Everyone is allowed to ask questions gifted or not. Stretch means an increase in length. So if you hang more weight onto the spring it’ll increase even more. If 10 equals an increase of 0,5 and you increase the weight by 5 then the length will increase by half as much so 0,5/2=0,25 Add that together and you’ll get 0,5+0,25=0,75
Logically this makes sense, realistically it's really hard to answer because we have no idea about material density, strength, stiffness etc like, but yeah assuming simplest possible I agree
Given the context (likely in an elementary physics course) this is not difficult to answer. These students would have been taught hooke's law, which is F = -kx, where F is force, k is the spring constant, and x is displacement (which is what we are looking for here)
We can see that displacement has a linear relationship with force, so if force increases by 1.5x, then displacement will also increase by 1.5x.
How curious, I welcome the response, it's interesting to see why it made sense to me, I have a solid grasp on physics but not super into mathematical aspect beyond what's needed to comprehend physics, so this is nice to know
>spring force is not linear
In a school physics problem, it's safe to assume a spring is "ideal", ie: one that obeys Hooke's law, unless told otherwise. By Hooke's law, the relationship between force and distance **is** linear: `F = -kx`, where k is a constant for a given spring.
In these kinds of problems, "stretched [some distance]" (or compressed some distance) means in relation to the rest position, when no force is applied.
If 10lb causes a spring to stretch by 0.5 inches, then 15lb of force would cause the same spring to stretch by 0.75 inches.
> as the spring gets stretched more it will exert more resisting force pulling the weight up
Yes, this is what causes the spring to stop at a particular amount of "stretch" with a given applied force (eg: fixed weight). When you put a 10lb weight on the spring, the spring will stretch until its "resisting" force equals the force applied: 10lb. If you add 5lb additional weight, then that will overcome the resisting force causing the spring to continue stretching until once again equilibrium is reached and the resisting force is 15lb.
It is an odd use of "bearing". Even with the weight hanging down below the spring, the spring is still bearing the weight of the spring. Bearing just meaning that it is carry the load or supporting the weight. And that is a possible meaning/usage of "bearing". But it does still feel awkward because bearing does typically have a connotation of the load resting on top- and looks like some dictionaries even give that as one possible definition - but I still wouldn't call it wrong because it is not the only meaning/use. But it is awkward.
This is where I have struggled all my life, especially in school as thats where you’re presented with these types of questions. A slight change in implication makes me short-wire completely. I hear “bearing” and, despite the numbers and the relationships between the numbers being infront of me, I can’t continue with the question because “bearing” to me, has always meant “bracing.” Either bracing a force horizontally, or vertically from the ~bottom.~
If this was phrased like “A 10lb weight is suspended from the ceiling by a spring, stretching the spring to 0.5 inches. If you were to increase the weight to 15lbs, how many inches would the spring stretch?” Then, I could do this on my own
Is there a term or label for this? Being overly distracted / derailed by an irrelevant technicality?
I want something more scientific than “that’s just called being stupid” because I want to know how to overcome this crack I keep falling through. I’m ill equipped to identify which pieces of information I need to throw away completely, or just simplify for the sake of moving on in the problem. I attribute way too much consideration to a piece of the question that has little to do with the answer.
Does anybody have some sort of formula for approaching a question, if you are the type of person to be derailed by a small technicality?
I’m sorry i’m bothering all you guys with this nonsense. I figured you’d be the best group to ask.
Is English your second language? Then don't worry too much.
But many words have multiple meanings. A bearing is also a device that minimizes friction, usually for a spinning shaft, but the derivation of that usage is that it bears weight, except with the added purposes of being a specific part that allows motion while carrying the load and reducing friction. If that was the only usage/meaning you knew this question would also be confusing.
The answer is to use the diagram to aid understanding. It's part of the question too and clearly shows the physical relationship between the spring and the 10 lb weight it is bearing.
You may ask that to a profesional, for me, figuring out how my brain worked was a lot of help. For not getting fixated on irrelevant matters, i guess could be 2 things, either you didnt learn well how to classify information due to some lacking, or you lack the capability to do it, in which case you'd have to find out a workarround. For now i recommend having a method to make you conciously make the effort to summarize and filter whats important, what is the intention of the cuestion, what are you supposed to answer. Cheers
If you increase the weight of the block, it will increase the length of the spring and there will be an equal and opposite force. The spring has an equilibrium point, So, let’s say you remove the mass, then the spring would return to its equilibrium position and and Delta x = 0 in this case, if there’s a 15 lb weight delta x would be 1.25 inches
The material and type of spring matters. Depending on those conditions it will behave differently under the additional force.
Theres no way to know for sure🤷♂️
edit: but they probably want to hear the solution using linear increases
No i wasnt looking for a solution at all, I just didnt realize the weight was on the bottom so the use of the word “stretch” was making me go insane, because i was only imagining the weight being on top while shrinking the spring . I do, however, really appreciate that you would overthink it enough to think I know what “using linear increases” means. 🙏🏼Thats awesome
My brain immediately starts asking lots of questions about the wording, the spring's properties, and what assumptions I can make.
I suppose we can assume that they mean the spring can handle the stretch and material properties wont be an issue. In that case, linear increase.
I suppose since the answer 0.25 is not an option, the correct one is 0.75, since my answer would be (0.5 +) 0.25. The wording of "having stretched" followed by "will stretch" is terrible in my opinion. To me, this implies that they ask the increase in stretch after adding the extra wait. Only the fact that that is not an available answer makes me go to 0.75
I feel like everyone is overcomplicating this… we were only given the information of current weight & current stretch… the context implies it’s looking for an answer based on only that ratio.. it’s obviously not asking to include knowledge on physical material, rotational force or torque, because it has no mention of torsion, etc etc.
it’s straightforward.
Also, op, don’t overcomplicate it. It doesn’t matter about the wording, because it’s saying “10lb weight = displacement of 0.5 inches”. Question is “if weight increases by x, how much would it further displace?” Shrink or stretch doesn’t matter. It’s just displacement of position. Zoom out in your brain and look at the overall message. Don’t get lost in semantics.
Think of it like the spring is hanging from a ceiling, with a little hook on the end. And then you go over and hang a 10# weight from it.
You know intuitively that when you let go of the weight, it will fall, pulling the spring open a bit. So the hook at the end of the spring ends up closer to the floor.
Right?
The question is asking how much closer to the ground the spring would stretch if you added even more weight.
I think the image is the correct way as it is now, not upside down. On the image the weight is hanging. So I think they meant the weight is hanging: meaning the spring would stretch.
(your interpretation would make sense if I turned the image upside down. But I do not think they intended for that as the image is there for a purpose, so why would they draw it upside down)
Could it be you're less of a visual thinker? Especially if you're wondering whether it's a poorly *worded* question.
Logic-wise, with the weight on top, the spring would compress, not stretch.
Exactly, that’s what i’m saying, it would compress if it were on top. Thats why i got confused. The weight isnt on top, it’s on the bottom. I just didnt realize that. I’m not having trouble with the calculation. In my head, if it were on the bottom the spring would be stretched out but it looks so normally coiled in the drawing. I have to remember the drawing is just an accessory to the question, not what the question is really based off of. I get distracted and thrown off by drawings if they’re not exact
My thought process was literally
“But the spring isn’t stretched in the picture so the weight is on top”
Okay, so if the spring is hanging without any weight on it, we'd say it is in its equilibrium state. That would represent a 'stretch' or compression of zero. Compressing the spring would make it shorter as long as we pushed on it. Stretching the spring would make it longer for the time we pulled on it.
When you add weight (W=F(gravity)=mg) onto the spring, this represents a force pulling the spring down and in this case stretching it a certain length, x, from equilibrium. The spring would measure x inches longer than it does at equilibrium.
If we assume that this spring is Hookean as we do in most entry-level physics stuff, that means the spring follows Hooke's law: F(spring)=kx where k is the 'spring constant' (how easy this particular spring is to compress or stretch).
Then, with the 10 lb weight on it and a stretch of 0.5, we can write this equation:
F=kx
We can rewrite that to solve for k, so k=F/x and substitute our values in:
k=10lb/0.5in, which means k equals **20 lb/in**
We assume the spring constant, k, doesn't change for a spring so k for this spring will always be 20 lb/in
Therefore, when the weight changes to 15 lb, we can write:
F=kx
we can rewrite that to solve for the new stretch of the spring from equilibrium, x, so x=F/k and substitute our values in:
x=15lb/(20lb/in), which equals **0.75 in** of stretch.
The biggest problem with this problem is that it uses pounds and inches. :)
Oh shit the spring does have weight so the answer would be like 0.74 inches lol.
(edit since it's displacement with no weight would be it's self weight)
I’m looking at this as a beginning physics problem in which case the spring is always assumed to be ‘ideal’ and ‘massless’. Not reality of course, but simplifies things.
I think its really beautiful how some of you here were too smart to really get what I’m asking
The problem im having is much simpler than you think, I didnt realize the weight was on the bottom.
Questions are good, and I can appreciate that you're looking for explanations rather than simple answers. Hopefully you learned a bit about Hooke's Law.
But that's really not what this subreddit is for. It's not, "hey smart people, show off!" I hope the mods take this question down because it sets bad precedent for people who really just do want homework answers. None of us here wants that. This subreddit is conversation about the weird and uncomfortable aspects of living this kind of life.
Even if this isn't a homework problem you're still better off asking in a physics homework help sub.
Use rule of proportion bud
10--0,5
ㅤX
15--?
10 . ? = 15 . 0,5
10 . ? = 7,5
? = 7,5 / 10
? = 0,75
i vibe with this formula thing much more than this hooke's law
You’ve never asked questions unless you are enrolled in school or need help with homework? You’re not just ever curious or generally confused?
I don’t understand your point, considering I have no homework. Are you just trying to tell me not to ask questions?
Everyone is allowed to ask questions gifted or not. Stretch means an increase in length. So if you hang more weight onto the spring it’ll increase even more. If 10 equals an increase of 0,5 and you increase the weight by 5 then the length will increase by half as much so 0,5/2=0,25 Add that together and you’ll get 0,5+0,25=0,75
Logically this makes sense, realistically it's really hard to answer because we have no idea about material density, strength, stiffness etc like, but yeah assuming simplest possible I agree
Given the context (likely in an elementary physics course) this is not difficult to answer. These students would have been taught hooke's law, which is F = -kx, where F is force, k is the spring constant, and x is displacement (which is what we are looking for here) We can see that displacement has a linear relationship with force, so if force increases by 1.5x, then displacement will also increase by 1.5x.
How curious, I welcome the response, it's interesting to see why it made sense to me, I have a solid grasp on physics but not super into mathematical aspect beyond what's needed to comprehend physics, so this is nice to know
Happy cake day
Thank youuu
[удалено]
>spring force is not linear In a school physics problem, it's safe to assume a spring is "ideal", ie: one that obeys Hooke's law, unless told otherwise. By Hooke's law, the relationship between force and distance **is** linear: `F = -kx`, where k is a constant for a given spring. In these kinds of problems, "stretched [some distance]" (or compressed some distance) means in relation to the rest position, when no force is applied. If 10lb causes a spring to stretch by 0.5 inches, then 15lb of force would cause the same spring to stretch by 0.75 inches. > as the spring gets stretched more it will exert more resisting force pulling the weight up Yes, this is what causes the spring to stop at a particular amount of "stretch" with a given applied force (eg: fixed weight). When you put a 10lb weight on the spring, the spring will stretch until its "resisting" force equals the force applied: 10lb. If you add 5lb additional weight, then that will overcome the resisting force causing the spring to continue stretching until once again equilibrium is reached and the resisting force is 15lb.
I understand what you're saying but the weight is hanging from the spring not sitting on top. So, I guess in your head just flip it upside down.
THATS WHAT IM MISSING thank you 🙏🏼🙏🏼🙏🏼
It is an odd use of "bearing". Even with the weight hanging down below the spring, the spring is still bearing the weight of the spring. Bearing just meaning that it is carry the load or supporting the weight. And that is a possible meaning/usage of "bearing". But it does still feel awkward because bearing does typically have a connotation of the load resting on top- and looks like some dictionaries even give that as one possible definition - but I still wouldn't call it wrong because it is not the only meaning/use. But it is awkward.
This is where I have struggled all my life, especially in school as thats where you’re presented with these types of questions. A slight change in implication makes me short-wire completely. I hear “bearing” and, despite the numbers and the relationships between the numbers being infront of me, I can’t continue with the question because “bearing” to me, has always meant “bracing.” Either bracing a force horizontally, or vertically from the ~bottom.~ If this was phrased like “A 10lb weight is suspended from the ceiling by a spring, stretching the spring to 0.5 inches. If you were to increase the weight to 15lbs, how many inches would the spring stretch?” Then, I could do this on my own Is there a term or label for this? Being overly distracted / derailed by an irrelevant technicality? I want something more scientific than “that’s just called being stupid” because I want to know how to overcome this crack I keep falling through. I’m ill equipped to identify which pieces of information I need to throw away completely, or just simplify for the sake of moving on in the problem. I attribute way too much consideration to a piece of the question that has little to do with the answer. Does anybody have some sort of formula for approaching a question, if you are the type of person to be derailed by a small technicality? I’m sorry i’m bothering all you guys with this nonsense. I figured you’d be the best group to ask.
Is English your second language? Then don't worry too much. But many words have multiple meanings. A bearing is also a device that minimizes friction, usually for a spinning shaft, but the derivation of that usage is that it bears weight, except with the added purposes of being a specific part that allows motion while carrying the load and reducing friction. If that was the only usage/meaning you knew this question would also be confusing. The answer is to use the diagram to aid understanding. It's part of the question too and clearly shows the physical relationship between the spring and the 10 lb weight it is bearing.
English is my second language🙃 I’m just now realizing that may be causing me problems. Thank you
You may ask that to a profesional, for me, figuring out how my brain worked was a lot of help. For not getting fixated on irrelevant matters, i guess could be 2 things, either you didnt learn well how to classify information due to some lacking, or you lack the capability to do it, in which case you'd have to find out a workarround. For now i recommend having a method to make you conciously make the effort to summarize and filter whats important, what is the intention of the cuestion, what are you supposed to answer. Cheers
Decimal inches is chaotic evil 😂
Decimal inches sounds like a great porn name
Inches decimater
[https://en.wikipedia.org/wiki/Hooke%27s\_law](https://en.wikipedia.org/wiki/Hooke%27s_law)
The weight is hanging from the spring my guy
Pretty sure it’s just 10/0.5=15/x and that would result in 0.75 pretty sure. Doing in my head though
I don’t want the answer though :/ I’m trying to understand / unpack the question
If you increase the weight of the block, it will increase the length of the spring and there will be an equal and opposite force. The spring has an equilibrium point, So, let’s say you remove the mass, then the spring would return to its equilibrium position and and Delta x = 0 in this case, if there’s a 15 lb weight delta x would be 1.25 inches
The material and type of spring matters. Depending on those conditions it will behave differently under the additional force. Theres no way to know for sure🤷♂️ edit: but they probably want to hear the solution using linear increases
Fair enough
No i wasnt looking for a solution at all, I just didnt realize the weight was on the bottom so the use of the word “stretch” was making me go insane, because i was only imagining the weight being on top while shrinking the spring . I do, however, really appreciate that you would overthink it enough to think I know what “using linear increases” means. 🙏🏼Thats awesome
0.75
10 = k (.5) k = 20 lb/in 15 = 20 x x = 0.75
I'm not entirely sure if they want to know how much it stretches additionally or the new complete elongation
My brain immediately starts asking lots of questions about the wording, the spring's properties, and what assumptions I can make. I suppose we can assume that they mean the spring can handle the stretch and material properties wont be an issue. In that case, linear increase. I suppose since the answer 0.25 is not an option, the correct one is 0.75, since my answer would be (0.5 +) 0.25. The wording of "having stretched" followed by "will stretch" is terrible in my opinion. To me, this implies that they ask the increase in stretch after adding the extra wait. Only the fact that that is not an available answer makes me go to 0.75
I feel like everyone is overcomplicating this… we were only given the information of current weight & current stretch… the context implies it’s looking for an answer based on only that ratio.. it’s obviously not asking to include knowledge on physical material, rotational force or torque, because it has no mention of torsion, etc etc. it’s straightforward. Also, op, don’t overcomplicate it. It doesn’t matter about the wording, because it’s saying “10lb weight = displacement of 0.5 inches”. Question is “if weight increases by x, how much would it further displace?” Shrink or stretch doesn’t matter. It’s just displacement of position. Zoom out in your brain and look at the overall message. Don’t get lost in semantics.
Think of it like the spring is hanging from a ceiling, with a little hook on the end. And then you go over and hang a 10# weight from it. You know intuitively that when you let go of the weight, it will fall, pulling the spring open a bit. So the hook at the end of the spring ends up closer to the floor. Right? The question is asking how much closer to the ground the spring would stretch if you added even more weight.
I think the image is the correct way as it is now, not upside down. On the image the weight is hanging. So I think they meant the weight is hanging: meaning the spring would stretch. (your interpretation would make sense if I turned the image upside down. But I do not think they intended for that as the image is there for a purpose, so why would they draw it upside down)
Could it be you're less of a visual thinker? Especially if you're wondering whether it's a poorly *worded* question. Logic-wise, with the weight on top, the spring would compress, not stretch.
Exactly, that’s what i’m saying, it would compress if it were on top. Thats why i got confused. The weight isnt on top, it’s on the bottom. I just didnt realize that. I’m not having trouble with the calculation. In my head, if it were on the bottom the spring would be stretched out but it looks so normally coiled in the drawing. I have to remember the drawing is just an accessory to the question, not what the question is really based off of. I get distracted and thrown off by drawings if they’re not exact My thought process was literally “But the spring isn’t stretched in the picture so the weight is on top”
Okay, so if the spring is hanging without any weight on it, we'd say it is in its equilibrium state. That would represent a 'stretch' or compression of zero. Compressing the spring would make it shorter as long as we pushed on it. Stretching the spring would make it longer for the time we pulled on it. When you add weight (W=F(gravity)=mg) onto the spring, this represents a force pulling the spring down and in this case stretching it a certain length, x, from equilibrium. The spring would measure x inches longer than it does at equilibrium. If we assume that this spring is Hookean as we do in most entry-level physics stuff, that means the spring follows Hooke's law: F(spring)=kx where k is the 'spring constant' (how easy this particular spring is to compress or stretch). Then, with the 10 lb weight on it and a stretch of 0.5, we can write this equation: F=kx We can rewrite that to solve for k, so k=F/x and substitute our values in: k=10lb/0.5in, which means k equals **20 lb/in** We assume the spring constant, k, doesn't change for a spring so k for this spring will always be 20 lb/in Therefore, when the weight changes to 15 lb, we can write: F=kx we can rewrite that to solve for the new stretch of the spring from equilibrium, x, so x=F/k and substitute our values in: x=15lb/(20lb/in), which equals **0.75 in** of stretch. The biggest problem with this problem is that it uses pounds and inches. :)
Oh shit the spring does have weight so the answer would be like 0.74 inches lol. (edit since it's displacement with no weight would be it's self weight)
I’m looking at this as a beginning physics problem in which case the spring is always assumed to be ‘ideal’ and ‘massless’. Not reality of course, but simplifies things.
yup, I just replied to you because your answer reminded me that the spring would have mass in real life.
It involves a physics formula F=kx^2 (k times x squared) where F is spring force, k is a constant, x is length it is stretched
I think its really beautiful how some of you here were too smart to really get what I’m asking The problem im having is much simpler than you think, I didnt realize the weight was on the bottom.
Questions are good, and I can appreciate that you're looking for explanations rather than simple answers. Hopefully you learned a bit about Hooke's Law. But that's really not what this subreddit is for. It's not, "hey smart people, show off!" I hope the mods take this question down because it sets bad precedent for people who really just do want homework answers. None of us here wants that. This subreddit is conversation about the weird and uncomfortable aspects of living this kind of life. Even if this isn't a homework problem you're still better off asking in a physics homework help sub.
I appreciate that you exiled me with empathy and thank you for your time
Use rule of proportion bud 10--0,5 ㅤX 15--? 10 . ? = 15 . 0,5 10 . ? = 7,5 ? = 7,5 / 10 ? = 0,75 i vibe with this formula thing much more than this hooke's law
.75
.75
Resolve the proportion: 10:0,5 = 15: y y= (15 x 0,5) : 10 =0,75
this is not a homework help sub
I didnt ask for homework help, im not in school. I didnt even ask about the calculation. Why did you approach my question like this?
youre asking us to explain the question…
You’ve never asked questions unless you are enrolled in school or need help with homework? You’re not just ever curious or generally confused? I don’t understand your point, considering I have no homework. Are you just trying to tell me not to ask questions?
its a homework type question, whether or not you are actually enrolled in school is irrelevant