[Kalle]

Could you guys explain this please: Assume an upright spring with a box on top of it. Two forces act on the spring, the box's weight G and the earth's support F. Why is the displacement of the spring calculated F/-k not F+G/-k ? Thank you very much!

[Unknown008]

I don't quite understand you.

Suppose there is a spring, hanging down from a roof.
The extension is at first zero, since there is no resultant force acting on it.
Once you put a mass there, the mass exerts a force on the spring, elongating the latter to a certain extension, due to the gravitational pull acting on the mass.

[Kalle]

I don't quite understand you.

Suppose there is a spring, hanging down from a roof.
The extension is at first zero, since there is no resultant force acting on it.
Once you put a mass there, the mass exerts a force on the spring, elongating the latter to a certain extension, due to the gravitational pull acting on the mass.

According to statics, as I understand it, there are two forces acting on the spring, the weight of the mass and the support of the roof. Why is only one of these forces taken into account when calculating the extension of the spring?

[Unknown008]

Because the force supporting the spring to the root is created as a reaction force. However, in these conditions, you don't consider reaction forces.

It's like pulling on a rope. If you tie the rope to a wall and you pull at it, it will be the same as someone pulling the rope at the other end instead of the wall with equal force as you apply. The tension in the rope remains the same in both cases.

You pulling from wall = 500N
You pulling from someone = 500N ; The person pulling = 500N

The tension in both cases is 500N and not 1000N for the other one.

[Kalle]

Because the force supporting the spring to the root is created as a reaction force. However, in these conditions, you don't consider reaction forces.

It's like pulling on a rope. If you tie the rope to a wall and you pull at it, it will be the same as someone pulling the rope at the other end instead of the wall with equal force as you apply. The tension in the rope remains the same in both cases.

You pulling from wall = 500N
You pulling from someone = 500N ; The person pulling = 500N

The tension in both cases is 500N and not 1000N for the other one.

So as a rule one could say that only half of the forces are taken into account?

[Unknown008]

Not half, but only those that are sort of 'active', or caused by an external factor, not caused by the system itself as a result of an exterior factor.

[Kalle]

And can the roof pulling on the spring be considered a reaction force at all because it is directed at the same object? Should forces and their reactions not be directed at different objects?

[Unknown008]

The post I made earlier was referring to the roof and the spring as one whole, since the spring was connected to the roof.

I'm not sure if my approach of it is the best one. Maybe you can wait for someone else to answer your question.

[Kalle]

The post I made earlier was referring to the roof and the spring as one whole, since the spring was connected to the roof.

I'm not sure if my approach of it is the best one. Maybe you can wait for someone else to answer your question.

Ok, but thanks for trying:)

[ebaines]

Maybe this will help - the formula F = kx assumes that the spring is in equilibrium. In order for the spring to be in equilibrium the compressive force from the weight of the box must equal the opposing compressive force from the support. So the value of k that you are given is determned under the condition of two equal and opposite compression forces on the spring.

If the reaction force from the support was 0, then that would mean the spring is not in equilibrium, but is actually accelerating (under F = ma). Under this condition the spring accelerates, and some compresion occurs that is dependent on the distributed mass of the spring as it opposes the acceleration - quite a different (and more complicated) problem.