# Suppose there is a weighing balance.In one hand of weighing balance there is weight of 1 KG which is perfectly balancing flask full of water which is on the other hand. Now if i immerse finger in water, will the equillibrium be disturbed?

Stercus

• Member Since Dec-2007 | Aug 17th, 2009

Akash, your FBD won’t work if you leave off external contacts – they need to be converted to force arrows. The buoyed object provides one such force: the buoyant reaction force.

You seem to agree that there is a force acting upwards on the test tube, appearing on the spring balance acting against the weight. As the test tube and the bowl of water are stationary, not accelerating, there must be an equivalent reaction force acting downwards on the body of water, imparted by the object. To go into more detail, the surface of the object experiences a pressure distribution from the water, gradated downwards due to increasing pressure within the water body. The net result from the pressure distribution is that the water pushes up on the object (buoyant force). To avoid being crushed, the object pushes back against this pressure in all directions – so again you get net a reaction force in the opposite direction, downwards.

So the FBD of the water body needs a reaction force to account for buoyancy. This reaction force acts downwards, from the centre of the displaced volume and is equal to the weight of the water displaced. The weight of the object immersed doesn’t need to be considered as this is supported but the weight of the displaced volume of water does.

#### Stercus Profile Answers by Stercus

• Jan 2nd, 2008

Yes - by displacing water you're getting a buoyant force (upwards) on your finger, equivalent to the weight (as a force, mind you - not the mass) of the water displaced by your finger. If Newton's third law is still holding true, the reaction force (downwards) on the flask of water will upset the balance of the scales, even though you may be holding your finger still.

This might be read as a trick question - if you've got a 1kg weight in one hand and the water in the other, how are you going to get your finger free to dip it into the flask?! :)

#### mailmisterk Profile Answers by mailmisterk

• Mar 30th, 2009

I think it doesn't get disturbed. because according to archimedes principle, the mass or weight of the object immersed is equal to the volume of the liquid displaced

#### Stercus Profile Answers by Stercus

• Mar 30th, 2009

Fair enough - it's good to question what doesn't seem right. Give it a go in a real experiment.

Find an accurate scale, put a glass of water (for the sake of scale,  use a regular glass, the water volume be ~200mL), zero the scale with the glass of water on it and immerse your finger, as far as you can go without touching the glass. You should see the reading rise by a few grams.

This is Archimedes' buoyant force - the force is equal to the weight (ie. mass x gravitational acceleration) of the volume of water displaced. (Note: this isn't the same as the weight of your finger)

If this doesn't work, or you haven't got access to scales that can be zeroed, find a set of regular bathroom scales, a bucket of water and an inflated balloon. Immerse the balloon in the bucket and you should see a result - it's the same principle at work.

#### akash.2929 Profile Answers by akash.2929

• Aug 7th, 2009

NO! No way is it going to disturb. Forget the water and Archimedes principle stuff. Consider a bowl of water as a simple weight which is equal to 1 kg. Now simply dipping your finger into the bowl wont effect anything, it would be the same as just touching a 1kg substance. You are NOT exerting any force on the balance, the weight of your finger is supported by your body. However if you put your finger-ring inside the bowl, there will be a disturbance as the finger-ring would exert a force on a balance :)

#### Stercus Profile Answers by Stercus

• Aug 9th, 2009

Interesting assumption Akash - forget Archimedes' Principle because... well, just because.  Maybe it's all a bit too tricky to imagine?

Hows this then - a standard test tube, about the same size and shape as your finger, right? It's hollow, it weighs less and can be easily rigged up to a Newtonian spring scale or a second balance - or the thing will probably just happily float of it's own accord (wondering why it floats? :D ).  Any which way you will observe a force acting to support the tube... it'll be small but it will be there.

You don't notice the weight of your own finger, because your body is quite used to supporting it, has been for a number of years if you can type. You do, however need to get of the computer and try some actual experimentation at times to get a feel of how things work (rather than a think of how things work).

Anyhow, please don't forget Archimedes. He's never done anything to offend you, and the shipping industry will be very upset if his principle gets forgotten.

You're half right about the ring part, though. If the ring was to drop and fall to the bottom of the bowl, it's own weight would add to the scale, but that force would be reduced by the buoyant force imparted by the volume of water the ring had already displaced.

#### akash.2929 Profile Answers by akash.2929

• Aug 10th, 2009

The buoyantic force has nothing to do here. Everyone knows when this buoyantic force acts and how.
The thing is that here buoyantic acts or not is not our concern. See, the force of the whole water bowl is acting on the weight pan, so buoyantic force in
the bowl is an INTERNAL force within the bowl and wont affect the weight in pan and similarly if
you take a test tube and allow it to float on water you will observe a change in weight of the pan.
However if you HANG the tube to a spring balance, you will see that the test tube's weight
reduces but that reading will be of the SPRING BALANCE which directly holds the test tube.
Here a buoyantic force comes into play as it is acting on the test tube, but
that reading will be of the SPRING BALANCE which directly holds the test tube.
The pan would NOT show any change because the test tube's weight is not acting on the pan.
So internal forces do not affect a different system. Just draw free body diagrams and
see what are the forces acting ON THE PAN if you insert a finger or say HANG a tube in the water.
Hope you get my point.

#### Stercus Profile Answers by Stercus

• Aug 17th, 2009

Akash, your FBD won’t work if you leave off external contacts – they need to be converted to force arrows. The buoyed object provides one such force: the buoyant reaction force.

You seem to agree that there is a force acting upwards on the test tube, appearing on the spring balance acting against the weight. As the test tube and the bowl of water are stationary, not accelerating, there must be an equivalent reaction force acting downwards on the body of water, imparted by the object. To go into more detail, the surface of the object experiences a pressure distribution from the water, gradated downwards due to increasing pressure within the water body. The net result from the pressure distribution is that the water pushes up on the object (buoyant force). To avoid being crushed, the object pushes back against this pressure in all directions – so again you get net a reaction force in the opposite direction, downwards.

So the FBD of the water body needs a reaction force to account for buoyancy. This reaction force acts downwards, from the centre of the displaced volume and is equal to the weight of the water displaced. The weight of the object immersed doesn’t need to be considered as this is supported but the weight of the displaced volume of water does.

#### akash.2929 Profile Answers by akash.2929

• Aug 18th, 2009

stercus, u r correct!! :) thnx :)

#### PRASHANT

• Oct 16th, 2011

It will not change the reading.....

Lets take a simple case... A water flask is putted on a weight balance and a rod is immersed in it which is tied to ceiling with help of a rope. The tension in the rope will decrease but there is no effect in the reading of the weight machine reading.. only the center of gravity will rise slightly....

Or take more simpler case.... A flask has diameter 20 mm has filled with water... and there is a another flask with 10mm diameter but there is same volume of water filled in both then weight of both will be same.... It is compared to this case because if we place a rod of 10mm diameter (attached to ceiling with rope) in the 20mm flask it will result only the case of of 10mm diameter flask....