This is really nice. You can follow this up by asking the same question, this time with the crane that is holding the mass placed on the same enlarged balance. Or even have the crane on a separate balance. Another great question is to consider a tennis ball in a beaker of water. The ball is being held entirely below the water level by a string that is attached to the bottom of the beaker. The system is placed on a balance, the reading is recorded. The string is then cut. What happens to the reading? Another one was given to a student that had an interview at the University of York: A sand timer (hourglass) has all its sand in one chamber, this sand timer allows only one grain of sand through at a time, and all grains of sand are the same mass (they are totally uniform). You upturn the sand timer and place it on a balance, so that the empty chamber now sits at the bottom. Describe and explain what happens to the reading on the balance.
stay the same, because the fluid will just move out of the way of the volume that is being displaced. The weight of the object will be countered by the thing that is holding it up, so the crane here. Fluid is displaced, after displacement, there will be no change observed
It would stay the same since the upthrust provided would decrease the tension on the string, leading the mass balance to decypher incorrectly that the weight would decrease, adding to my statement that it would stay the same. :)
Ok, I'm only 1 minute into the video, but here's my answer: The scale will increase in mass equivalent to the water-mass equivalent of the submerged volume. My initial thought was that the mass would stay the same, but then I realized the buoyant force even on something that sinks in water. Another way of looking at it is to use a tension scale on the suspended mass: this scale will reflect the loss of weight due to the buoyant force. That weight's gotta go somewhere. Still another way of looking at it is to ask what would happen if the suspended weight was just a water balloon. Once fully submerged, the tension in the suspending string would be near zero, and the beaker will have a water level reflecting the additional volume of water in the balloon.
Drop unattached weight in water : weight increase. Drop attached weight in water : weight increases but less. How much less depends of the tension left in the cable, which depends of Archimedes principle. Now if you start thinking about Archimedes and pressure you might get confused and end up thinking the balance will read the same. Should a teacher really value the student who gets lost in the thinking process because he lost touch with reality ?
I would think that the reading increases. By Archimedes principle, the water exerts an upward buyont force on the mass. Hence, by Newton's Third Law, the mass exerts an equal and opposite force on the water, and so this force is in the downward direction. This downward force will cause the reading on the balance to increase.
If we consider the whole system of the weight with the line and the beaker with liquid as two separate systems in such a way that the beaker with water has the forces of normal reaction from scale (up) weight of liquid and beaker (down) and the upthrust (down)....then R ( reaction from the scale to the beaker with water)= mg+(weight of displaced fluid) > mg (where m is the collective mass of beaker and liquid) so as the weight is lowered the reading on scale increases initially at a higher rate then at a lower rate and finally stay constant( due to the sudden reduction in cross sectional area and the object being completely lowered)....just as the reading on the scale increases the tension in the string reduces by the same amount...
Before entering the water, the downward force from the mass (m.g) is balanced, in equilibrium, with the ( upward ) tensile load in the cable, and ultimately the crane. Entering the water , the upthrust produced simply reduces the tensile load in the cable. The water in the beaker would rise, but its mass does not change. Therefore there would be no change on the scales. It would be totally different if the cable was cut because the mass of the “system” would increase. I’m confident I’m correct, but I’ve retired, so it doesn’t matter 🎃🤗
@@californiadreamin8423 Allah be merciful on you! You talk about the upthrust, the force exerted by the liquid on the load. According to Newton's third law, an equal and opposite reaction force is acting on the liquid by the load. What happens to that force ?
@@barneycasting8331 When the object enters the water, it displaces the water upwards. That mass of water wants to return to its original level, because it subject to gravitational acceleration , and in effect “pushes back” against the object making it apparently lighter….reducing the downward force of the object. Why doesn’t the object descend ? It cannot descend because it is supported by an equal and opposite upward force produced by the cable, which creates the equilibrium. That cable force is less than it would have been if there was no water in the jar. If the object was lowered further until it rested on the bottom of the jar, then the force on the scales would increase because the cable would go slack, with the bottom of the jar applying an equal and opposite upward force. If it couldn’t do this, then the jar would break. That is my interpretation of what is going on.
It gets bigger proportionate to the amount of water displaced or to be exact, the density of the liquid times the volume of the liquid displaced by the heavier object. There are multiple ways to look at this but probably the easiest way is think of the height of the water column. As the level of liquid goes up the pressure on the bottom increases and the total force is pressure times area.
This is really nice. You can follow this up by asking the same question, this time with the crane that is holding the mass placed on the same enlarged balance. Or even have the crane on a separate balance.
Another great question is to consider a tennis ball in a beaker of water. The ball is being held entirely below the water level by a string that is attached to the bottom of the beaker. The system is placed on a balance, the reading is recorded. The string is then cut. What happens to the reading?
Another one was given to a student that had an interview at the University of York:
A sand timer (hourglass) has all its sand in one chamber, this sand timer allows only one grain of sand through at a time, and all grains of sand are the same mass (they are totally uniform). You upturn the sand timer and place it on a balance, so that the empty chamber now sits at the bottom. Describe and explain what happens to the reading on the balance.
Normally really like your videos, but the title of this one is both click baity / insulting to our profession.
stay the same, because the fluid will just move out of the way of the volume that is being displaced. The weight of the object will be countered by the thing that is holding it up, so the crane here. Fluid is displaced, after displacement, there will be no change observed
🤗. The upthrust would simply reduce the tension in the crane cable.
It would stay the same since the upthrust provided would decrease the tension on the string, leading the mass balance to decypher incorrectly that the weight would decrease, adding to my statement that it would stay the same. :)
Ok, I'm only 1 minute into the video, but here's my answer: The scale will increase in mass equivalent to the water-mass equivalent of the submerged volume. My initial thought was that the mass would stay the same, but then I realized the buoyant force even on something that sinks in water. Another way of looking at it is to use a tension scale on the suspended mass: this scale will reflect the loss of weight due to the buoyant force. That weight's gotta go somewhere.
Still another way of looking at it is to ask what would happen if the suspended weight was just a water balloon. Once fully submerged, the tension in the suspending string would be near zero, and the beaker will have a water level reflecting the additional volume of water in the balloon.
Drop unattached weight in water : weight increase.
Drop attached weight in water : weight increases but less.
How much less depends of the tension left in the cable, which depends of Archimedes principle.
Now if you start thinking about Archimedes and pressure you might get confused and end up thinking the balance will read the same.
Should a teacher really value the student who gets lost in the thinking process because he lost touch with reality ?
It’s interesting reading all the comments and arguments!
I would think that the reading increases.
By Archimedes principle, the water exerts an upward buyont force on the mass. Hence, by Newton's Third Law, the mass exerts an equal and opposite force on the water, and so this force is in the downward direction. This downward force will cause the reading on the balance to increase.
Not too sure what happened with the description sirrreeeee
If we consider the whole system of the weight with the line and the beaker with liquid as two separate systems in such a way that the beaker with water has the forces of normal reaction from scale (up) weight of liquid and beaker (down) and the upthrust (down)....then R ( reaction from the scale to the beaker with water)= mg+(weight of displaced fluid) > mg (where m is the collective mass of beaker and liquid) so as the weight is lowered the reading on scale increases initially at a higher rate then at a lower rate and finally stay constant( due to the sudden reduction in cross sectional area and the object being completely lowered)....just as the reading on the scale increases the tension in the string reduces by the same amount...
Inreases. upthrust is exerted by the water on the mass. tension on the string would decrease
Before entering the water, the downward force from the mass (m.g) is balanced, in equilibrium, with the ( upward ) tensile load in the cable, and ultimately the crane. Entering the water , the upthrust produced simply reduces the tensile load in the cable.
The water in the beaker would rise, but its mass does not change. Therefore there would be no change on the scales.
It would be totally different if the cable was cut because the mass of the “system” would increase.
I’m confident I’m correct, but I’ve retired, so it doesn’t matter 🎃🤗
@@californiadreamin8423 Allah be merciful on you! You talk about the upthrust, the force exerted by the liquid on the load. According to Newton's third law, an equal and opposite reaction force is acting on the liquid by the load. What happens to that force ?
@@barneycasting8331 When the object enters the water, it displaces the water upwards. That mass of water wants to return to its original level, because it subject to gravitational acceleration , and in effect “pushes back” against the object making it apparently lighter….reducing the downward force of the object. Why doesn’t the object descend ? It cannot descend because it is supported by an equal and opposite upward force produced by the cable, which creates the equilibrium. That cable force is less than it would have been if there was no water in the jar.
If the object was lowered further until it rested on the bottom of the jar, then the force on the scales would increase because the cable would go slack, with the bottom of the jar applying an equal and opposite upward force. If it couldn’t do this, then the jar would break.
That is my interpretation of what is going on.
Mr. Wizard did this about 40 years ago. Weight will increase.
YESSSS DR PALFREYMAN‼️‼️‼️‼️‼️
It gets bigger proportionate to the amount of water displaced or to be exact, the density of the liquid times the volume of the liquid displaced by the heavier object. There are multiple ways to look at this but probably the easiest way is think of the height of the water column. As the level of liquid goes up the pressure on the bottom increases and the total force is pressure times area.
Big up palfreyman