How Do You Incorporate Drag Force in Free Body Diagrams for Multiple Objects?

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To incorporate drag force in free body diagrams for multiple objects, it is essential to start by drawing a separate free body diagram for each object, in this case, the ducks. Assign unknown forces consistently between the objects, particularly focusing on the drag force exerted by duck 3 on the first two ducks. After establishing the diagrams, apply Newton's second law (F=ma) to each duck, ensuring that all forces, including drag, are accounted for in the equations. This method allows for a systematic approach to analyze the interactions and forces at play. Understanding these steps is crucial for accurately modeling the dynamics of the system.
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Homework Statement
Three toy ducks on wheels are attached to the same rope and a boy begins pulling one end of the rope with a force of 10 N. The first duck (duck 1) has a mass of 1.0 kg, the middle duck (duck 2) has a mass of 5.0 kg, and the third duck in line (duck 3) has a mass of 2.0 kg. I already calculated the acceleration of the three ducks in part a of this question (1.25 m/s^2). The next thing is to find the tension force in the rope connecting ducks 1 and 2. I don't understand how to go about doing this: any help is appreciated!
Relevant Equations
F = ma
I don't even know how to begin this. I know that I need to somehow account for the drag force that duck 3 is causing on the first 2, but I don't know how to deal with that. I am asking for someone to help me get started, not to give me the answer.
 
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There can be shortcuts, but the standard procedure is to draw a free body diagram for each duck and assign unknowns to the forces between them in a consistent manner.
Then write the F=ma equation for each duck.
 
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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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