Calculating Tension and Motion in a Three-Box and Pulley System

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In a three-box and pulley system with boxes A, B, and C, the tension in the cord connecting boxes B and C is calculated to be 60.1 N when released from rest. Box A, which is on a frictionless table, moves 0.202 m in the first 0.250 seconds. The discussion emphasizes the need to calculate the tension in the string connecting A to B and C, noting that the system's motion complicates the relationship between tension and gravitational force. The net force on box C is determined using Newton's second law, factoring in the gravitational force and tension. The conversation draws parallels to a similar physics problem, highlighting the importance of understanding acceleration in the system.
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Three boxes (A, B, and C) are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. Box A is on a frictionless table, boxes B and C hang over the side of the table.
The masses are mA = 30.0 kg, mB = 40.0 kg, mC = 18.0 kg.

(a) When the assembly is released from rest, what is the tension in the cord that connects boxes B and C?
60.1 N
(b) How far does box A move in the first 0.250 s (assuming it does not reach the pulley)?
0.202 m

Answers given are the book's answer's. (Actually, part b used to be my answer also, but I can't duplicate it...)

I'd like to start by calculating tension in the string connecting A to B&C, and then say that for A in the x direction, \Sigma\ F=T=m_Aa, and then use a to find delta x. Unfortunately, since the system is in motion, T\neq\ (m_B+m_C)g, and I don't know how to find it.
 
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All the blocks accelerate at the same magnitude, so first find that acceleration.

You know that net force on block C = McA (by Newton's second law) and you also know that net force on C=(Mc)(g)-T.
 
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