Guest Posted January 8, 2009 Report Share Posted January 8, 2009 There is a glass of water, standing on a scale, showing a digital weigth of 300.00 grams. You hold a toothpick in your hand and submerge it in to the water. You don't touch the glass, and no water is overflowed from the glass. Does the number on the scale changes? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 There is a glass of water, standing on a scale, showing a digital weigth of 300.00 grams. You hold a toothpick in your hand and submerge it in to the water. You don't touch the glass, and no water is overflowed from the glass. Does the number on the scale changes? Yes, It decrese the amount of water the toothpick absorbs. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Yes, It decrese the amount of water the toothpick absorbs. Sorry, it's my fault. Assume the toothpick is made of plastic and doesn't abserbes a drop of water. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 It doesnot change Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Imediatly after, yes, the number will change corresponding to the force the toothpick impress to the water, relative to the speed you insert it (maybe this part its just preciosity =)). Just after that, the balance will remain 300.00 since your hand are holding the toothpick, not the water.Sorry, it's my fault. Assume the toothpick is made of plastic and doesn't abserbes a drop of water. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 I would say that any effect of down-pressure from inserting/holding the toothpick in the water would be canceled by the effects of bouyancy. Alternately, if the needle were made of a material denser than water, like steel, then it's simply a case of gravity canceled by holding the needle up. Either way, the only change in the scale would be a slight temporary bump as you insert the needle and the water displaces to adjust, and that only if you do it quickly. The sum of the rest of the forces involved would equal zero. Interesting riddle. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 LOST THE TOOTH PIC DEEPendS if YOU ARE IN EUROPE dot V coma Nb ..300.00 v 300,00 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Answers are not correct yet. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Just a guess I think it does change, the mass of the needle displaces "some" (probably very little depending on the size) water, so if this happens doesn't it mean that the mass that's actually on the scale is higher, therefore the weight increases? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 (edited) My thoughts... I think it does change. If you were to replace the needle with, say, an inflatable ball that just fits inside a really big glass, then you would need to exert some downward pressure on the ball to submerge it partially. The more you want it submerged, the more downward pressure you have to apply. If you are exerting downward pressure, then this means that the reading on the weighing scales would go up. To see this in practice, you could stand on a separate set of scales at the same time. As you exert downward pressure on the ball, you are transferring some of your weight onto the ball. This would make your own scales go down, which must be countered by an equal increase in the scales of the glass. This means the answer depends on the buoyancy of the object being submerged and how much it is submerged. In the case of the ball, if you let go then the scales will clearly go up by the weight of the ball (or the weight of the water displaced by the ball above the original water level). If you push the ball down, the scales will go up by the weight of the extra water that is displaced by pushing the ball down. If you hold the ball in the water but above the level it would ordinarily float at, then the scales will show an increase less than the weight of the ball (equal to the weight of the water displaced by the ball above the original water level). In any of these cases the scales go up by the weight of the water displaced above the original water level. Now, if the ball was made of a lead weight then, if you let go, clearly the scales will go up by the weight of the ball. However, if you don't let go, then you will still be supporting part of the weight of the ball. Assuming the ball is fully submerged, you will be supporting the weight of the ball minus the weight of the amount of water displaced by the ball, which is equal to the weight of an amount of water equal to the volume of the ball. If you are supporting that amount less, then the scales of the glass are supporting that amount more, which means that they go up by the weight of water displaced by the ball. Putting these two cases together (I think) we see that the scales under the glass increase by the weight of the water displaced by the ball above the original water level, unless the ball sinks and is dropped in which case the scales go up by the weight of the ball. Moving back to our original problem, the toothpick may be smaller than a ball, but will have exactly the same effect. As we know you are still holding the toothpick (i.e. you haven't dropped it), the scales will show an increase in weight equal to the weight of the water displaced by the toothpick above the original water level. In reality for a toothpick, this will be very little water, so whether your scales are accurate enough to measure it or not I can't say! Edited January 8, 2009 by neida Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Just a guess I think it does change, the mass of the needle displaces "some" (probably very little depending on the size) water, so if this happens doesn't it mean that the mass that's actually on the scale is higher, therefore the weight increases? you are holding the needle. Wouldn't that nullifying any effect gravity would have on it. But since I've already been told I'm incorrect, I'll be interested in seeing what the OP has to offer. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 you are holding the needle. Wouldn't that nullifying any effect gravity would have on it. But since I've already been told I'm incorrect, I'll be interested in seeing what the OP has to offer. No, because... the weight of the needle/toothpick is being partially supported by the water as described in my post. In fact, if it's a particularly bouyant toothpick (unlikely, but then it is made of plastic), you may even have to be applying downward pressure in order to submerge it in the water. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 (edited) The weight of the water is a force (mass*gravity) working downward. If you put a toothpick in that is only partially submerged, you have the bouyant force acting upward which much be countered by an additional force downward according to Mr. Newton. However since you have to hold the toothpick to keep it partially submerged, the force applied by you on the toothpick is counteracting the bouyancy therefore there is no change in weight because you are exerting the force that would have to be exerted by gravity in order to effect the weight. So while the mass in the beaker increases the volume of the water * its density is still the only weight read by the scale. To simplify the thinking, neglect the toothpick and put your finger in the water partially. There is no change in weight because you are holding the weight of your finger, not the scale. Or something like that Edited January 8, 2009 by HokieKen Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 (edited) Or something like that The weight of the water is a force (mass*gravity) working downward. If you put a toothpick in that is only partially submerged, you have the bouyant force acting upward which much be countered by an additional force downward according to Mr. Newton. However since you have to hold the toothpick to keep it partially submerged, the force applied by you on the toothpick is counteracting the bouyancy therefore there is no change in weight because you are exerting the force that would have to be exerted by gravity in order to effect the weight. So while the mass in the beaker increases the volume of the water * its density is still the only weight read by the scale. To simplify the thinking, neglect the toothpick and put your finger in the water partially. There is no change in weight because you are holding the weight of your finger, not the scale. I understand what you mean As neida states, the water is supporting part of the wieght because of the buoyancy... Isn't this why it should feel lighter in the water, because as the mass pushes the water down, the water pushes the mass upwards (bouyancy)??? So this would make the water hold "some" of the weight making the scale change. Just a thought *EDIT* Typos Edited January 8, 2009 by JarZe Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 No Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 (edited) My thoughts... I think it does change. If you were to replace the needle with, say, an inflatable ball that just fits inside a really big glass, then you would need to exert some downward pressure on the ball to submerge it partially. The more you want it submerged, the more downward pressure you have to apply. If you are exerting downward pressure, then this means that the reading on the weighing scales would go up. To see this in practice, you could stand on a separate set of scales at the same time. As you exert downward pressure on the ball, you are transferring some of your weight onto the ball. This would make your own scales go down, which must be countered by an equal increase in the scales of the glass. This means the answer depends on the buoyancy of the object being submerged and how much it is submerged. In the case of the ball, if you let go then the scales will clearly go up by the weight of the ball (or the weight of the water displaced by the ball above the original water level). If you push the ball down, the scales will go up by the weight of the extra water that is displaced by pushing the ball down. If you hold the ball in the water but above the level it would ordinarily float at, then the scales will show an increase less than the weight of the ball (equal to the weight of the water displaced by the ball above the original water level). In any of these cases the scales go up by the weight of the water displaced above the original water level. Now, if the ball was made of a lead weight then, if you let go, clearly the scales will go up by the weight of the ball. However, if you don't let go, then you will still be supporting part of the weight of the ball. Assuming the ball is fully submerged, you will be supporting the weight of the ball minus the weight of the amount of water displaced by the ball, which is equal to the weight of an amount of water equal to the volume of the ball. If you are supporting that amount less, then the scales of the glass are supporting that amount more, which means that they go up by the weight of water displaced by the ball. Putting these two cases together (I think) we see that the scales under the glass increase by the weight of the water displaced by the ball above the original water level, unless the ball sinks and is dropped in which case the scales go up by the weight of the ball. Moving back to our original problem, the toothpick may be smaller than a ball, but will have exactly the same effect. As we know you are still holding the toothpick (i.e. you haven't dropped it), the scales will show an increase in weight equal to the weight of the water displaced by the toothpick above the original water level. In reality for a toothpick, this will be very little water, so whether your scales are accurate enough to measure it or not I can't say! I stand corrected.Your solution is dependent on the density of the toothpick relative to the water. A toothpick made of anyting denser than water (though decidedly dangerous to use) would, in fact, need to be held up and not pushed down, making the second half of my original post correct. I'm still not totally convinced that bouyancy dosn't push back and cancel that force out in the first half, but your ball analogy does make a convincing argument. Edited January 8, 2009 by Grayven Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 (edited) No, because... the weight of the needle/toothpick is being partially supported by the water as described in my post. In fact, if it's a particularly bouyant toothpick (unlikely, but then it is made of plastic), you may even have to be applying downward pressure in order to submerge it in the water Moving back to our original problem, the toothpick may be smaller than a ball, but will have exactly the same effect. As we know you are still holding the toothpick (i.e. you haven't dropped it), the scales will show an increase in weight equal to the weight of the water displaced by the toothpick above the original water level. . That Neida was basically correct. 1) if you simply place the toothpick on top of the water, the scale increases by the weight of the toothpick. 2) In this state, some water has been displaced, equal to the (slight) weight of the toothpick. 3) to submerge the toothpick, you have to apply enough force to displace the water for parts of the toothpick that are not already submerged. 4) The net force you exert, which will be reflected on the scale as weight, is the sum of these two, which is equal to the weight of the water that the entire toothpick displaces. Whether or not the scale can detect this is separate question. Edited January 8, 2009 by xucam Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Maybe this will clarify things... Neida's answer, while a bit convoluted, is correct. See attached sketch. The subscript "w" signifies the water, and the superscript "*" the immersed object. Notice that the solution for the force supporting the beaker ( R ) is independent of the weight, w, of the immersed object (or its density), provided the force, F, required to keep the immersed object stationary comes from something other than the beaker or water (such as a finger, string, etc...). This means that F could point upwards (as shown) for negatively buoyant objects (sinking things), or downwards for positively buoyant objects (floating things). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 That Neida was basically correct. 1) if you simply place the toothpick on top of the water, the scale increases by the weight of the toothpick. 2) In this state, some water has been displaced, equal to the (slight) weight of the toothpick. 3) to submerge the toothpick, you have to apply enough force to displace the water for parts of the toothpick that are not already submerged. 4) The net force you exert, which will be reflected on the scale as weight, is the sum of these two, which is equal to the weight of the water that the entire toothpick displaces. Whether or not the scale can detect this is separate question. Maybe this will clarify things... Neida's answer, while a bit convoluted, is correct. See attached sketch. The subscript "w" signifies the water, and the superscript "*" the immersed object. Notice that the solution for the force supporting the beaker ( R ) is independent of the weight, w, of the immersed object (or its density), provided the force, F, required to keep the immersed object stationary comes from something other than the beaker or water (such as a finger, string, etc...). This means that F could point upwards (as shown) for negatively buoyant objects (sinking things), or downwards for positively buoyant objects (floating things). Now I stand corrected. What would the Den do without someone like me to point out the painfully obvious wrong answers all the time? I applaud my shameless courage in the face of almost certain failure. (ooohh, I think I might have a new sig...) Thanks for the clarification xucam & d3k3 & neida. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 To simplify the thinking, neglect the toothpick and put your finger in the water partially. There is no change in weight because you are holding the weight of your finger, not the scale. It may seem that way because your finger is very light. However, if it's possible at all to imagine putting the entire top half of your body in the water (obviously of a very large and funny shaped glass) then the water would support the weight of the top half of your body. Maybe a way to simulate this is to think of a swimming pool with a set of scales on the side. When you are in the pool you are floating and the pool (if it had a giant set of scales underneath it) would be taking your weight. You now put your hands on the scales on the side and start to lift yourself out of the water. As your body starts to lift out of the water, the scales on the side will show that your hands are gradually taking on more and more weight, and so the pool is carrying less and less. When you finally pull yourself entirely out of the pool, your entire weight is now on the set of scales. So, whilst everything from your waste down is in the water, it is supported by the water. Whilst everything from your knees down is in the water, it is supported by the water. Whilst only your big toe is in the water, it is supported by the water. Now replace "big toe" by "finger" in the puzzle and we have exactly the same situation. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 It may seem that way because your finger is very light. However, if it's possible at all to imagine putting the entire top half of your body in the water (obviously of a very large and funny shaped glass) then the water would support the weight of the top half of your body. Maybe a way to simulate this is to think of a swimming pool with a set of scales on the side. When you are in the pool you are floating and the pool (if it had a giant set of scales underneath it) would be taking your weight. You now put your hands on the scales on the side and start to lift yourself out of the water. As your body starts to lift out of the water, the scales on the side will show that your hands are gradually taking on more and more weight, and so the pool is carrying less and less. When you finally pull yourself entirely out of the pool, your entire weight is now on the set of scales. So, whilst everything from your waste down is in the water, it is supported by the water. Whilst everything from your knees down is in the water, it is supported by the water. Whilst only your big toe is in the water, it is supported by the water. Now replace "big toe" by "finger" in the puzzle and we have exactly the same situation. Tact is the art of making a point without making an enemy. - Howard W. Newton Neida, you have quite a talent for illustrating your argument effectively and concisely, without making one feel lectured to. It is rare and appreciated. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Maybe this will clarify things... Neida's answer, while a bit convoluted, is correct. See attached sketch. The subscript "w" signifies the water, and the superscript "*" the immersed object. Notice that the solution for the force supporting the beaker ( R ) is independent of the weight, w, of the immersed object (or its density), provided the force, F, required to keep the immersed object stationary comes from something other than the beaker or water (such as a finger, string, etc...). This means that F could point upwards (as shown) for negatively buoyant objects (sinking things), or downwards for positively buoyant objects (floating things). I stand corrected I hastily jumped to the conclusion that F=w+B, but as you have shown, F=w and B effects weight. I acquiese to d3k3 and neida and will be enrolling in a elementary fluid mechanics course at my local community college Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 The weight inreases by the amount of the toothpick that is in the water (total toothpick less the amount out of the water). That's assuming that the fingers are not in the water. If the fingers are in the water, I would increase this answer to include the weight of the fingers in the water. It seems to me that the water medium transfers the weight to the scale, regardless of buoyancy. For example, if there is a pool on a scale and I jump in the pool, the weight of the pool will increase by my weight, regardless of whether I am floating or submerged. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 I stand corrected I hastily jumped to the conclusion that F=w+B, but as you have shown, F=w and B effects weight. Uh, no. w+F+B=0, or in terms of absolute values, F=w-B, or F=B-w, depending on whether the object floats or sinks. What I probably wasn't clear about is that there are two separate balances to consider: 1. The entire system is in static equilibrium, so the sum of external forces (the forces supporting the glass of water and the toothpick/submerged object, R and F) and the total weight of all the objects (W + w) is zero. 2. The submerged object is also in equilibrium, so the sum of the forces supporting the submerged object (F), the buoyancy force, (B) and the object's weight (w) is also zero. D. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Uh, no. w+F+B=0, or in terms of absolute values, F=w-B, or F=B-w, depending on whether the object floats or sinks. What I probably wasn't clear about is that there are two separate balances to consider: 1. The entire system is in static equilibrium, so the sum of external forces (the forces supporting the glass of water and the toothpick/submerged object, R and F) and the total weight of all the objects (W + w) is zero. 2. The submerged object is also in equilibrium, so the sum of the forces supporting the submerged object (F), the buoyancy force, (B) and the object's weight (w) is also zero. D. Exactly. I apologize for my misstatement. My intention was to say that the weight will increase by the amount of the buoyant force (B) which is the volume (V*) submerged times the density of water. So when taking an overall balance, F and w will not effect the reading on the scale. Correct? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 8, 2009 Report Share Posted January 8, 2009 Exactly. I apologize for my misstatement. My intention was to say that the weight will increase by the amount of the buoyant force (B) which is the volume (V*) submerged times the density of water. So when taking an overall balance, F and w will not effect the reading on the scale. Correct? Precisely. I took your statement to mean that buoyancy affects the weight of the submerged object, but I realize now you meant the weight reading on the scale. Quote Link to comment Share on other sites More sharing options...
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There is a glass of water, standing on a scale, showing a digital weigth of 300.00 grams.
You hold a toothpick in your hand and submerge it in to the water. You don't touch the glass, and no water is overflowed from the glass.
Does the number on the scale changes?
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