Why does dipping a stick into water on a balance cause a decrease in weight?

[zorro]

I watched this activity on T.V. and performed it:

I placed a beaker containing water on the pan of a balance and dipped a stick into the water without touching the wall or bottom of the beaker and observed that the reading of the spring balance decreased.

I could not figure out why this happened. Since the stick is held without touching any thing except water, there should not have been any change in weight. What is the reason behind this?

 

[Andy Resnick]

This is a great demonstration- FWIW, it stumped a roomful of physics professors.

Hint- what are the forces acting on the stick?

 

[zorro]

The forces acting on the stick are-
1)Buoyant force of water
2)Contact force between stick and our hand.

 

[Doc Al]

I placed a beaker containing water on the pan of a balance and dipped a stick into the water without touching the wall or bottom of the beaker and observed that the reading of the spring balance decreased.

Really?

 

[zorro]

Yes, you read it right.

 

[Doc Al]

Yes, you read it right.

So you claim that putting the stick in the beaker causes the scale to read a lower force?

 

[jgm340]

If the stick is very heavy, then another force acting on the stick is your hand. As you are keeping the stick still, this would mean the buoyant force ought to be independent of the weight of the stick. This leaves the shape/volume of the stick, and how far down in the water you stick it.

If you were to hold the stick right at the surface of the water, I'd imagine the capillary action would decrease the reading on the balance, since the force of buoyancy would be 0. If it breaks the surface, however, the force of buoyancy would surely counteract the capillary force pulling the water up the stick, right?

The balance ought to read higher, not lower...

EDIT: In other words, instead of a stick and beaker of water, imagine you use a balloon filled with air and a beaker filled with a very, very, very dense liquid instead. In order to put the balloon in the water, you actually have to PUSH it down (in order to displace the liquid). After that, you can think of it as a closed system, right? Then the sum of the forces pointing downwards (your hand + gravity) must equal the sum of the forces pointing upwards (the scale).

 

[zorro]

Yes, you are right. Even I was surprised with such an observation.
May be the balance in my lab is faulty

 

[JDługosz]

EDIT: In other words, instead of a stick and beaker of water, imagine you use a balloon filled with air and a beaker filled with a very, very, very dense liquid instead. In order to put the balloon in the water, you actually have to PUSH it down (in order to displace the liquid).

So, in the case where the stick is not buoyant, the opposite occurs?

Hmm, Consider holding a stick, you are lifting a certain weight. If the stick is partly submerged, it will feel lighter since, although it doesn't float, there is some buoyant force. If you are applying less force to lift the stick, what's holding it up? It shows up on the scale.

 

[rcgldr]

Regardless of the density of the stick, if it displaces water, then the water level rises, pressure at the bottom increases, and reading on the scale increases.

If the stick absorbs more water than it displaces, then the water level lowers, pressure at the bottom decreases, and the reading on the scale decreases.

 

[zorro]

Good explanations!