How Does Submerging a Sphere Affect Scale Readings?

[Doc Al]

I understand that the scale will not change its reading, so W_2= 1N, but how would you find the weight of the ball? I thought it would be:
m(ball)*g - rho_f*V*g

That would be the apparent weight of the submerged ball. Meaning: If you hung the ball from a scale, then dunked it into the fluid, that's what the scale would read.

 

[Vanessa23]

So the question asks "What is the weight W_b of the ball?" and 2.94-(890*0.00006*9.81)=2.42 is incorrect, then they are simply asking for the weight=mg?

 

[Doc Al]

So the question asks "What is the weight W_b of the ball?" and 2.94-(890*0.00006*9.81)=2.42 is incorrect, then they are simply asking for the weight=mg?

That's what I would assume.

 

[Vanessa23]

Apparently I am making the whole problem more difficult than it actually is. I want to make sure that I understand everything about this kind of situation. I think that because the string is rigid it is throwing me off. Could you let me know if I have the concept right?
Normally buoyancy force comes into play when a mass is suspended by a spring into the fluid. Here I believe that the ball is held in place and the scale does not change, so there is no buoyancy force.
The only thing keeping the ball where it is in the fluid is the rigid rod, so the force applied by the rod is equal and opposite to the force applied by the ball since there is no buoyancy force. Therefore, the force applied by the rod would just be negative the weight.
So basically in this whole situation you treat the beaker of fluid as one system and the ball+rod as a separate system because inserting the ball has no affect on either system. By that I mean that the scale doesn't change and the fluid does not affect the ball and rod.
It just seems odd that buoyancy wouldn't factor in...

 

[Doc Al]

Normally buoyancy force comes into play when a mass is suspended by a spring into the fluid. Here I believe that the ball is held in place and the scale does not change, so there is no buoyancy force.

Not so. There is definitely a buoyant force acting on the submerged ball, as usual. It equals rho_f*V*g, just like you would expect.

The question is whether the scale reading changes. But to answer that, you must take everything into account.

The only thing keeping the ball where it is in the fluid is the rigid rod, so the force applied by the rod is equal and opposite to the force applied by the ball since there is no buoyancy force. Therefore, the force applied by the rod would just be negative the weight.

No. The forces on the submerged ball are:
(1) Its weight, mg down.
(2) The buoyant force, rho_f*V*g up.
(3) The force from the rod, which must be mg - rho_f*V*g up.

These three forces must add to zero, since the ball is in equilibrium.

So basically in this whole situation you treat the beaker of fluid as one system and the ball+rod as a separate system because inserting the ball has no affect on either system. By that I mean that the scale doesn't change and the fluid does not affect the ball and rod.
It just seems odd that buoyancy wouldn't factor in...

There are two ways to solve this problem. An easy way (if you spot it) and a "hard" way. Let's do the hard way first.

We start with the original weight of the beaker of fluid: W
Add the weight of the ball: mg
Subtract the weight of the fluid that over flowed: -rho_f*V*g
Subtract the upward force exerted by the rod on the system: -(mg - rho_f*V*g)

Add them all up and you get W (thus no change in the scale weight), an interesting result.

The easy way is just to realize that the fluid pressure at the bottom of the beaker doesn't change when you add the ball: The fluid height remains the same. The amount of fluid lost just exactly balances the added force of the ball on the fluid (the buoyant force). Interesting!

 

[mochigirl]

Well, in that case the W of the contents would be 1.0N as before right? But that's not the correct answer.

 

[Doc Al]

Well, in that case the W of the contents would be 1.0N as before right?

No, the weight of the contents is greater, since the ball is more dense than the fluid it displaced. But the scale reading is the same, which is what the question asks about. (Realize that the ball is being partly supported by the rod.)

 

[mochigirl]

sorry. that is what I meant. the weight that the scale displays should be 1 N. but the answer says that it is not. So I'm still a bit confused. Because the hint says:

"It makes no difference how the ball and fluid are arranged inside the beaker. If there are no external forces, the weight reading would be equal to the total weight of the beaker and its contents."

So in that case, the weight of the ball should be counterbalanced by the tension and buoyancy forces right? And then I thought that since they had asked that I calculate the weight of the displaced water=0.524N then, the new weight displays would be 0.476. But that is wrong too...what am I doing wrong?

 

[Doc Al]

sorry. that is what I meant. the weight that the scale displays should be 1 N. but the answer says that it is not. So I'm still a bit confused.

Try 1.00 N.

Because the hint says:

"It makes no difference how the ball and fluid are arranged inside the beaker. If there are no external forces, the weight reading would be equal to the total weight of the beaker and its contents."

But there are external forces: The tension in the rod pulls up on the ball.

So in that case, the weight of the ball should be counterbalanced by the tension and buoyancy forces right?

True, but having the ball in the fluid changes things. After all, if you then removed the ball, the fluid level will lower.

Realize that whatever force the fluid exerts upward on the ball (the buoyant force), the ball exerts downward on the fluid (Newton's 3rd law). Thus the ball exerts a downward force on the fluid, exactly equal to the weight of the displaced fluid, thus compensating for the spilled fluid.

And then I thought that since they had asked that I calculate the weight of the displaced water=0.524N then, the new weight displays would be 0.476. But that is wrong too...what am I doing wrong?

That "hint" seems a bit misleading, since the scale reading is not simply the weight of the contents.