})}(t) and has fallen to rest at time t+\delta{t}. Its mass is \delta{m}^{(\delta{t}}.
Now, we split up our chain in two sub-systems, S_{U+E}^{(\delta{t})}, that consists of a)+c), and S_{L}^{(\delta{t})}, that consists of b).
We may now state how Newton's laws are approximately given for the two subsystems:
S_{U}^{(\delta{t})}:
F^{(\delta{t})}_{U+E}-m_{U}^{(\delta{t})}(t)g\approx\frac{m_{U}^{(\delta{t})}(t)(v_{U}^{(\delta{t})}(t+\delta{t})-v_{U}^{(\delta{t})})}{\delta{t}}-\frac{\delta{m}^{(\delta{t})}v_{U}^{(\delta{t})}}{\delta{t}}=m_{U}^{(\delta{t})}(t)\frac{v_{U}^{(\delta{t})}(t+\delta{t})-v_{U}^{(\delta{t})}(t)}{\delta{t}}-\frac{\delta{m}^{(\delta{t})}}{\delta{t}}v_{U}^{(\delta{t})}(t)
where F_{U+E}^{(\delta{t})} express the forces from either the lower chain or from the scale at an instant within the time interval were looking at.
S_{L}^{(\delta{t})}:
F^{(\delta{t})}_{L}-m_{L}^{(\delta{t})}(t)g=0
We may now, of course, sum together these expressions to get rid of any stresses within the chain and proceed, but I choose to keep them in the following in order to illustrate something later on.
Now is the time for a rather convoluted limiting process!
By going to the limit \delta{t}\to{0}, what we are actually doing, is to go through DISTINCT PARTITIONS P^{(\delta{t})}.
So, by going through these different chain-partitions, we get to the material system for which only an infinitesemal part changes its velocity abruptly.
We define: a_{U}(t)=\lim_{\delta{t}\to{0}}\frac{v_{U}^{(\delta{t})}(t+\delta{t})-v_{U}^{(\delta{t})})}{\delta{t}}
and we have \lim_{\delta{t}\to{0}}\frac{\delta{m}^{(\delta{t})}}{\delta{t}}=\frac{M}{L}|v_{u}(t)| where M is the total chain's mass, L the total length of the chain, and v_{U}(t)=\lim_{\delta{t}\to{0}}v_{U}^{(\delta{t})}(t)
With x(t) being the length on the table, we therefore gain Newton's second laws:
F_{U+E}-\frac{M}{L}(L-x(t))g=-\frac{M}{L}|v_{U}(t)|v_{U}(t)+\frac{M}{L}(L-x)a_{U}(t)(1)
F_{L}-\frac{M}{L}xg=0(2)
Summing together removes the possible stress within the chain, and remembering that v_{U}(t)\leq0 we gain the expression for the normal force:
F_{N}(t)=Mg+\frac{M}{L}v_{U}^{2}+M(1-\frac{x(t)}{L})a_{U}=M(g+(1-\frac{x}{L})a_{U})+\frac{M}{L}v_{U}^{2}(3)
It can be seen that the only resulting difference between between the suggested approach and my own is that my method allows for a_{U}\neq-g
Finally, a word about the possible stress in the chain:
Let us set F_{U+E}=F_{U,S}+F_{C} with (U,S) and C subscripts for scale force on the system and stress force from the rest of the chain.
Similarly, we write F_{L}=F_{L,S}-F_{C} using Newton's 3.law.
Since we have F_{U,S}=0 when the whole chain lies on the scale, we get from (1), F_{C}=0, that is, once the chain is lying wholly on the scale, any stress that may have been has vanished..
, that is in fact equal to mg.
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