How much does the ball weight in water

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    Ball Water Weight
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Homework Help Overview

The problem involves an iron ball suspended in air and submerged in water, with a focus on understanding the forces acting on the system and how they affect the reading on a balance scale. The subject area pertains to fluid mechanics and forces.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the ball and the water, including tension, weight, and upthrust. There are attempts to derive equations and clarify the equilibrium of forces. Questions arise regarding the assumptions made about the system and the implications of Newton's Third Law.

Discussion Status

The discussion is ongoing, with various interpretations being explored regarding the forces at play. Some participants have provided guidance on considering the equilibrium of forces, while others express confusion about the implications of these forces on the balance reading.

Contextual Notes

Participants are grappling with the assumptions related to the forces acting on the ball and the water, particularly in the context of equilibrium and the effects of buoyancy. There is a noted lack of consensus on the correct interpretation of these forces.

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Homework Statement


An iron ball is suspended in the air by a string attached to the ceiling. This iron ball is then submerged into the a beaker of water on top of a measuring balance, such that the ball is not in contact with the base of the beaker. When the beaker of water(without anything in it) is measured with the balance, the reading on the balance is W. What is the reading on the balance with the suspended ball in it?


Homework Equations


Nil


The Attempt at a Solution


T=mg-u ,
T= tension of the string
mg=weight of ball
u=upthrust due to the weight of ball displaced by the water
Since this system is in equilibruim, it can be taken that nothing has been added into the beaker of water. So, the reading on the balance with the suspended ball in it should be W.
But the answer given is W+U. Can anyone tell me what i have done wrong? Is there anything wrong with my assumption? Or is the answer really wrong? Help~~~~~
 
Last edited:
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write the eqn for beaker and ball as a system and note that the only force acting on them is the force of tension in string(connected to ball)and Normal(the weight measured)

Find tension using eqm of ball
 
Hi there,

You should forget about the gravity, otherwise you would have an unbalanced equation.

Cheers
 
What force does the water exert on the ball?
Now apply Newton's Third Law.
 
oh man i still don't understand...
 
davieddy said:
What force does the water exert on the ball?
Now apply Newton's Third Law.

there is upthrust. So do u mean the there is a equal but opposite force which is equal to the magnitude of upthrust acting on the ball? if it is so wouldn't the tension equals to the weight of the ball ie. T=mg?
 
Last edited:
Hi there,

No that's no correct. But since the ball suspended is montionless (I suppose), then the force have come to an equilibrium. \sum \vec{F} = 0

Then, you need to see what are the forces acting on this system.
 
fatra2 said:
Hi there,

No that's no correct. But since the ball suspended is montionless (I suppose), then the force have come to an equilibrium. \sum \vec{F} = 0

Then, you need to see what are the forces acting on this system.

I can only think of the weight of the beaker of water...
so do u mean that the balance's reading is the same regardless of whether the suspended ball is present, since the force have come to equilibrium?
 
The water exerts an upthrust on the ball.
Newton's Third Law tells us that the ball exerts a downthrust on the
water equal in magnitude.
 
  • #10
oooo..i have forgotten the downthrust
thx to all that have help me... i understand now..
 

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