How Does Submerging a Sphere Affect Scale Readings?

[jaded18]

http://session.masteringphysics.com/problemAsset/1011221/19/SFL_ap_6a.jpg

A cylindrical beaker of height 0.100 {m} and negligible weight is filled to the brim with a fluid of density rho = 890 {kg/m}^3. When the beaker is placed on a scale, its weight is measured to be 1.00 {N}. View Figure

A ball of density rho_b = 5000 {kg/m}^3 and volume V = 60.0 {cm}^3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81

------> What is the reading W_2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale.
________________________________________________________________
I know that the weight of the ball is 2.94 N, and that the beaker is 1 N... And that pressure= F/A and pressure = p_0 + density of liquid(gh) ... Help??

 

[Doc Al]

How does the water pressure change when the ball is submerged?

Another way to analyze this setup is to examine all the forces acting on the beaker and its contents. (Start by analyzing the forces on the ball. What must they add to?)

 

[jaded18]

I don't understand ... Pressure force on the bottom is greater in order to balance gravity...

 

[learningphysics]

What is the pressure at the bottom of the beaker before the ball is submerged?

What is the pressure at the bottom after the ball is submerged?

 

[jaded18]

Ok, so pressure at the bottom of the beaker before the ball is submerged ... well P=F/A and A of the beaker I don't know, but I know that the weight of the beaker+fluid = 1N. A=density(V) = 890(V) ... I'm stuck

 

[jaded18]

after the ball is submerged, P=F/A and A of the ball is just (5000(60/(100^3)))=0.3m^2 ?? and force is just the weight which equals 2.94? so pressure is then 98 pa?? That sounds .. wrong

 

[learningphysics]

Ok, so pressure at the bottom of the beaker before the ball is submerged ... well P=F/A and A of the beaker I don't know, but I know that the weight of the beaker+fluid = 1N. A=density(V) = 890(V) ... I'm stuck

Another way of putting it... does the pressure change at the bottom of the beaker, from before and after?

The force on the bottom of the beaker is the pressure due to the water*area...

The normal force upward on the beaker = force due to pressure + weight of beaker...

this normal force is what the scale measures...

how do either of these 2 things change... "force due to pressure" or "weight of the beaker"

do they change at all?

 

[jaded18]

Ah, ok, no, replacing the fluid volume with the object does not change the fluid ... so pressure force stays the same... By the way, how do you calculate the pressure?

 

[learningphysics]

Ah, ok, no, replacing the fluid volume with the object does not change the fluid ... so pressure force stays the same...

exactly. everything's the same. it's a trick question.

 

[learningphysics]

Ah, ok, no, replacing the fluid volume with the object does not change the fluid ... so pressure force stays the same... By the way, how do you calculate the pressure?

you don't need it... you know the pressure force is the same... the weight of the beaker is the same... so the scale measures the same force... 1N.

 

[jaded18]

I just looked it up and the answer key said 873 pa and i was wondering where the heck that came from. But what you're saying makes sense...

 

[learningphysics]

I just looked it up and the answer key said 873 pa and i was wondering where the heck that came from. But what you're saying makes sense...

yeah, that's just rho*g*h = 890*9.81*0.1 = 873 pa

 

[jaded18]

question: when is this p_0 101.3 kpa and when is it 0 ...

 

[learningphysics]

question: when is this p_0 101.3 kpa and when is it 0 ...

it depends on what exactly the question was asking for... did the question ask for the pressure at the bottom of the beaker, from inside...

was it asking for the difference in pressure between inside and outside?

 

[jaded18]

Well when you were calculating the pressure at the bottom of the beaker before the ball was submerged and after the ball was submerged, you used p_0 = 0 in the equation p=p_0+rho(g)(h) for both cases

 

[learningphysics]

Well when you were calculating the pressure at the bottom of the beaker before the ball was submerged and after the ball was submerged, you used p_0 = 0 in the equation p=p_0+rho(g)(h) for both cases

yeah... I should use p_0 = 101kpa both times. we also need to remember that there's a pressure outside the beaker acting upwards on the bottom of the beaker of 101kpa.

The 101kpa inside and outisde cancel... so that's why I didn't include it...

When you consider the force on the beaker...

Fnormal - (pressure inside)*area + (pressure outside)*area - (weight of beaker) = 0.

Fnormal - (101kpa + rho*g*h)*area + (101kpa)*area - weight of beaker = 0

see now that the 2 101kpa's cancel

Fnormal - rho*g*h*area - (weight of beaker) = 0

giving the same equation as before...

 

[get_physical]

I am confused about this question. So what is the answer?
Do I use the formula, W= rho * V* g?
I know everything except for V... how can I find V when i only have the height?

Thank you

 

[Doc Al]

So what is the answer?

What do you think the answer is?

Do I use the formula, W= rho * V* g?
I know everything except for V... how can I find V when i only have the height?

You don't need that to answer the question.

 

[get_physical]

ohh the answer is 2.94N because nothing changes.

 

[get_physical]

nevermind. that is wrong. i just tried it.

hmm looks like I'm really not understanding buoyancy! =(

 

[get_physical]

anybody able to explain this again? :shy:

 

[Doc Al]

If you can explain just how you are trying to solve the problem, perhaps we can point out where you are going wrong. Have you read the earlier posts in this thread? They contain hints as to how to solve this.

 

[get_physical]

I'm trying to find the force of tension on the ball by the rod.

I know that...
Ft = Fb - mg
and Fb= density of fluid*V*g
V would be the volume of the ball (6x10^-5)m

I get an answer of 2.29.
But its not correct.
Somebody tell me where I'm going wrong?

 

[get_physical]

Here is the information given from the question. Also the same as the first post.

A cylindrical beaker of height 0.100m and negligible weight is filled to the brim with a fluid of density rho = 890kg/m^3. When the beaker is placed on a scale, its weight is measured to be 1.00N.

A ball of density rho_b = 5000kg/m^3 and volume V = 60.0cm^3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81

 

[Doc Al]

I'm trying to find the force of tension on the ball by the rod.

I know that...
Ft = Fb - mg
and Fb= density of fluid*V*g
V would be the volume of the ball (6x10^-5)m

That's fine. (Except that Ft = mg - Fb.)

I get an answer of 2.29.
But its not correct.
Somebody tell me where I'm going wrong?

Show exactly what you plugged in and maybe we can spot your error.

 

[Vanessa23]

It seems as though the answer to "what is the weight of the ball" should be [m(ball)*g-rho_f*V*g]. Why is that not right?

 

[Doc Al]

It seems as though the answer to "what is the weight of the ball" should be [m(ball)*g-rho_f*V*g]. Why is that not right?

Please state the exact question you are trying to answer.

 

[Vanessa23]

In answering the original question:
A cylindrical beaker of height 0.100 {m} and negligible weight is filled to the brim with a fluid of density rho = 890 {kg/m}^3. When the beaker is placed on a scale, its weight is measured to be 1.00 {N}.

A ball of density rho_b = 5000 {kg/m}^3 and volume V = 60.0 {cm}^3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81

------> What is the reading W_2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale.

It seems as though the answer should be:
W_2=m(ball)*g - rho_f*V*g
Why is that not right?

 

[Doc Al]

In answering the original question:
A cylindrical beaker of height 0.100 {m} and negligible weight is filled to the brim with a fluid of density rho = 890 {kg/m}^3. When the beaker is placed on a scale, its weight is measured to be 1.00 {N}.

A ball of density rho_b = 5000 {kg/m}^3 and volume V = 60.0 {cm}^3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81

------> What is the reading W_2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale.

OK.

It seems as though the answer should be:
W_2=m(ball)*g - rho_f*V*g
Why is that not right?

That would be the increase in scale reading if the ball were just dropped in and sunk to the bottom. But it's being held in place by a rod.

 

[Vanessa23]

I misread the question, sorry. 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

 

[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!