Force required to stretch training resistance bands

  • #1
dlucaswood
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1
Hi all!

I hope I'm posting in the correct section! I’m trying to work out the force required to stretch resistance training bands for two particular set ups shown in the pictures. I’m not a physics student or anything like that (I'm many years out of school!), however I am pretty good at maths. I’m just not quite sure of the equations I need to work these problems out, so any help much appreciated!
Set ups.jpg
Set up 1:

With the first set up, I would be standing in the same position, with my feet anchoring the band a certain distance apart, and my hands holding the band a certain distance up, as shown.

I’m guessing the force required on each side is based on the length of exposed band from my left hand to left foot, and the same on the right. Is that correct? What I’m confused about is that the smaller that distance, the more force required to pull the band. So, pulling the band 5 inches when there is a lot of band exposed is easier than pulling it 5 inches when only a small amount of band is exposed. Why is this?

How might I put this into an equation that relates the three variables of force, length of required elongation and exposed band length for each band in question (I have 5 bands, all with different cross-sectional areas, all made of latex rubber)?

Set up 2:

With the next set up, the band is now anchored and pulled at a certain angle:

In this instance, the band is anchored on a vertical surface, and the is held a certain distance from that anchor, leaving a certain length of band exposed.

How might I put this scenario into an equation that relates the four variables of force, angle of pull, length of required elongation and exposed band length for each band in question?

The bands are loop bands as shown below:

Loop bands picture.jpg

I’m guessing that using them doubled over vs. just one side means that the force required in the above scenarios is just multiplied by 2?

Really appreciate your help, apologies if this all seems very basic! It’s been a few decades since school!!

Darren
 
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  • #2
dlucaswood said:
What I’m confused about is that the smaller that distance, the more force required to pull the band. So, pulling the band 5 inches when there is a lot of band exposed is easier than pulling it 5 inches when only a small amount of band is exposed. Why is this?
This is, roughly speaking, Hooke's Law.

Hooke's Law is an idealization that applies to many real materials. It says that the force required to maintain a deflection (e.g. compress a spring or stretch a rubber band) is proportional to the deflection. The more you deflect something, the more force you have to use to keep it deflected.

If you pull a given band by only 2.5 inches, you'll need less force than if you pulled it a full 5 inches.

Suppose that you a little band and you tie it to the end of another little band. You pull the combination out the full 5 inches. Each little band has only expanded by only 2.5 inches. Each is only providing a weak force.

The total force that you have to apply is only as much as for a 2.5 inch expansion of a little band. It is kind of a "chain is no stronger than its weakest link" thing.
 
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  • #3
dlucaswood said:
What I’m confused about is that the smaller that distance, the more force required to pull the band. So, pulling the band 5 inches when there is a lot of band exposed is easier than pulling it 5 inches when only a small amount of band is exposed. Why is this?
Because the force depends on the realtive elongation (strain) ΔL/L0:
https://en.wikipedia.org/wiki/Young's_modulus#Force_exerted_by_stretched_or_contracted_material

When L0 is made small, the same ΔL of 5 inches means a greater strain ΔL/L0, and thus greater force.
 
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  • #4
jbriggs444 said:
This is, roughly speaking, Hooke's Law.

Hooke's Law is an idealization that applies to many real materials. It says that the force required to maintain a deflection (e.g. compress a spring or stretch a rubber band) is proportional to the deflection. The more you deflect something, the more force you have to use to keep it deflected.

If you pull a given band by only 2.5 inches, you'll need less force than if you pulled it a full 5 inches.

Suppose that you a little band and you tie it to the end of another little band. You pull the combination out the full 5 inches. Each little band has only expanded by only 2.5 inches. Each is only providing a weak force.

The total force that you have to apply is only as much as for a 2.5 inch expansion of a little band. It is kind of a "chain is no stronger than its weakest link" thing.
Ok, yes, I think I get this. Thanks
 
  • #6
dlucaswood said:
How might I put this scenario into an equation that relates the four variables of force, angle of pull, length of required elongation and exposed band length for each band in question?
Just as much as one cannot push a rope, one cannot pull on a rubber band at an angle.
The tins rubber band allows an axial force, but not a transverse ( ie to the side ).

The force on the rubber band will be whatever has produced its extension.

the 'force' on body parts, ie arm, back, forearm, legs - is a different story, and can get complicated to analyze.
 
  • #7
A.T. said:
Because the force depends on the realtive elongation (strain) ΔL/L0:
https://en.wikipedia.org/wiki/Young's_modulus#Force_exerted_by_stretched_or_contracted_material

When L0 is made small, the same ΔL of 5 inches means a greater strain ΔL/L0, and thus greater force.
Ah OK. So, based on the below diagram:

Screenshot 2024-07-10 142347.png

Imagining that I am holding the band with my right hand at the red dot (or indeed even stepping on the band with a foot), and pulling with my left, the force the band exerts at each point that I pull along the band (green dots) can be worked out using the following (just Googling around, hope I’ve rearranged it correctly!):

F = E(∆L * A)/Lo

where

F = force

E = Modulus of elasticity of latex rubber

∆L = length being pulled

A = cross-sectional area of band

Lo = original length (i.e. distance from my left hand to right hand)



Have I got this right?!
 
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  • #8
I haven't taken the time do the experiment but it was my understanding that rubber (and probably latex) bands don't follow Hooke's Law all that well. There is a change in the elasticity with stretching, a sensitivity to rate of applied elongation, and some amount of hysteresis. I guess for talking about the general concept using Hooke's Law is close enough but if you were trying to predict a force you might be disappointed in the accuracy in this case.

Or am I wrong?
 
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  • #9
JT Smith said:
I haven't taken the time do the experiment but it was my understanding that rubber (and probably latex) bands don't follow Hooke's Law all that well.
You are right.

However, even if it is non-linear, the force needed will vary with the percentage expansion rather than with the raw amount of expansion.
 
  • #10
JT Smith said:
I haven't taken the time do the experiment but it was my understanding that rubber (and probably latex) bands don't follow Hooke's Law all that well. There is a change in the elasticity with stretching, a sensitivity to rate of applied elongation, and some amount of hysteresis. I guess for talking about the general concept using Hooke's Law is close enough but if you were trying to predict a force you might be disappointed in the accuracy in this case.

Or am I wrong?
I'm not too worried about it being 100% accurate. A general representation/estimation would be fine. It's only to track progression in load for training purposes as opposed to absolute values.
 
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  • #12
dlucaswood said:
I'm not too worried about it being 100% accurate. A general representation/estimation would be fine. It's only to track progression in load for training purposes as opposed to absolute values.

In that case all you really need to know is that pulling it farther will produce a larger force.

But it's possible that the non-linearity is more than trivial. Consider the Theraband elastic exercise bands. They are very popular. Their force/elongation values were measured (and compared to the manufacturer's values) in this paper. As you can see stretching one 100% beyond its initial length results in a force only about 40% greater than stretching 50%.

Theraband.png

(values are in kg force)
 
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  • #13
dlucaswood said:
Ah OK. So, based on the below diagram:


Imagining that I am holding the band with my right hand at the red dot (or indeed even stepping on the band with a foot), and pulling with my left, the force the band exerts at each point that I pull along the band (green dots) can be worked out using the following (just Googling around, hope I’ve rearranged it correctly!):

F = E(∆L * A)/Lo

where

F = force

E = Modulus of elasticity of latex rubber

∆L = length being pulled

A = cross-sectional area of band

Lo = original length (i.e. distance from my left hand to right hand)



Have I got this right?!
Yes, that is right, for an ideal elastic band.
 

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