Four different tensions on a cord with four different disks attached.

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The discussion revolves around a system of four disks suspended by cords, with specific tensions measured for each disk. The top cord exerts a force of 98N, while the tensions for the shorter cords are T1=56.6 N, T2=50.8 N, and T3=8.1 N. To determine the mass of each disk, it is essential to consider the tensions acting on both sides of each disk, as the system is in equilibrium. The net force on each disk must equal zero, meaning all forces, including the tensions from the cords, must be accounted for in the calculations. Understanding this equilibrium condition is crucial for solving the problem effectively.
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The figure shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a pulley and pulls with a force of magnitude 98N on the wall to which it is attached. The tensions in the shorter cords are T1=56.6 N, T2=50.8 N, and T3=8.1 N. What is the mass of disk A,B,C,D?

Unfortunately I can't put the figure on here so hopefully the question is descriptive enough. What's throwing me off on this question is that there are four different tensions. Everything is vertical and I'm wondering if in order to calculate the mass of each disk do the tensions on both sides of the disk need to factored in? For example disk A is at the top and has a cord above with a tension of 98N and has a cord attached below with tension of 56.6N that attaches to disk B. This is what's confusing me, any help is appreciated.
 
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bearhug said:
Everything is vertical and I'm wondering if in order to calculate the mass of each disk do the tensions on both sides of the disk need to factored in? For example disk A is at the top and has a cord above with a tension of 98N and has a cord attached below with tension of 56.6N that attaches to disk B.
You need to take advantage of the fact that the disks are in equilibrium, which means that the net force is zero on each. If two ropes pull on a disk, you must include both tensions in calculating the net force.
 
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