Calculate the Vertical Acceleration

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SUMMARY

The discussion focuses on calculating the vertical acceleration of a 70-lb cylinder in two scenarios involving a 122-lb force. The user successfully calculated the acceleration for part b as 23.92 ft/s² using the equation ƩFy = may = 122lbf - 70lbf. However, confusion arises regarding the impact of the 122-lb block's acceleration on the 70-lb block in part a. The key insight is that the tension in the cable remains constant on both sides, allowing for the establishment of two equations with two unknowns to solve the problem accurately.

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Northbysouth
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Homework Statement


Calculate the vertical acceleration a (positive if up, negative if down) of the 70-lb cylinder for each of the two cases illustrated. Neglect friction and the mass of the pulleys.

I have attached an image of the problem

Homework Equations





The Attempt at a Solution



I think I've managed to calculate part b:

ƩFy = may = 122lbf - 70lbf

ay = (122lbf - 70lbf)/(70/32.2)

ay = 23.92 ft/s2

But I'm unsure with part a. Initially I has thought they were the same scenario, but after looking at the practice example (the same scenario with the final answers but the numbers are different) I can see that ay is different in both scenarios. I can see how in situation a that the 122 lb force has an acceleration which is not present in situation b but I don't see how the the acceleration in the 122 lb block would impact the 70 lb block.

Help is appreciated
 

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Northbysouth said:
I can see how in situation a that the 122 lb force has an acceleration which is not present in situation b but I don't see how the the acceleration in the 122 lb block would impact the 70 lb block.
You could think of it as the 122lb mass is 'using' some of the 122lbf acting on it for its own acceleration. So it is not all transmitted through the cable.
Let the tension in the cable be T. This is the same both sides (yes?). The acceleration is also the same both sides. That gives you two equations with the same two unknowns.
 

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