I calculating the buoyancy force

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  • #1
Molly
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Homework Statement


A ball with a radius of 1.22 cm is dropped in a mixed drink (specific gravity = 0.910) so that it is submerged. What is the magnitude of the buoyant force acting on the ball?

Homework Equations


FB=Pfluid*g*Vsubmerged[/B]


The Attempt at a Solution


I am confused about what specific gravity means. I attempted to calculate the density of the ball by doing .910/1000 and got 910. I don't know how to continue with the problem because I don't know how to calculate density of the fluid or the volume submerged from what the problem has given.
 

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  • #2
phinds
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I am confused about what specific gravity means.
And what effort have you made to find out what it means? It IS, after all, a VERY simple, straightforward definition.
 
  • #3
RPinPA
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I am confused about what specific gravity means.
Specific gravity is its density relative to water. A specific gravity of 0.910 means the density is 0.910 times the density of water.

I attempted to calculate the density of the ball by doing .910/1000 and got 910.
910 would be a correct number if you use 1000 for the density of water. That would be correct in some units. You didn't say what the units are. I don't know about your teacher but I consider numbers without units to be incorrect.

I don't know how to calculate density of the fluid or the volume submerged from what the problem has given.
Well you just told us what the density of the fluid is, so I'm not sure what you mean by not knowing how to calculate the density of the fluid.
The volume submerged is the volume of a sphere of radius 1.22 cm. That's all the information you need to calculate the volume. Write down the formula for volume of a sphere and you'll see that it only depends on the radius, which you know.

Make sure everything is in consistent units though. If you leave the radius in cm, the volume is in ##\text{cm}^3##. Are the units of your density using ##\text{cm}^3##? What units will you use for the value of ##g##? Everything has to be in the same system of units.
 
  • #4
phinds
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Specific gravity is its density relative to water. A specific gravity of 0.910 means the density is 0.910 times the density of water.
@RPinPA I see you are relatively new to PF so perhaps you are not aware but when a poster has been asked a specific question it is bad form to answer it for him/her. This dis-incentivizes independent research.
 
  • #5
Molly
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And what effort have you made to find out what it means? It IS, after all, a VERY simple, straightforward definition.
I meant I am confused about how it correlates to the problem. I looked it up in my textbook and it says specific gravity=average density/ the density of water. So I tried to solve that for the average density and got 910. But I don't know if that works for this problem or what to do next.
 
  • #6
phinds
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I meant I am confused about how it correlates to the problem. I looked it up in my textbook and it says specific gravity=average density/ the density of water. So I tried to solve that for the average density and got 910. But I don't know if that works for this problem or what to do next.
Well, re-read RPinPA's post
 
  • #7
RPinPA
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@RPinPA I see you are relatively new to PF so perhaps you are not aware but when a poster has been asked a specific question it is bad form to answer it for him/her. This dis-incentivizes independent research.

I didn't think answering such a basic definitional question was outside of the bounds of homework help. But OK, I'll accept the reprimand.
 
  • #8
phinds
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I didn't think answering such a basic definitional question was outside of the bounds of homework help. But OK, I'll accept the reprimand.
It's NOT out of bounds at all, but my point was that I had specifically asked HIM to find out on his own.
 
  • #9
RPinPA
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I meant I am confused about how it correlates to the problem. I looked it up in my textbook and it says specific gravity=average density/ the density of water. So I tried to solve that for the average density and got 910. But I don't know if that works for this problem or what to do next.

Why don't you know if that works?
You need the density of the fluid.
You have a formula for the density of the fluid.
You used it to find the density of the fluid, which you needed. Why do you still doubt if the density of the fluid is going to help you with a formula that requires the density of the fluid?

As for what to do next, you have a formula for the buoyant force. What is your question about using that formula?
 
  • #10
RPinPA
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I think the issue here is confidence as much as anything else. You have a formula. It has three terms in it which are to be multiplied.

What does each of those terms mean?
What value will you use for that term or how will you find that term?

What is Pfluid?
What is g?
What is the submerged volume?

If you know all three of those things, then put them in.

You may have an issue as I said with units, if some things are in meters and others are in cm. You will have to be careful about that.
 
  • #11
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@Molly What research have you done on this so far? Have you ever heard of Archimides Principle? If so, please state this principle and tell us your understanding of it.
 

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