Two masses attached to a fixed vertical spring

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SUMMARY

The discussion centers on a physics problem involving two masses (5 kg each) attached to a fixed vertical spring with a spring constant of 1000 N/m. The key conclusion is that the displacement from equilibrium of Spring A is determined solely by the combined weight of both masses, regardless of the presence of Spring B. The reasoning is based on Hooke's Law and the force of gravity, which indicates that Spring B does not alter the total force acting on Spring A, as it merely transmits the force from Mass B to Mass A without providing additional resistance to gravity.

PREREQUISITES
  • Understanding of Hooke's Law and its application in spring systems
  • Knowledge of gravitational force calculations (F=mg)
  • Familiarity with concepts of equilibrium in mechanical systems
  • Basic principles of force transmission in connected systems
NEXT STEPS
  • Explore advanced applications of Hooke's Law in multi-spring systems
  • Study the effects of damping in spring-mass systems
  • Learn about the principles of static and dynamic equilibrium in physics
  • Investigate the role of atomic forces in material science and their analogy to macroscopic springs
USEFUL FOR

Physics students, educators, and anyone interested in understanding mechanical systems involving springs and forces, particularly in the context of equilibrium and force transmission.

JayB
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Homework Statement


Two mass-less springs with spring constant k = 1000 N/m each have 1 block attached (Spring A is fixed to the ceiling and is attached to a 5 kg Mass A, Spring B is attached and below the 5 kg Mass A and is attached to another 5 kg Mass B at the other end; this system is vertical).
When the masses and springs are resting freely, how far from equilibrium is Spring A extended?

Homework Equations


Hooke's Law: Fspring=(k)(-Δd)
Force of gravity: F=mg

The Attempt at a Solution


Finding the solution is straightforward: you ignore Spring B, make the force of gravity on both masses equal to the force exerted by the spring on both masses, and solve for Δd.

I'm having trouble understanding the solution conceptually. I don't understand why Spring B doesn't contribute to the question. Spring B is attached to Mass B so doesn't it help Spring A resist the pull of gravity on the two masses? I thought that the amount of displacement from equilibrium of Spring A would be less with the inclusion of Spring B than without Spring B.

However, according to the solution, having both Mass A and B attached directly to Spring A would yield the same amount of displacement from equilibrium of Spring A as having Spring B in between Spring A and Mass B. I don't understand why and would greatly appreciate if something could clear this up!
 
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Technically Spring B does contribute since it transmits the force from Mass B to Mass A, doesn't it?

If you think about it, what is Mass A composed of but tiny atoms joined together by atomic forces that are essentially linear springs? But these numerous tiny springs don't enter into the solution either (other than transmitting their individual tiny forces from their masses).
 
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Yes you're right that Spring B transmits the force from Mass B to Mass A and in that sense contributes. I meant "contribute" as in "helping" Spring A to resist the pull of gravity.

And thank you for that explanation. I didn't think of it that way: of matter being composed of numerous springs, which we ignore in such questions.

Thank you!

paisiello2 said:
Technically Spring B does contribute since it transmits the force from Mass B to Mass A, doesn't it?

If you think about it, what is Mass A composed of but tiny atoms joined together by atomic forces that are essentially linear springs? But these numerous tiny springs don't enter into the solution either (other than transmitting their individual tiny forces from their masses).
 

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