# What is the tension force at the top of a spring?

• Cloruro de potasio
In summary, the tension force at the bottom of the pier is equal to the restoring force, but the force at the top is in the opposite direction.
Cloruro de potasio
Good afternoon,

I have some doubts about the tension force suffered by a spring to which a mass is hung and which is making a simple vertical armoin movement. My doubt lies in the fact that at the bottom of the pier (where the mass hangs), the spring exerts the restoring force that is given by Hooke's Law. However, at the top of the dock (at the point where it is fixed), what is the tension force that appears? Is it the same as the restoring force but in the opposite direction?

Ask yourself what it is in equilibrium: the restoring force is zero (at least that is my perception of a restoring force ...) but the suspendion point has to exert an upward force !

And when the mass is oscillating up and down, you first ask if the spring is massless. If it is, a simple force balance tells you the answer to your question !

Cloruro de potasio

What I mean by restoring force is the force given by Hooke's Law, in the system we are studying, that force would be given in the opposite direction to elongation, what I don't understand very well is the force that has to It appears at the point of contact between the spring and the surface, I have measured it experimentally with a PASCO sensor, and I get that it is equal to the restoring force but in the opposite direction, why does this happen?

This is very hard to decipher. Can you possibly show the equations?

Cloruro de potasio said:
why does this happen?
Apparently -- according to your observations -- the spring mass can be ignored. The whole thing stays in place (a=0), so there must be an equilibrium between the external forces.

Were your measurements done while the mass was going up and down, or in equilibrium situations ?

Cloruro de potasio
Chestermiller said:
This is very hard to decipher. Can you possibly show the equations?

The question is that the only equation I can think of is Hooke's Law, but I don't understand why the force that reads the sensor is exactly the same as the restoring force but in the opposite direction (as the sensor is at the point of contact of the spring with the surface, this must be the force that feels the point of contact of the spring with the surface, something similar to the tension in a system in which a body hangs on a rope) there must be some consideration relative to the third Law of Newton, or something like that escapes me ...

BvU said:
Apparently -- according to your observations -- the spring mass can be ignored. The whole thing stays in place (a=0), so there must be an equilibrium between the external forces.

Were your measurements done while the mass was going up and down, or in equilibrium situations ?

In both ways, I studied both the static and the dynamic method, in both cases the same thing happened, the force determined by the sensor was in the same direction as the elongation ...

Cloruro de potasio said:
I get that it is equal to the restoring force but in the opposite direction
Cloruro de potasio said:
the force determined by the sensor was in the same direction as the elongation
What did the sensor measure ? The force exerted by the suspension, or thte force exerted by the spring on the suspension point ?

vanhees71, Cloruro de potasio and Chestermiller
Thanks again, the sensor mediates the force exerted by the spring on the suspension. If we think about it slowly, it makes sense that this force goes in the opposite direction than the restoring force given by Hooke's Law, but I don't know very well how to explain it from a formal point of view ...I enclose the same diagram you show in your response with the force to which I refer to...

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Cloruro de potasio said:
but I don't know very well how to explain it from a formal point of view
It's simple Newton: Your forces exerted by the spring are equal and opposite to the forces on the spring (Newton 3). And if they were not equal in magnitude, the spring would accelerate away (Newton 2) -- which it doesn't do. A case of ##F = ma## (##m## the mass of the spring itself) with ##a## going very big if ##m## is very small.

Take a rubber band and hold the ends. You can't pull with different magnitude forces. Same with a rope or a rod (##k \approx \infty##).

vanhees71 and Cloruro de potasio
Okay, thank you very much, I think I understand something better now, what I have not seen is the reason why the forces have to balance, after all the lower part of the pier if it is suffering an acceleration .. .

Is the diagram that I attached with the value of the different forces correct?

Cloruro de potasio said:
Okay, thank you very much, I think I understand something better now, what I have not seen is the reason why the forces have to balance, after all the lower part of the pier if it is suffering an acceleration .. .
If the mass m is oscillating, then so is the center of mass of the spring. So if you don't assume a mass-less spring, the forces on the spring are not balanced.
Cloruro de potasio said:
Is the diagram that I attached with the value of the different forces correct?
View attachment 254595
It's not clear which force acts on which body.
It's not clear if the spring has any mass.
The red force is missing its Newtons 3rd Law partner.
The 2 forces at the upper attachment must satisfy Newtons 3rd Law (equal but opposite), but in a dynamic situation there is no reason to assume -mg acting at the ceiling.

Thanks again, the mass of the pier is not zero, but if it is very small compared to the masses that are placed on the pier, I still don't quite understand the reason why the force exerted by the spring on the surface is equal to the restoring force given by Hooke's Law, but in the opposite direction ...

I have tested it experimentally both in the static and dynamic case and in both cases the same happens (attached a graph obtained from force and position sensors)

This is the position:

This is the force:

Hm, that's not very surprising since from
$$\ddot{x}=-\omega^2 x$$
the general solution is
$$x(t)=x_0 \cos(\omega t) + \frac{v_0}{\omega} \sin(\omega t),$$
and the force is thus proportional to the elongation,
$$F=m \ddot{x}=-m \omega^2 x.$$
Since the force is in phase (not shifted by ##\pi##, i.e., not with the sign) you obviously measure not the force acting on the body but the reaction force on the suspension.

Yes, thank you, I understand that measure the reaction force, it is something logical by the way in which the sensor was placed, what I cannot understand from a physical point of view is the reason why this force is the same but in the opposite direction to Hooke's Law

Cloruro de potasio said:
what I cannot understand from a physical point of view is the reason why this force is the same but in the opposite direction to Hooke's Law
Newton's 3rd Law.

But... how do you apply Newton's Third Law to this system??

Cloruro de potasio said:
Yes, thank you, I understand that measure the reaction force, it is something logical by the way in which the sensor was placed, what I cannot understand from a physical point of view is the reason why this force is the same but in the opposite direction to Hooke's Law
That's Newton's 3rd postulate.

Cloruro de potasio said:
But... how do you apply Newton's Third Law to this system??
force_by_spring_on_ceiling = - force_by_ceiling_on_spring
force_by_spring_on_mass = - force_by_mass_on_spring

Cloruro de potasio said:
I still don't quite understand the reason why the force exerted by the spring on the surface is equal to the restoring force given by Hooke's Law
It's a tautology. When you make the assumption that Hooke's Law is valid for the spring you assume Hooke's Law gives you the force exerted by the spring.

Cloruro de potasio

## What is tension force in a spring?

Tension force in a spring is the force that is exerted on the spring when it is stretched or compressed. It is the force that is responsible for the spring's ability to return to its original shape after being deformed.

## How is tension force related to the spring constant?

Tension force is directly proportional to the spring constant, which is a measure of the stiffness of the spring. This means that a higher spring constant will result in a higher tension force for the same amount of deformation.

## What factors affect the tension force in a spring?

The tension force in a spring is affected by the material of the spring, the length and diameter of the spring, and the amount of deformation it undergoes. It is also affected by external forces acting on the spring, such as gravity or additional weights.

## How is tension force measured in a spring?

Tension force in a spring can be measured using a spring scale or a force gauge. These tools use the principle of Hooke's law, which states that the force exerted by a spring is directly proportional to its deformation.

## What are some real-life applications of tension force in springs?

Tension force in springs is used in many everyday objects, such as door hinges, trampolines, and car suspensions. It is also used in more specialized applications, such as in mechanical watches and shock absorbers.

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