Upthrust/ how does the weight of a sphere change in water?

In summary, the homework statement states that the weight of the sphere, partially submerged in water, is its "usual" weight, out of water, minus the weight of the water displaced.
  • #1
C0balt
7
0

Homework Statement


See image attached. Oh it's part c by the way.

Homework Equations


maybe upthrust=weight of water displaced... None really relevant.

The Attempt at a Solution


I thought the balance would initially go up as the sphere entered the water, (but maybe slightly less than the weight of the beaker+sphere because up thrust is acting on the sphere) because the weight of the sphere would be greater than any opposing forces i.e upthrust because the sphere is accelerating. Then when the ball was rising to the surface I would assume the balance reading would be just the weight of the beaker as upthrust will be greater than the weight of the sphere ( or maybe you could calculate the upthrust then take that away from the weight of the beaker?)Then when the sphere is floating on the surface the balance would read the weight of the sphere+beaker. Is any of this sort of right?
 

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  • #2
C0balt said:
ecause the weight of the sphere would be greater than any opposing forces i.e upthrust because the sphere is accelerating.
Do you mean before or after the sphere has become fully immersed?
 
  • #3
haruspex said:
Do you mean before or after the sphere has become fully immersed?
Um before
 
  • #4
I think I would break this into stages, eg. ball out of water, ball entering water, ball falling in water, ball stationary in water, ball rising in water, ball floating on water, (*)
At each stage consider at first only the main forces such as weight and buoyancy to establish the general pattern.
Then you can add considerations of acceleration to see how they affect it, perhaps qualitatively at first, then calculate some values if you can.

There may be other factors you could think about (* and stages), but I'll not mention them unless you do.
 
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  • #5
The "weight" of sphere, partially submerged in water, is its "usual" weight, out of water, minus the weight of the water displaced.
 
  • #6
Show us what you did in parts a and b. Your conclusion regarding the final state in part c is correct. It seems to me, the missing piece of the puzzle is doping out the situation at the instant that the ball has come to a stop under the water. Once you have that, you should be able to fill in all the blanks.

Chet
 

Related to Upthrust/ how does the weight of a sphere change in water?

1. What is upthrust?

Upthrust, also known as buoyancy, is the upward force exerted by a fluid (such as water) on an object that is partially or fully submerged in it. It is caused by the difference in pressure on the top and bottom of an object in a fluid.

2. How is upthrust related to the weight of a sphere in water?

Upthrust is directly related to the weight of a sphere in water. As a sphere is submerged in water, the fluid exerts an upward force on it, which reduces the effective weight of the sphere. The greater the upthrust, the less the weight of the sphere will be in water.

3. Why does the weight of a sphere change in water?

The weight of a sphere changes in water due to the upthrust force exerted by the water. As the sphere is submerged, it displaces the water around it, causing an upward force to be exerted on it. This reduces the effective weight of the sphere.

4. How does the weight of a sphere in water compare to its weight in air?

The weight of a sphere in water will be less than its weight in air due to the presence of upthrust. In air, the only force acting on the sphere is its weight, but in water, the upthrust force counteracts some of its weight, resulting in a lower overall weight.

5. Does the weight of a sphere in water change with depth?

Yes, the weight of a sphere in water will change with depth. As the sphere is submerged deeper into the water, the pressure on it increases, causing the upthrust force to also increase. This results in a further reduction in the effective weight of the sphere.

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