# Does the weight of a steel ball changes when submerged in water? [duplicate]

Will there be a difference in the weight of just the ball under water Vs in normal atmospheric pressure conditions i.e. a weighing scale with the steel ball on it Vs in vacuum assuming earth's gravity is same in all three cases

Now that I THINK I better understand your question, thanks to @David White and @JMac

For the three cases:

1. The weight will be a minimum under water

2. The weight will be a maximum out of the water in a vacuum.

3. The weight out of the water under normal atmospheric temperature and pressure will be very, very slightly less than the weight in a vacuum.

For case 2, the only force acting on the ball is the downward force of gravity, $$mg$$. That’s the maximum force the scale will experience.

For case 3, in addition to the downward force of gravity the ball will experience an extremely small upward buoyancy force due to atmospheric air. That upward force will equal the weight of the air displaced by the ball, which would be the density of the air at standard atmospheric temperature and pressure times the volume of the ball. This would be a very, very small force. It would take an extremely sensitive scale to read this difference between the downward force of gravity and the upward buoyant force of air. If the upward buoyant force were greater than the downward force of gravity, such as for a helium inflated balloon, there would be no reading at all as the ball will rise.

For case 1, the upward buoyant force equals the weight of the water displaced by the ball. This upward force is much, much greater than case 3. The scale will read a minimum.

Hope this helps.

• There is in fact a very small buoyant force on the ball when it is weighed in air, equal to the weight of the air that was displaced by the ball. If this force didn't exist, helium balloons wouldn't float in air. May 22, 2019 at 19:59
• "That displacement, in turn, will depend on the density of the ball." Only if the ball is floating. For the steel ball case in this question, the displacement only cares about the total volume of the ball. It just so happens that the water displaced by the ball's full volume still weighs less than the ball (i.e. steel is more dense than water).
– JMac
May 22, 2019 at 20:07
• Bob D, it's not "critical" for this discussion, but the OP asked about weighing the steel ball in water, in the atmosphere, and in vacuum. With a sufficiently accurate and precise scale, there would be a slight difference in weight between weight in air and weight in vacuum. May 22, 2019 at 20:15
• @DavidWhite To be fair, the question does a poor job IMO making that clear. I only noticed it when re-reading after seeing your comment. It mentions the ball under water vs atmospheric conditions, then uses i.e. and seems to compare water vs atmosphere to water vs vacuum. The only reason it's clear to me that he meant all 3 was because he says "all 3 cases" at the end, but that's easy to miss.
– JMac
May 22, 2019 at 20:20
• @JMac, I agree ... the problem statement could have been somewhat clearer. May 22, 2019 at 20:25

If you weigh the steel ball in something close to a vacuum,then weigh it again immersed in water,its weight will be its weight in a vacuum less the weight of the volume of water it displaces,so it will weigh less. Its mass will remain the same in water as it was in a vacuum,however.

Whether there will be a difference in the ball's weight actually depends on your definition of weight. Weight is defined differently by different people. Some define it as a measure of the force of gravity upon the object (gravitational weight). If you use this definition, the object will weigh the same in all cases. Some define it to be a measure of the force exerted to support the object (operational weight). In this case, the force of gravity and the forces of bouyancy are counfounded, so the force supporting the object will be less if an object is immersed in a dense liquid.

Your question appears to assume the latter case. However, if the results seem counter intuitive, this may be because you're asking about the "operational weight," but your intuition is trying to make sense of the results in terms of "gravitational weight."