ArtW Posted July 27, 2011 Share Posted July 27, 2011 A small balloon with a stone, stably suspending in the water, gently push it down, then release, it will: A. suspend in the position where it is released; B. go up and suspend at the original height; C. sink to the bottom of the container; D. float to the surface of water. The right answer is __ . I think the correct answer is B, but the most Chinese people believe the answer is C. My reason is that the balloon will go back to the balanced point. I put the question on many forums in China, and I am not persuaded. This question is famous because it often appears in the important exams, for example, the BIG exam for university entrance. Unfortunately, answer C is often chosen as the correct answer even though it is wrong. It is difficult to explain, especially hard to explain for the people who know something about physics. I want to know how the people in the other countries think. This is a question about System Science, so I hope true scientists in this area give us the right answer, and show it in a test. This is a hot question now in China, so take a part in it and have fun. Thank you. If you understand Chinese, you could go to: http://www.tianya.cn...1/1893401.shtml http://www.tianya.cn...09.shtml#Bottom http://bbs.cntv.cn/t...810381-1-1.html Link to comment Share on other sites More sharing options...
CaptainPanic Posted July 27, 2011 Share Posted July 27, 2011 Is the balloon at the surface, or really hanging below the surface? If you push down, the pressure becomes higher. The gas in the balloon will be compressed, the balloon will have less volume, and it will logically displace less water. Link to comment Share on other sites More sharing options...
ArtW Posted July 27, 2011 Author Share Posted July 27, 2011 I mean, the balloon is below the surface, and beyond the bottom, and stops somewhere. Link to comment Share on other sites More sharing options...
swansont Posted July 27, 2011 Share Posted July 27, 2011 Why do you think C is wrong? Link to comment Share on other sites More sharing options...
ewmon Posted July 27, 2011 Share Posted July 27, 2011 If you push down, the pressure becomes higher. The gas in the balloon will be compressed, the balloon will have less volume, and it will logically displace less water. Yes, as any submariner will tell you, buoyancy involves positive feedback such as the Captain has described. The answer is C, the balloon/stone will sink to the bottom of the container. The same positive feedback holds true if the object is nudged upward. Scuba divers see this with air bubbles, which continuously become larger and faster. Maintaining constant depth in a motionless submarine requires work, and some forward motion allows them to use the diving planes to aid in maintaining depth. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 27, 2011 Share Posted July 27, 2011 A small balloon with a stone, stably suspending in the water, gently push it down, then release, it will: A. suspend in the position where it is released; B. go up and suspend at the original height; C. sink to the bottom of the container; D. float to the surface of water. The right answer is __ . I think the correct answer is B, but the most Chinese people believe the answer is C. My reason is that the balloon will go back to the balanced point. I put the question on many forums in China, and I am not persuaded. This question is famous because it often appears in the important exams, for example, the BIG exam for university entrance. Unfortunately, answer C is often chosen as the correct answer even though it is wrong. It is difficult to explain, especially hard to explain for the people who know something about physics. I want to know how the people in the other countries think. This is a question about System Science, so I hope true scientists in this area give us the right answer, and show it in a test. This is a hot question now in China, so take a part in it and have fun. Thank you. If you understand Chinese, you could go to: http://www.tianya.cn...1/1893401.shtml http://www.tianya.cn...09.shtml#Bottom http://bbs.cntv.cn/t...810381-1-1.html I see only two obvious stable positions for the system, stone on the bottom and balloon at the surface. In between I don't see how you can easily tune the weight/volume to a stable condition, though you could tune the temperature and temperature gradient of the water to get that affect. Link to comment Share on other sites More sharing options...
imatfaal Posted July 27, 2011 Share Posted July 27, 2011 B would only be the correct answer if the balloon/air inside was as incompressible as water. Glass spheres of a predetermined denisty can be used to ascertain temperature of a fluid - Galilean Thermometer Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 27, 2011 Share Posted July 27, 2011 B would only be the correct answer if the balloon/air inside was as incompressible as water. Glass spheres of a predetermined denisty can be used to ascertain temperature of a fluid - Galilean Thermometer B could be correct if the balloon was originally stably at the surface, by enough to keep the average density of the balloon/stone system above that of the water even after the "gentle" compression. Link to comment Share on other sites More sharing options...
ArtW Posted July 28, 2011 Author Share Posted July 28, 2011 Usually, the two obvious stable positions, at the surface and on the bottom, exist. Here, this question means that the system, including the balloon and air inside, the stone and water, is at a STABLE and BALANCED point. This STABLE and BALANCED point may exist, even if it is difficult to tune. At the ideal point, all forces get to the balanced values, the balloon with the stone will suspends at this point. I give you an example alike. The story happened the last winter. My daughter’s balloon quietly suspended in the air inside my house, did not touch the ceiling and kept the distance of 1 meter, and stayed there. I remembered this question I just put here today, and she met this question in her test. I asked her how the balloon would go if I pulled it down, and she gave me the answer as the most people, YES, just SINK! To the floor! She believes these two questions are the same, I also believe. What happened later? I pulled it down by the string under the balloon, and held it for several feet, and for several seconds, then release it. IT RETURNED TO THE ORIGINAL POSITION! YES! This is a real story! OK, now I make a little bit change about the question. The new one is: ================================================================================ A balloon with a string, stably suspending in the air inside a house, gently pull it down, then release, it will: •A. suspend in the position where it is released; •B. go up and suspend at the original height; •C. sink to the floor; •D. go up and stop at the ceiling. The right answer is __ . ================================================================================ Now, I have told the REAL answer, tell me why please. Thank you a lot! Link to comment Share on other sites More sharing options...
ArtW Posted July 28, 2011 Author Share Posted July 28, 2011 Why do you think C is wrong? Air in the balloon will resist the compression. The balloon suspends somewhere means all forces are balanced. The bigger the compression force from water is, the bigger the resisting force from air is. The bigger resisting force makes the balloon biger, and then the floating force becomes bigger, and the balloon goes up, renturns to the old place. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 28, 2011 Share Posted July 28, 2011 Usually, the two obvious stable positions, at the surface and on the bottom, exist. Here, this question means that the system, including the balloon and air inside, the stone and water, is at a STABLE and BALANCED point. This STABLE and BALANCED point may exist, even if it is difficult to tune. At the ideal point, all forces get to the balanced values, the balloon with the stone will suspends at this point. I give you an example alike. The story happened the last winter. My daughter’s balloon quietly suspended in the air inside my house, did not touch the ceiling and kept the distance of 1 meter, and stayed there. I remembered this question I just put here today, and she met this question in her test. I asked her how the balloon would go if I pulled it down, and she gave me the answer as the most people, YES, just SINK! To the floor! She believes these two questions are the same, I also believe. What happened later? I pulled it down by the string under the balloon, and held it for several feet, and for several seconds, then release it. IT RETURNED TO THE ORIGINAL POSITION! YES! This is a real story! OK, now I make a little bit change about the question. The new one is: ================================================================================ A balloon with a string, stably suspending in the air inside a house, gently pull it down, then release, it will: •A. suspend in the position where it is released; •B. go up and suspend at the original height; •C. sink to the floor; •D. go up and stop at the ceiling. The right answer is __ . ================================================================================ Now, I have told the REAL answer, tell me why please. Thank you a lot! I assume this is a helium filled balloon in this case? Probably one that has lost some helium and absorbed some air? Or tuned to be just the right weight? Assuming that: Air is cooler and denser near the floor and the temperature gradient and resulting density gradiant is sufficient to give the balloon (which as a system is denser than the ceiling air yet lighter than air at the floor) a net buoyant force near the floor, and net sinking force near the ceiling. Unlike the situation with the balloon in the water the pressure gradient is not sufficient to change the balloon enough to overcome the difference and the balloon finds equilibrium at it's original height or position. When I say "unlike" I am comparing it to a typical case...a water balloon/stone set up could also do this in the right conditions as per your quote that I bolded and as I mentioned in an earlier post. But typically this water balloon/stone would not be a stable setup and any displacement up or down would send it in that direction. Air in the balloon will resist the compression. The balloon suspends somewhere means all forces are balanced. The bigger the compression force from water is, the bigger the resisting force from air is. The bigger resisting force makes the balloon biger, and then the floating force becomes bigger, and the balloon goes up, renturns to the old place. As the air resists the increased compression at greater depth, what happens to the volume? With less volume, what happens to the buoyant force? Link to comment Share on other sites More sharing options...
ArtW Posted July 28, 2011 Author Share Posted July 28, 2011 I assume this is a helium filled balloon in this case? Probably one that has lost some helium and absorbed some air? Or tuned to be just the right weight? Assuming that: Air is cooler and denser near the floor and the temperature gradient and resulting density gradiant is sufficient to give the balloon (which as a system is denser than the ceiling air yet lighter than air at the floor) a net buoyant force near the floor, and net sinking force near the ceiling. Unlike the situation with the balloon in the water the pressure gradient is not sufficient to change the balloon enough to overcome the difference and the balloon finds equilibrium at it's original height or position. When I say "unlike" I am comparing it to a typical case...a water balloon/stone set up could also do this in the right conditions as per your quote that I bolded and as I mentioned in an earlier post. But typically this water balloon/stone would not be a stable setup and any displacement up or down would send it in that direction. As the air resists the increased compression at greater depth, what happens to the volume? With less volume, what happens to the buoyant force? The increasing speed of the force from the air inside the balloon, is faster, than the increasing speed of the force of compression from the water, as the depth becomes greater and greater. The increasing speed of the volume, is faster, than the reducing speed of the volume, as the depth becomes greater and greater. Air is cooler and denser near the floor, yes! BUT, water is cooler and denser near the bottom, too. You know, consider the case in sea. We can assume the container is deep enough, so a stable balanced point can exist. Link to comment Share on other sites More sharing options...
swansont Posted July 28, 2011 Share Posted July 28, 2011 Air in the balloon will resist the compression. The balloon suspends somewhere means all forces are balanced. The bigger the compression force from water is, the bigger the resisting force from air is. The bigger resisting force makes the balloon biger, and then the floating force becomes bigger, and the balloon goes up, renturns to the old place. The balloon is not rigid. The volume will vary; PV = constant at constant T As P goes up, V goes down. Bouyancy force varies with V 1 Link to comment Share on other sites More sharing options...
John Cuthber Posted July 28, 2011 Share Posted July 28, 2011 When it comes down to it, the experience of many divers shows that Artw is just plain wrong. He has failed to understand that the rise in air pressure inside the balloon is a consequence of a reduction in volume. That volume is what provides the buoyancy. Reducing it will reduce the buoyancy and the system will sink. Quite what the Chinese have to do with this I'm not sure, but it look like most of them got it right. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 28, 2011 Share Posted July 28, 2011 When it comes down to it, the experience of many divers shows that Artw is just plain wrong. He has failed to understand that the rise in air pressure inside the balloon is a consequence of a reduction in volume. That volume is what provides the buoyancy. Reducing it will reduce the buoyancy and the system will sink. Quite what the Chinese have to do with this I'm not sure, but it look like most of them got it right. Originally I thought it was going to be a trick question, where our up (for those of us on the opposite side of the Earth) is their down. Link to comment Share on other sites More sharing options...
ArtW Posted July 29, 2011 Author Share Posted July 29, 2011 This is a simple question, but it is saying a principle for "SYSTEM", BALANCE. This SYSTEM is originally at a balanced point, even though to get this point is difficult. Now, nothing is changed but the downwards pushing force, and then the force disappears. The system will return into BALANCE, which means the balloon will go to the original position. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 29, 2011 Share Posted July 29, 2011 This is a simple question, but it is saying a principle for "SYSTEM", BALANCE. This SYSTEM is originally at a balanced point, even though to get this point is difficult. Now, nothing is changed but the downwards pushing force, and then the force disappears. The system will return into BALANCE, which means the balloon will go to the original position. If you take a ballon to the bottom of a pool will it: 1. stay the same size? 2. get bigger-increase in volume? 3. get smaller-decrease in volume? Link to comment Share on other sites More sharing options...
ArtW Posted July 29, 2011 Author Share Posted July 29, 2011 中国的果壳网也在直播讨论 Guokr in China discussing... http://www.guokr.com/post/56593/ If you take a ballon to the bottom of a pool will it: 1. stay the same size? 2. get bigger-increase in volume? 3. get smaller-decrease in volume? NO DOUBT, SMALLER! but do you still remember that your hand is pulling the balloon! Your hand makes it smaller, even if indirectly. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 29, 2011 Share Posted July 29, 2011 中国的果壳网也在直播讨论 Guokr in China discussing... http://www.guokr.com/post/56593/ NO DOUBT, SMALLER! but do you still remember that your hand is pulling the balloon! Your hand makes it smaller, even if indirectly. So it's smaller, less buoyant and attached to a weight that made it neutral buoyant when it was larger. With that weight included it now as a system is heavier than water. When I let it go, why would it rise? Link to comment Share on other sites More sharing options...
ArtW Posted July 29, 2011 Author Share Posted July 29, 2011 So it's smaller, less buoyant and attached to a weight that made it neutral buoyant when it was larger. With that weight included it now as a system is heavier than water. When I let it go, why would it rise? Just at this moment, the resisting force from air inside baloon is bigger the pressure from water. The balloon wants to be bigger. Link to comment Share on other sites More sharing options...
ArtW Posted July 29, 2011 Author Share Posted July 29, 2011 We are fighting! Hahahaha We are fighting at http://www.guokr.com/post/56593/ Come on! Poeple in the world, go to China to fight, but you have to understand Chinese at first. Please consider, if you pull the balloon up and release, what will happen? Link to comment Share on other sites More sharing options...
csmyth3025 Posted July 29, 2011 Share Posted July 29, 2011 Just at this moment, the resisting force from air inside balloon is bigger the pressure from water. The balloon wants to be bigger. The "resisting force from air inside the balloon" (pressure) is not bigger than the pressure of the surrounding water - it exactly matches the pressure of the surrounding water. The balloon will not want to be bigger unless the pressure of the surrounding water is reduced. It may be easier to envision the effect that increasing pressure will have as depth increases if you think of a rock suspended from a plastic bag containing an air bubble. If this set-up is in a submerged state - but suspended in equilibrium, it will remain where it is. This state of equilibrium is balanced on a knife-edge, though. The slightest nudge up or down will cause it to continue its motion at an escalating rate. If it rises, the reduced water pressure will allow the bubble of air to expand and thus displace more water - which means that it will be more buoyant. Conversely, if it sinks even slightly, the increased water pressure will compress the air bubble and the bubble will therefore displace less water. It will become less buoyant. Chris Link to comment Share on other sites More sharing options...
ArtW Posted July 29, 2011 Author Share Posted July 29, 2011 The "resisting force from air inside the balloon" (pressure) is not bigger than the pressure of the surrounding water - it exactly matches the pressure of the surrounding water. The balloon will not want to be bigger unless the pressure of the surrounding water is reduced. It may be easier to envision the effect that increasing pressure will have as depth increases if you think of a rock suspended from a plastic bag containing an air bubble. If this set-up is in a submerged state - but suspended in equilibrium, it will remain where it is. This state of equilibrium is balanced on a knife-edge, though. The slightest nudge up or down will cause it to continue its motion at an escalating rate. If it rises, the reduced water pressure will allow the bubble of air to expand and thus displace more water - which means that it will be more buoyant. Conversely, if it sinks even slightly, the increased water pressure will compress the air bubble and the bubble will therefore displace less water. It will become less buoyant. Chris When you push it down, and do not release, resisting force from air inside the balloon = pressure from water + elastic force from the skin of the balloon + pressure from your hand it is bigger than ...? Link to comment Share on other sites More sharing options...
swansont Posted July 29, 2011 Share Posted July 29, 2011 When you push it down, and do not release, resisting force from air inside the balloon = pressure from water + elastic force from the skin of the balloon + pressure from your hand it is bigger than ...? When you push the balloon down, the pressure from the water increases according to pgh. The volume of the air will decrease until the pressure equalizes. Link to comment Share on other sites More sharing options...
John Cuthber Posted July 29, 2011 Share Posted July 29, 2011 This is heading into troll territory. If ArtW wants to continue to believe something silly even after it has been explained why it's wrong, I don't think it's our problem. Link to comment Share on other sites More sharing options...
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