Tension of strings in an elevator

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SUMMARY

The discussion focuses on calculating the tension in two strings connected to a 4kg mass and a 6kg mass suspended in an elevator under various conditions. Key scenarios include the elevator being stationary, moving upwards or downwards at a constant speed of 3 m/s, and accelerating upwards or downwards at 3 m/s². The primary formula for tension is derived from Newton's second law, F=ma, and considerations of net force acting on the masses. Participants emphasize the importance of understanding the direction of forces and applying Newton's laws appropriately.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of forces and tension in physics
  • Familiarity with mass and weight calculations
  • Concept of acceleration and its effects on forces
NEXT STEPS
  • Study the application of Newton's second law in different scenarios
  • Learn about tension in strings and cables in physics
  • Explore the effects of acceleration on forces in moving systems
  • Practice problems involving forces in elevators and similar contexts
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Students studying physics, educators teaching mechanics, and anyone interested in understanding forces and tension in dynamic systems.

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Two masses are connected by a string and are hanging from the ceiling of an elevator like this:

[string 1]
|
[4kg mass]
|
[string 2]
|
[6kg mass]

Show the tension of each string in the following situations:
i) Stationary
ii) Moving up at 3m/s
iii) Moving down at 3m/s
iv) Moving up at 3m/s/s
v) Moving down at 3m/s/s

Ok, I am kinda stumped. Where do I start? Is there a formula for tension? Is it F=ma? Or F=m(9.8-a) or something?

I'm just so horrible at physics, I'll be so grateful if someone could explain how this works..

Thanks! :)
 
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You just need to look at the net force on each string. For example, you have two forces on the first string (weights of the hanging bodies). Other one is even easier. Remember the 1st Newton's law. That's when the elevator is stationary.

The formula F=ma can be used in part (iv) and (v). It's the inertial force. Just be aware of its direction.
 

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