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Dean ![]() Special Collaborator ![]() ![]() Retired Admin and Amateur Layabout Joined: May 13 2007 Location: Europe Status: Offline Points: 37575 |
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Yes, exactly. You could replace the steel ball and string with a ping pong ball and rigid rod and the right-side will still go down.
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Gerinski ![]() Prog Reviewer ![]() ![]() Joined: February 10 2010 Location: Barcelona Spain Status: Offline Points: 5154 |
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His conclusion was correct, the scale will tip to the right, but I'd say that the reasoning was not correct: "Assuming the strings are mass less, I would imagine that the right hand side would fall. It should be independent on the weight of the two balls, only depending on their volume and the fact that the left hand side will float while the right hand side will not. Essentially what's pressing on the scales on both sides is the weight of the water and the pressure of the displaced water by the balls. However, on the left we will have an opposing contribution due to the tension in the string since it is connected to the scale. The left hand side should be lighter. " 1-it is not the fact that one ball floats and the other not which is relevant, as I said the ping pong ball would balance the scale if it was pushed submerged from outside, having exactly the same tendency to float. 2-the displaced water does not contribute to the weight of the left, only on the right. If it did, the scale would stay balanced. 3-the tension in the string does not "reduce the weight of the left side" since they for part of the same system. No matter how big a bag full of air you would submerge (tied to the bottom of the container) it will never lift up its own container.
Edited by Gerinski - August 14 2014 at 06:42 |
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Equality 7-2521 ![]() Forum Senior Member ![]() ![]() Joined: August 11 2005 Location: Philly Status: Offline Points: 15784 |
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1) I see the tension in the string keeping the ball in equilibrium to be an important part of the analysis.
EDIT: Also mean to say, you are correct the displaced water does not factor on the left. EDIT Again: A google image searched turned up a solution where someone was nice enough to draw the diagram and write out the equations. So definitive answer I suppose here. Edited by Equality 7-2521 - August 14 2014 at 06:57 |
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"One had to be a Newton to notice that the moon is falling, when everyone sees that it doesn't fall. "
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Dean ![]() Special Collaborator ![]() ![]() Retired Admin and Amateur Layabout Joined: May 13 2007 Location: Europe Status: Offline Points: 37575 |
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As I love to disagree, I shall continue to do so. ![]() Both objects have the same tendency to float by the same degree, both are exerting the same downward force on the scale. That is what Pat has said in the second sentence you have highlighted The steel ball does not float because the force of gravity acting upon its mass is far greater than the buoyancy force pushing it up. That is what Pat has said in the first sentence you have highlighted. The tension in the string that keeps the ping pong ball submerged is exerting an equal and opposite force on the ping-pong ball to counter-act the weight of water the ball displaces. That is what he has said in the third highlighted sentence. The left-hand side is lighter than the right, it is not lighter because it is trying to pull itself up by its own bootlaces (which is what you have interpreted).
1. The ping-pong ball floating is wholly relevant because of the tension in the string. The tension of the string counter-acts the buoyancy so the net result is equilibrium. This is the same as taking the ball and string out of the water and placing it alongside the beaker on the scale - only then does "floating" become irrelevant. 2. Correct, but weight is a force. The displaced water contributes equally in both instances but in the left-hand beaker there is an opposing force that negates this increased "weight" 3 - Yes it does. The total "weight" is the mass of the (beaker + water + string + ping-pong-ball) x g plus the buoyancy of the displaced water minus the tension (force) in the string. ...plus whatever Pat wrote while I was typing this.
Edited by Dean - August 14 2014 at 07:12 |
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Gerinski ![]() Prog Reviewer ![]() ![]() Joined: February 10 2010 Location: Barcelona Spain Status: Offline Points: 5154 |
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OK, perhaps the wording was not clear enough to me. Here is the actual experiment, long preparation, you can jump directly to the 10 min mark |
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Dean ![]() Special Collaborator ![]() ![]() Retired Admin and Amateur Layabout Joined: May 13 2007 Location: Europe Status: Offline Points: 37575 |
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This was a fun puzzle Gerard. ![]()
The first thing you will notice is the reading on the scales should increase as the object is lowered into the water. Once the object is fully submerged the reading should not change until the object hits the bottom of the jug. What you should find is
from these results:
Because 1ml of water weighs exactly 1gm we can then calculate the predicted weight of the displaced water simply by multiplying the volume (B) by the constant 1 gm/ml, therefore: F: (B) x 1gm/ml from those calculations you should find:
...there will be a slight error in the last one because the graduations on kitchen jugs aren't very accurate compared to the accuracy of the scales. |
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Equality 7-2521 ![]() Forum Senior Member ![]() ![]() Joined: August 11 2005 Location: Philly Status: Offline Points: 15784 |
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You responded on my behalf much more eloquently than I did. Thank you. |
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"One had to be a Newton to notice that the moon is falling, when everyone sees that it doesn't fall. "
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Gerinski ![]() Prog Reviewer ![]() ![]() Joined: February 10 2010 Location: Barcelona Spain Status: Offline Points: 5154 |
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Alright here's another one, this one is easy but believe me, I have seen a couple of people scratching their heads for a couple of minutes...
GRAVITY AT THE CENTER OF THE EARTH (for simplicity we assume that the Earth is perfectly spherical and that at any given distance from the center towards the surface it's average density the same in every direction) Obviously being on the surface, the Earth's gravity pulls us towards its center. Now imagine, hypothetically of course, that you sit at the very center of the Earth. What will you feel? a) since the Earth's mass is now surrounding you in every direction, its gravity will pull you outwards, you will feel stretched out in every direction towards the surface. Since the pull is the same in every direction you will not move anywhere but your body will be stretched outwards. b) gravity pulls you outwards towards the surface, but since the force is the same in every direction they all cancel out so you will not feel anything at all, you will feel weightless but "not stretched". c) even if the Earth's mass is now surrounding you in every direction, you will still be crushed to the very center to a ball the size of a pea if not less
Edited by Gerinski - August 14 2014 at 09:15 |
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Dean ![]() Special Collaborator ![]() ![]() Retired Admin and Amateur Layabout Joined: May 13 2007 Location: Europe Status: Offline Points: 37575 |
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My pleasure. As you answered the puzzle first I thought you deserved the credit for giving the correct reason too.
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Dean ![]() Special Collaborator ![]() ![]() Retired Admin and Amateur Layabout Joined: May 13 2007 Location: Europe Status: Offline Points: 37575 |
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I would feel very hot and short of breath. Other than that, none of the above, but b) is the closest.
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Equality 7-2521 ![]() Forum Senior Member ![]() ![]() Joined: August 11 2005 Location: Philly Status: Offline Points: 15784 |
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Yeah with the obvious simplifying assumptions why would anyone think anything except (b).
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"One had to be a Newton to notice that the moon is falling, when everyone sees that it doesn't fall. "
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Gerinski ![]() Prog Reviewer ![]() ![]() Joined: February 10 2010 Location: Barcelona Spain Status: Offline Points: 5154 |
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Alright, nobody else posted so I guess that we can crack this one. As is often the case with puzzles, the key to this one lies in that the subject is phrased in a deliberately misleading and incomplete way. At first the question seems to be about gravity, but then what is actually asked is "what will you feel" or "what will happen to you" which is an altogether different question. Without any additional information, and ignoring the effects of the extreme temperature believed to be around 6000 Celsius at the inner core, if you would magically suddenly appear at the center of the Earth you would be immediately crushed into nothing by the huge pressure, even if the net gravitational force there is indeed zero, same as if you dive too deep in the ocean you will be crushed by the hydrostatic pressure even if the gravitational force at that depth is less than at the surface. So the correct answer is "c". Only if we were told that you were sitting at the center inside an hypothetical protective spherical shell capable of supporting the immense pressure and holding atmospheric pressure in its interior, then would "b" be the correct answer. |
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Dean ![]() Special Collaborator ![]() ![]() Retired Admin and Amateur Layabout Joined: May 13 2007 Location: Europe Status: Offline Points: 37575 |
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Oh no. 'Hypothetical' implies that, as Pat said, certain simplifying assumptions are made and the bounds of the answer are contained within the limiting conditions set by the question. The primary assumption inherent in the formulation of the question is that the environmental conditions can, hypothetically, support the subject of the experiment. The thought-experiment is only concerned with the effects of gravity. Under those predefined conditions and implied assumptions (b) gives the correct answer but not the exact reason.
![]() Edited by Dean - August 15 2014 at 07:58 |
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Gerinski ![]() Prog Reviewer ![]() ![]() Joined: February 10 2010 Location: Barcelona Spain Status: Offline Points: 5154 |
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"for simplicity we assume that the Earth is perfectly spherical and that at any given distance from the center towards the surface it's average density the same in every direction"
Where do the "simplifying assumptions" talk about hyperstrong shells? Why would "c" be listed as an option then?
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Dean ![]() Special Collaborator ![]() ![]() Retired Admin and Amateur Layabout Joined: May 13 2007 Location: Europe Status: Offline Points: 37575 |
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Because every multiple choice has a silly answer?
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Dean ![]() Special Collaborator ![]() ![]() Retired Admin and Amateur Layabout Joined: May 13 2007 Location: Europe Status: Offline Points: 37575 |
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Also, if the average density in all directions is the same then you have assumed that pressure is also constant through-out the sphere. The water density at the bottom of the sea is greater than at the surface.
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Gerinski ![]() Prog Reviewer ![]() ![]() Joined: February 10 2010 Location: Barcelona Spain Status: Offline Points: 5154 |
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The assumption said that "at any given distance from the center towards the surface, the average density is the same in every direction". This is not saying that density is the same at any given distance, but only that there are no noticeable concentrations of mass which could pull you in some direction more than to others, that the Earth is formed of concentrical shells of a certain density each. But unless explicitly stated otherwise, logic says that density increases with depth.
At any rate, c'mon don't be a bad loser, this one is not about actual scientific knowledge which I know you have in spades, it is about the tricky phrasing, about spotting it or not. If you spot it then it's as simple as replying something like "net gravity at the center is zero but unless you are shielded from the pressure the answer to "what will happen to you" is (c) = you will be crushed into nothing, which has nothing to do with gravity". In any case I'm curious as to why did you say in your first reply that "none of the above" and in a later one "(b) gives the correct answer but not the exact reason".
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Equality 7-2521 ![]() Forum Senior Member ![]() ![]() Joined: August 11 2005 Location: Philly Status: Offline Points: 15784 |
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I'm sorry, but no. You can ask hypothetical physical situations for 3 reasons: Pure fun, to clarify understanding of a physical system, to be a smartass. If you're doing the third, then it has no place anywhere really. Phrasing the problem as a scientific question precludes the third. |
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"One had to be a Newton to notice that the moon is falling, when everyone sees that it doesn't fall. "
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Dean ![]() Special Collaborator ![]() ![]() Retired Admin and Amateur Layabout Joined: May 13 2007 Location: Europe Status: Offline Points: 37575 |
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If you must keep stretching the point about density then you cannot in the same breath ignore the effects of temperature, part of which is a consequence of the pressure, nor can you chose to ignore the composition of the core - when embedded in molten iron you would not survive long enough to feel anything, nor would your body survive long enough to be crushed to the size of a pea. All 'hypothetical' experiments make simplified assumptions. Even with the trick phrasing of "at any given distance" you have made assumptions. As to why "none of the above" and "not the exact reason": the force isn't "the same in all directions"...
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Gerinski ![]() Prog Reviewer ![]() ![]() Joined: February 10 2010 Location: Barcelona Spain Status: Offline Points: 5154 |
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Alright I see that you didn't like this one, but you can still try it with some of your friends if you want to.
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