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UghboyancyI don't get it.

  • Thread starter MD2000
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  • #1
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1. A 210 N object floats with three-fourths of its volume beneath the surface of the water. What is the buoyant force on the object? Newtons

Is this a trick question? I thought that in case like this.. Fb - W = 0..thus Fb = W..so shouldn't the Fb simply equal 210?

2. A lead block is suspended from the underside of a 0.2 kg block of wood of density of 716 kg/m3. If the upper surface of the wood is just level with the water, what is the mass of the lead block?(The density of lead is 11340 kg/m3)

In this case I know you can figure out the V of the wood block..would you once again use a buoyancy equation?

FB - W1 - W2 = 0..? Any help on where I should start
 

Answers and Replies

  • #2
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You should go back and read what buoyancy is all about. I don't think you understand it quite yet.

For number 1, what is the definition of the buoyancy force?
 
  • #3
nrqed
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cyrusabdollahi said:
You should go back and read what buoyancy is all about. I don't think you understand it quite yet.

For number 1, what is the definition of the buoyancy force?
Err...maybe I am missing something but it looks to me as if MD2000 is completely right about that question. The buoyancy force is simply mg (and the information about the volume is irrelevant which, in that sense,makes it a "trick" question).

Am I missing something?
 
  • #4
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[tex] F_b = \gamma V [/tex]

where [tex]\gamma [/tex] is the specific weight of the fluid and V is the volume of displaced fluid.

Work from there.

Edit, I see what you're saying now. You are correct, I glanced over the problem too quickly. But the force is a function of the volume.

So you can conclude that the density of the rock is 3/4 that of the water.
 
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  • #5
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cyrusabdollahi said:
[tex] F_b = \gamma V [/tex]

where [tex]\gamma [/tex] is the specific weight of the fluid and V is the volume of displaced fluid.

Work from there.

Edit, I see what you're saying now. You are correct. But the force is a function of the volume.
I agree with all that but of they are asking the buoyant force and the object is floating and they give the weight of the object to be 210 N then there is no need to even know the fraction of the volume beneath the surface of the specific weight of the liquid! The answer is simply 210 N, period.

That seemed the reasoning of the OP and it seems completely right to me.

regards

Patrick
 
  • #6
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Yes, you both are correct. My apologies!! :tongue2:
 
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  • #7
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lol. ok. whew. i thought maybe i was just crazy. hehe..

what about the second one..am i on the right track as far as the FB - W1 - W2 = 0 equation?
 
  • #8
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MD2000 said:
lol. ok. whew. i thought maybe i was just crazy. hehe..

what about the second one..am i on the right track as far as the FB - W1 - W2 = 0 equation?
Yes, as long as you mean by F_B the total buoyancy force on both blocks.
So that gives
[tex] F_{B, lead} + F_{B, wood} - m_L g - m_W g =0 [/tex]

Now, using [itex] F_B = \rho_{water} V g [/itex], it should be easy to solve. (notice that you can replace the volume by mass over density, which is a useful thing to do in this problem...it will allow you to solve in one step).

Patrick
 
  • #9
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okay..thats what i figured..the only thing that confused me was whether i should use the volume for the respective blocks? for example..the v of the wood would is .2/716..and for the L is m/11340?
 
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  • #10
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MD2000 said:
okay..thats what i figured..the only thing that confused me was whether i should use the volume for the respective blocks? for example..the v of the wood would is .2/716..and for the L is m/11340?
Yes, exactly. Just use that for the two volumes and the equation I wrote earlier will now contain only one unknown: the mass of the lead block which you can now solve for.

Patrick
 
  • #11
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Thanks Pat..I got .0866 kg..looks reasonable
 
  • #12
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MD2000 said:
Thanks Pat..I got .0866 kg..looks reasonable
that seems right, Glad I could help!
 

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