Two masses attached to a fixed vertical spring

Question

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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!

Answer 1

Good question. Consider the following two scenarios.

1) Your friend fell out the window while trying to stop his refrigerator from falling out the window and you caught him. He is dangling out of the window and you are holding him by the ankle. He is holding the refrigerator in his hand. His arm which is supporting the fridge is essentially a spring.

2) Your friend fell out the window taped to a refrigerator and you caught him. He is dangling out the window and you are holding him by the ankle. The tape is very strong, rigid, and essentially massless.

The means by which your friend is attached to the refrigerator doesn't change the force you feel AT EQUILIBRIUM. In situation 1, the springiness of his arm will initially soften the force you feel while the refrigerator's motion is halted. However, after a little while once the system comes to static equilibrium, you will know longer be able to tell how your friend is attached to the refrigerator.

One way to think about this is with Newton's Third Law. One statement of Newton's Third Law is that every force has an equal and opposite reaction force. If your friend is pulling up with the force of the weight of the fridge ($W$), the fridge must also be pulling down on your friend with the force $W$. Since your friend is not accelerating downward, the force you exert on your friend must be equal to the sum of his weight and $W$, the weight of the fridge. In this way the forces on your friend are balanced and he is at static equilibrium. Try drawing a free-body diagram to corroborate this reasoning.

Answer 2

Replace spring B with a massless rod. Do you expect spring A to extend more, less or the same? Why?

You can treat the whole hanging system of the two masses and the middle spring as black box. As long the the total weight does not change, it does not matter how much spring B deflects (or not in case of a rod). It is going to deflect as much as it is needed in order to transfer the weight of the second mass to the first mass.