The ten year old really got me stumped on this one...
The problem with that is the reason your plug floats is because it is already surrounded by water. As for the mass of the water thing; yes; that is the issue. and the taller you make it the more the mass; but regardless you wont be able to displace more volume than the mass of the water so the ball entering the tank will be pushing against a greater force. Also the bouyancy doesnt fight gravity. it fights pressur. The deeper you go in water; the more pressure, so as the ball rises it goes to areas of deacreasing pressure. That is why the ball rises. BUT in the case of the lowest ball entering the tank the greater pressure is actually above it. not below. so it will be 'attempting to move' Downwards. Not upwards. And hence you are back to the volume displacement issue.
It's not completely clear to me if it is Archimedes' principle you disagree or something I wrote. I can't see your comments be directly related response to my 2'nd or 3'rd paragraph.
In my second paragraph I'm only speaking about the ball on it's way through the valve. And lets focus on the moment where it's exactly halfway through. At this particular moment the upper half is wet, inside the tank, and bottom half is dry outside the tank.
At that particular moment it experience two forces, air pressure from below at approx one atmosphere and water pressure from above depending on the depth of the water.
The water pressure at the bottom of the tank is the same no matter how many balls in there as long they are all making the tank empty. And that pressure pushes all directions down there.
About the ball at 1000 meters, yes it will rise as well. It's feels pressure from both above and below, and it's sides as well.
ha ha I said I agree with the first (Archimedes) statement.
I believe at that particular moment the ball experiences more forces than two.
1. air pressure from below
2. weight of water from above
3. gravity pulling down on it as well as the water.
4. pull from the balls above which are entirely in the water and being forced upwards.
Of these I see the fourth one as the most important. The larger the balls (or cylanders), the more water they displace and therefore the more force being applied in an upward direction. While very tiny ones might lift themselves, they would not add enough upward force to counter the forces 2 and 3 above, but as they increase in size the force they impart will increase as well to the point where the ball that is not yet fully in the water cannot stay where it is. It will be pulled upwards like pulling a drain plug in a sink. –and a drain plug can be pulled from a sink drain if enough upward force is applied.
I think this problem becomes easier to analyse if you first consider each ball by itself, and then assemble the forces acting on the whole system. One of the major problems with the series of diagrams used above is that V2, when driven downwards, is only being immersed in water that is at a lower surface pressure, whereas in the real experiment V2 is entering the tank at the point where the water pressure is significantly higher.
The ball that is partly out of the water at the top is subjected to a net upwards force which is composed of:• Average water pressure just below the surface multiplied by the effective area of the submerged part of the ball, acting upwards• Roughly constant atmospheric multiplied by the effective area of the non-submerged part of the ball, acting downwards• Weight of the ball acting downwards, which is practically nothing compared to the bouyancy forces if an appropriate ball is usedThis ball will experience maximum upwards force when halfway out of the water, where the effective areas above and below the surface are the same.
The balls that are submerged completely are affected in a similar way, being propelled upwards with a force composed of:• Average water pressure on the lower half of the ball, multiplied by the effective area of the lower half, acting upwards• Average water pressure on the upper half of the ball, multiplied by the effective area of the upper half, acting upwards• Weight of the ball acting downwardsOnce again the effective areas are equal to each other, and the net bouyancy force is a result of the difference in pressure between the water pushing the ball down from above and the water pushing the ball up from below.The water pressure is calculated as p = ρgh, where the density 'ρ' is a constant, gravitational accel 'g' is constant, and it is only the effective height of the two areas that determines the pressure difference. As the pressure increases uniformly with depth, all balls that are submerged completely will experience the same upwards force.
The ball that is partly immersed in water, coming into the water tank from the bottom is subjected to a similar set of forces:• Average water pressure at the bottom of the tank, multiplied by the effective area of the submerged part of the ball, acting downwards• Roughly constant atmospheric multiplied by the effective area of the non-submerged part of the ball, acting upwards• Weight of the ball acting downwardsThe difference in pressure between the water pushing the ball down and the air pushing the ball up is greater than for the ball at the top, because the water pressure has increased with depth.
Now if we look at the whole system we can simply a few things right away.• The number of balls on the left side of the pulleys is the same as the number on the right, and they all weight the same amount. This means that overall the weight of the balls can be ignored as the weight forces balance each other.• The air pressure pushing the top ball down generates the same force as the air pressure pushing the bottom ball up, so you can cancel those out.
The forces that are left over to drive the system are the following three:• Water pressure acting on the lower half of the top ball• Water pressure acting on the top half of the bottom ball• Differential water pressure acting on each submerged ball, multiplied by the number of balls
The downwards force on the bottom ball will naturally be greater than the upwards force on the top ball, so there is a net downwards force resulting from the combination of those two. The force of each submerged ball is now a result of the distance between the average height of the top and bottom areas, and the effective areas of the two halves of the ball, which are both determined by the radius.
If we look at the whole system now, we can represent the total force acting on it as:F = Pt*A - Pb*A + n*Pd*AWhere 'F' is the total upwards force, 'Pt' the the pressure of the water acting upwards on the top ball, 'Pb' is the pressure acting downwards on the bottom ball, 'Pd' is the difference in pressure acting on each half of the submerged balls, 'A' is the effective area of half of any of the balls, and 'n' is the number of totally submerged balls.
If we rearrange the formula like so: F = A*(Pt - Pb + n*Pd) then we can see that changing A will only scale the resultant force up or down, but it can never be negative, so it won't change the direction of F. Pt can never be greater than Pb, so Pt - Pb will always be a negative value.
So, what are we left with?The magnitude of Pt - Pb is directly proportional to the distance between the effective height of the lower half of the top ball and the upper half of the bottom ball.The magnitude of n*Pd is directly proportional to the distance between the effective height of the upper and lower halves of each submerged ball. If you increase the radius of each ball, Pd will increase as well. The other option to increased n*Pd is just to add more balls between the top and bottom balls.However, you can quickly see that no matter how many balls you add, or how big you make each ball, the distance between the effective upper and lower halves of the balls (which must be less than the diameter) multiplied by the number of balls cannot be greater than the distance between the top ball and the bottom ball. Due to this restriction, the Pt - Pb factor will always be a larger negative value than n*Pd can be positive, no matter how ideal the setup is.
I followed that all the way up to, but not including the very last conclusion sentence. After your other statements, why would they lead to the final conclusion? (That the Pt – Pb factor will always be a larger negative value than n*Pd can be positive). I believe there was a step left out. Include that and you may convince me.
Also remember that by Archimedes principle, quoted elsewhere on the page, the value of the force will relate to the volume of the displaced water, not its surface area. [Archimedes' original problem was to determine the relative density of two objects with different surface areas to determine the relative density of the two] If you use balls that will mean the volume of displaced water will be proportional to 4/3 * π * r³ [ with 4/3 and pi being constants, we can say it will be proportional to the radius cubed, thus doubling the radius or diameter would give 8 times the amount of water being displaced.] On the other hand, I was considering a cylander probably having a greater height than diameter. This may not matter much other than in the scope of the answer.
Hmmm, considering not just this entry, but some of the others as well, I am reminded of an anemometer. They could not possibly turn, since the surface area is the same on the front and the back, making the applied forces equal. If they turned in the wind that would be perpetual motion which is impossible, right?
[ Tongue-firmly-in-cheek] This cannot rotate just as pinwheels cannot spin. Air-bubble wheels in an aquarium cannot turn because the weight of the water is the same on both sides. Those little clam shells in an aquarium cannot open no matter how much air is inside, because of the weight of all that water above keeping them closed...
Archimedes was a clever guy and all, but his principle does not always hold true - it assumes the water always extends below the object. The ball at the bottom of the tank would be pushed downwards by the water pressure, as the air pressure below is weaker.
The easiest way to look at the different pressures is to divide the height of the tank up as follows:• The height of the water in the tank, from top to bottom is 'hw'• The distance from the top surface of the water to the effective height of the bottom half of the top ball is 'ht'If the top ball was an equivalent cylinder, with the same volume and radius, this would be half the height of the cylinder• The distance from the bottom surface of the water to the effective height of the top half of the bottom ball is 'hb'Likewise, if the bottom ball was an equivalent cylinder, this would be half the height of the cylinder• The distance between the effective heights of the top half and bottom half of each submerged ball is 'hd'This would be the total height of an equivalent cylinder
Based on the above definitions, we can see that hd = 2*ht = 2*hb.Also, there is a restrictions that hw >= ht + hb + n*hd, that is to say that the height of the immersed balls put together cannot be greater than the height of the water.
Because p = ρgh as above, we can state the following:• Pt = ρg*ht• Pb = ρg*(hw-hb) = ρg*hw - ρg*hb• Pd = ρg*hd
Going back to F = A*(Pt - Pb + n*Pd), we can substitute our new equations in to get:F = A*(ρg*ht - (ρg*hw - ρg*hb) + n*ρg*hd)Which can be rearranged to:F = A*(ρg*ht - ρg*hw + ρg*hb + n*ρg*hd)F = Aρg*(ht - hw + hb + n*hd)F = Aρg*(ht + hb + n*hd - hw)
Now we have a nice bundle of values that determines the direction of F, namely the (ht + hb + n*hd - hw) part of the equation.Here we can refer back to the hw >= ht + hb + n*hd relationship defined earlier, which tells us that all the positive parts of equation cannot be greater than the big negative value of hw.Therefore (ht + hb + n*hd - hw) is less than or equal to zero.Therefore Aρg*(ht + hb + n*hd - hw) is less than or equal to zero.Therefore F is less than or equal to zero, which means the net force is pulling the system down, or at best holding it stationary.
An anemometer does not turn unless there is wind. The wind hitting the anemometer's vanes/cups is slowed down, which means that the local air pressure at that point must increase (refer to Bernoulli's equation). One side of the vanes/cups is better at slowing down the air than the other (higher drag), so the pressure generated at the surfaces of the anemometer is not equal, which produces an unbalanced force, causing the anemometer to rotate.
Exactly but; those balls cannot displace more than the water in the container or the container would be empty... So the water should have enough force to keep the lowest ball (the one trying to enter the tank) from entering. And hence we come full circle to that 'magic valve' even some sort of low friction valve has the force of all that water above it. The air below it doesn't really factor in because the air in a normal environment should press equally on all sides with far less force than the water.
Again though; its a cool idea and some really nice and creative thinking. Props to Phoenix (i apalogize in advance for/if i spelled that incorrectly or got the name wrong).
That time you lost me as to what you are talking about... The balls/cylanders being so large they take up the majority of the tank? That would be counterproductive. The water being displaced yields the value of the upward force.
ok here you go.
You can place at the opening some kind of door, meaning two hemispheres that open like a salloon door using some kind
of light spings attached to the doors through each ball passes .
The thing is to overcome the power of the water that holds the doors closed .
This can be accomplished if you find the perfect combination of the gap between the balls and the material in the ball
if you do this the whole thing will work with the help of the springs .
So what do you think ?