Calculate Tension Forces in Rope for Two Barges

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SUMMARY

The discussion focuses on calculating the tension forces in ropes connecting a boat to two barges, A and B, with respective inertias of 2x10^5 kg and 3x10^5 kg. The resistive forces acting on barge A and barge B are 7x10^3 N and 9x10^3 N, respectively, while the common acceleration for the system is 0.4 m/s². The tension in the rope connecting the boat to barge A is calculated as 7.3x10^4 N, and the tension in the rope connecting the two barges is determined to be 1.11x10^5 N. The calculations remain consistent regardless of the order of the barges.

PREREQUISITES
  • Newton's Second Law of Motion
  • Basic principles of tension in ropes
  • Understanding of resistive forces in fluid dynamics
  • Knowledge of inertia and its impact on motion
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  • Study the effects of varying resistive forces on tension calculations
  • Explore advanced concepts in fluid dynamics related to barge movement
  • Learn about the impact of changing acceleration on tension in multi-body systems
  • Investigate the role of inertia in complex mechanical systems
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KatyM7
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A boat pulls two barges down the river. The barge connected to the boat (barge A) has an inertia of 2x10^5 kg, and the other barge (barge B) has an inertia of 3x10^5 kg. the resistive force between barge A and the water is 7x10^3 N, and the resistive fore of barge B to the water is 9x10^3 N. The common acceleration of all three boats is .4 m/s^2. Even thought the ropes are huge, the gravitational exerted on them is much smaller than the pulling forces.(a). What is the tension in the rope that connects the boat to barge A? (b). What is the tension in the rope that connects the two barges? (C). Repeat steps a and b for the case in which the order of the barges is reversed.

So for part (a) I took the contact force of the rope to barge A which is 2x10^5kg * .4m/s^2= 8x10^4 N and added the negative of the resistive force; 8x10^4N - 7x10^3N= 7.3x10^4N. I wasn't sure if I should also subtract the contact force of the second rope on barge A which would cancel out the two contact forces and the answer would be the resitive force of barge A and the water.
(b). Contact force of the second rope on barge B; 3x10^5kg * .4m/s^2 = 1.2x10^5N; subtract the resistive force of barge B to the water; 1.2x10^5N - 9x10^3N = 1.11x10^5N. Again not sure if I should add the contact force of the first rope on barge B.
(c). They would be the same unless I do subtract/add the parts I'm not sure about.
 
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Mark44 said:
Please use the template in future posts...
My apologies, I am very new to this site. I have reposted the question with the provided template. Thank you for letting me know!
 

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