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A water tank put on scale measuring $50 kg$ , inside the tank a fully submerged balloon tied with a thread to the tank bottom.

If the thread was cut, will there be a different reading on the scale (momentarily until the balloon reach the surface) ?

I believe that the forces downward are not only the mass but the water pressure at the tank bottom multiplied by the area. A balloon being submerged will cause displacement and raise water level thus total pressure of tank bottom, that can be observed during the process of submerging the balloon and until you tie it to the tank bottom, only then the buoyancy force will cancel the added pressure until that thread is cut .

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  • $\begingroup$ What exactly is your question? $\endgroup$ – sammy gerbil May 15 '17 at 10:47
  • $\begingroup$ will the force on the scale increase, despite the mass of the water being the same? $\endgroup$ – Khaled Salah May 15 '17 at 23:44
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This is not how it works. You don't sum up pressure to weight because the pressure water apply on the floor of the container IS its weight, and indeed if you calculate the total force ie you multiply the pressure $\rho gh$ for the surface of the base, suppose rectangular cuboid container you obtain $\rho hAg$ $h\cdot A$ is the volume $V$ and $\rho V$ is the mass so $mg$ with is indeed the weight. The right way to think about it is using dynamic principle and so impose that the sum of the force on a resting body is zero. In this case the only force the water is subjected is its weight, so the force the scale apply must even be equal to the weight. When you cut the thread the balloon accelerate up, so there is a force the water applies on it and the opposite force that the balloon apply on the water directed down (third law), so the weight measured on the scale will increase. The amount of the increase depends on the volume of the body : $\rho Vg$, where $\rho$ is the density of water and $V$ the volume of the balloon.

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  • $\begingroup$ i didn't add the weight of the water to the total pressure it causes on the floor. In my believe, they gonna be 2 different things in this case. if i force a balloon into the water, the scale will give higher reading despite there's no contact between the balloon and scale, the increase in reading is direct result of the water level rising . in other words , the force i exerted by my hand is translated into displacement and added to the reading $\endgroup$ – Khaled Salah May 15 '17 at 23:41
  • $\begingroup$ But it's not like this, for what you are saying if we change the shape of the container leaving the amount of water and the balloon unchanged the value on the scale would change, because if we take a container with a larger surface then the level of the water will increase less than with a container with a smaller base surface. The change in the scale will depends on the force you apply to keep the balloon underwater, which depends on his volume, and not on how much the water level increases. $\endgroup$ – Claudio P May 15 '17 at 23:46
  • $\begingroup$ the shape wont matter , little change in water level will be multiplied by big area or big change in water level will be multiplied by little area . make 2 examples and try it $\endgroup$ – Khaled Salah May 15 '17 at 23:53
  • $\begingroup$ Yes, that's true, and indeed it means that you are considering the volume and not the level (level increase multiplied per surface gives volume increase), the increasing in the volume it's equal to the volume of the balloons, which takes to the calculation in my answer. You can think like this if you want, but the most direct way of thinking about is through forces and weight and not through pressure, even because pressure is actually caused by weight. If you want to make the calculation through pressure than you need to know the shape, since as you said you need to multiply for it. $\endgroup$ – Claudio P May 15 '17 at 23:56
  • $\begingroup$ yes i'd like to do it through pressure and shape, i can choose a specific shape and start from there. weight is just mass x G and that wont change. but if we calculate through pressure it will show a change. think of 30 x 30 x height 30 cm water tank, pressure would be 30 gm /cm , total pressure 27,000 gm (27 kg). dip in a 900 cm balloon , the water level will rise 1 cm , pressure on floor will be 31 gm/cm . total pressure will be 28,000 gm . if it sounds odd then think of the concept that says centrifugal force is not real force yet it exists and affect lots of things $\endgroup$ – Khaled Salah May 16 '17 at 22:23
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The water tank has forces on it only of gravity and the scale (ignoring air pressure/buoyancy affects in the atmosphere). Since no mass leaves the tank, the total gravitational pull on it must be constant.

That leaves only the scale. Any changes in the normal force from it would accelerate the tank.

At the moment the string is cut the balloon starts to accelerate up. Presumably it would then reach some nearly-constant speed as drag increases until it reached the surface. But the space occupied by the balloon is replaced with water. The net result of these changes is that the center of mass of the tank moves downward while the balloon moves upward.

So momentarily after the string is cut, the scale will be slightly lighter as the COM accelerates downward. It will then return to the initial static weight as the balloon ascends at constant speed. When the balloon reaches the surface, the scale will show slightly higher as the COM decelerates to a stop.

I believe that the forces downward are not only the mass but the water pressure at the tank bottom multiplied by the area.

(Again, assuming we are ignoring air pressure affects here) the mass of the water is what generates the pressure. As long as we account for all the masses, we don't have to worry about pressure separately.

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  • $\begingroup$ The center-of-mass approach is a really nice, intuitive way to approach this problem. You can even invert the whole problem, casting it as releasing a balloon of water from the top of an air-filled tank, to see the same effect on the scale. This is may be even more intuitive, since it corresponds to the well-known sensation of being in an elevator that starts/stops - as the elevator starts to descend, you temporarily have a lower apparent weight. $\endgroup$ – Nuclear Wang Nov 22 at 20:18
  • $\begingroup$ I think it takes a minute to see it though. Dropping a water-filled balloon, it's not too difficult to grasp that there's a deceleration when it slams into the floor. It's very easy to overlook the same thing happening with the water in the tank as the air balloon rises and reaches the surface. $\endgroup$ – BowlOfRed Nov 22 at 20:26

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