Does Tying a Knot in a Plastic Bag Double its Weight?

[J23Smith]

I know this will seemingly sound stupid but I have had multiple people argue with me about it. They claim that if they tie a knot in a plastic baggy it doubles the weight from 1 gram to 2 grams. I tell them there is no way because the mass doesn't change. I've tried to bet them any amount of money they were willing to bet and I have had no takers. So is there any way the weight changes without changing the mass or gravity?

 

[rock.freak667]

W=mg...Putting the knot in a bag won't change m or g so the weight of the knot is the same.

Unless they meant that you weigh plastic bag that contains the knot.

 

[praharmitra]

W=mg...Putting the knot in a bag won't change m or g so the weight of the knot is the same.

Unless they meant that you weigh plastic bag that contains the knot.

maybe they mean that they allow air to enter the plastic bag, then tie the knot so as to trap air...hence weight of the "bag" increses

 

[J23Smith]

maybe they mean that they allow air to enter the plastic bag, then tie the knot so as to trap air...hence weight of the "bag" increses

They didn't intend on any air or anything else that might increase the weight entering the bag. They are just dumb as I suspected they were in this matter. Thanks guys.

 

[DaveC426913]

They didn't intend on any air or anything else that might increase the weight entering the bag.

Are you sure? Because this would actually work. I can see there being some intuitive confusion around this.

You weigh an open baggie on a scale, you tie a knot in it and you are weighing the baggie, plus any air trapped in it.

(A question arises: do you need to do the weighing in a vacuum chamber to get correct readings?)

 

[Redbelly98]

maybe they mean that they allow air to enter the plastic bag, then tie the knot so as to trap air...hence weight of the "bag" increses

If that actually works (and I'm not sure if it does), the 1 gram increase is in the right ballpark. Air's density is very roughly 1/1000 that of water, or 1 gram / liter. A 1 liter volume would contain about 1 gram of air.

What I do know is that a 1 liter bag of air (1 atm, room temperature), if weighed in a vacuum, would be about 1 gram heavier than an empty bag.

Edit:
Just occurred to me that this can be easily tested. If you weight 15 bags together, the increased weight (if it happens) would amount to 15 grams or about 1/2 ounce. A scale used for weighing food or post letters would show this difference.

 

[Taika]

Are you sure? Because this would actually work. I can see there being some intuitive confusion around this.

You weigh an open baggie on a scale, you tie a knot in it and you are weighing the baggie, plus any air trapped in it.

(A question arises: do you need to do the weighing in a vacuum chamber to get correct readings?)

This doesn't totally make sense to me. Air is obviously creating pressure on everything, all the time. Say you have a bag that can enclose 10cm³ of air. If the bag is laying flat, there's still that much air above it. The layers on the scale would be scale>layer of plastic>layer of plastic>miles of air. If you blow up the bag and set it on the scale, you have scale>layer of plastic>10cm³ air>layer of plastic>miles of air minus 10cm³air. Any way you look at it, the same amount of "weight" or pressure from the air is going to be showing on the scale.

Where am I going wrong here?

(I haven't had anything to do with physics beyond SciAm articles for a good five years now, so forgive layman's terminology and such)

 

[DaveC426913]

This doesn't totally make sense to me. Air is obviously creating pressure on everything, all the time. Say you have a bag that can enclose 10cm³ of air. If the bag is laying flat, there's still that much air above it. The layers on the scale would be scale>layer of plastic>layer of plastic>miles of air. If you blow up the bag and set it on the scale, you have scale>layer of plastic>10cm³ air>layer of plastic>miles of air minus 10cm³air. Any way you look at it, the same amount of "weight" or pressure from the air is going to be showing on the scale.

Where am I going wrong here?

(I haven't had anything to do with physics beyond SciAm articles for a good five years now, so forgive layman's terminology and such)

Agreed. I'm not sure either. That's why I qualified it with the last line: it would work in a vacuum chamber, because you'd eliminate the bouyancy factor.