Iron ball in water, reading on weighing machine

[unscientific]

What is the reading on the scale for the first one?? My friend mentioned that upthrust need not be considered as its an 'internal force', i wonder why??

Shouldnt' the net force dowards be (Weight of stone - pVg) + (Weight of water + pVg) as the water experiences an equal an opposite downthrust (exerted by the stone on water) due to it exerting upthrust on stone?

For the 2nd case there is a string,

the net force downwards be (Weight of stone - pVg - T) + (Weight of water + pVg), as the water experiences an equal an opposite downthrust (exerted by the stone on water) due to it exerting upthrust on stone?
Furthermore, since the ball is in static and dynamic equilibrium, T + pVg = mg, therefore (Weight of stone - pVg - T) = 0
So the reading no the scale would be = Weight of water + pVg(further workings)
The difference in scale readings would be (Weight of water + weight of iron ball) - (Weight of water + pVg) = Weight of iron ball - pVg = T

So the difference is just tension T.

 

[Doc Al]

What is the reading on the scale for the first one?? My friend mentioned that upthrust need not be considered as its an 'internal force', i wonder why??

Shouldnt' the net force dowards be (Weight of stone - pVg) + (Weight of water + pVg) as the water experiences an equal an opposite downthrust (exerted by the stone on water) due to it exerting upthrust on stone?

Note that what you call the "upthrust" is exactly canceled by the "downthrust"--this is no coincidence. These are the buoyant force between water and stone, thus they are internal forces to the system of "beaker + water + stone" and can be ignored. The weight measured by the scale is the just the weight of the contents, as one might expect. (But your analysis is perfectly correct.)

For the 2nd case there is a string,

the net force downwards be (Weight of stone - pVg - T) + (Weight of water + pVg), as the water experiences an equal an opposite downthrust (exerted by the stone on water) due to it exerting upthrust on stone?
Furthermore, since the ball is in static and dynamic equilibrium, T + pVg = mg, therefore (Weight of stone - pVg - T) = 0
So the reading no the scale would be = Weight of water + pVg


(further workings)
The difference in scale readings would be (Weight of water + weight of iron ball) - (Weight of water + pVg) = Weight of iron ball - pVg = T

So the difference is just tension T.

All good.

 

[unscientific]

Note that what you call the "upthrust" is exactly canceled by the "downthrust"--this is no coincidence. These are the buoyant force between water and stone, thus they are internal forces to the system of "beaker + water + stone" and can be ignored. The weight measured by the scale is the just the weight of the contents, as one might expect. (But your analysis is perfectly correct.)


All good.

YAY THANKS! :D (that made my day)

But I still have the problem on differentiating internal and external forces. Any way to go about doing that??

 

[Doc Al]

But I still have the problem on differentiating internal and external forces. Any way to go about doing that??

If a force is between two parts of the same system, then it is an internal force. In the example with the water and stone, the buoyant force is between water and stone--both part of the same system of stuff in the beaker--so it's an internal force. The string tension is a force between the string and the stone; since the string is not part of the system, the string force is an external force.