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?
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!