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Buoyancy and floating

by tigers4
Tags: buoyancy, floating
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tigers4
#1
Mar31-11, 12:54 PM
P: 24
1. The problem statement, all variables and given/known data

A block of wood has a mass of 3.84 kg and a density of 598 kg/m3. It is to be loaded with lead so that it will float in water with 0.89 of its volume immersed. The density of lead is 1.13 104 kg/m3.

a)What mass of lead is needed if the lead is on top of the wood?

b)What mass of lead is needed if the lead is attached below the wood?

2. Relevant equations
Fb=MG
mg=.89Fb



3. The attempt at a solution
I dont really know where to begin, .89Fb=mg. but Im confused on how to use density of the wood or the lead to help solve the equation? Any ideas on how to start off the problem?
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gneill
#2
Mar31-11, 03:10 PM
Mentor
P: 11,625
You know that the goal is to have 0.89 of the volume of wood underwater. Why not start by figuring out what the weight of the block of wood is, its volume, and how much of that volume will be below water. From there you should be able to tell what the buoyant force is (due to displaced water). Compare with the weight of the block.
tigers4
#3
Apr1-11, 11:19 AM
P: 24
d=m/v, so v=m/d.
3.84kg/598kg/m^3=.00642m^3=Vwood
.89v is under water, so the volume of displaced water=.89Vwood.
(density of displaced fluid)(V of displaced fluid)g=buoyancy force
.89*buoyancy force=(Mwood+Mlead)g?

gneill
#4
Apr1-11, 11:35 AM
Mentor
P: 11,625
Buoyancy and floating

Quote Quote by tigers4 View Post
d=m/v, so v=m/d.
3.84kg/598kg/m^3=.00642m^3=Vwood
.89v is under water, so the volume of displaced water=.89Vwood.
(density of displaced fluid)(V of displaced fluid)g=buoyancy force
.89*buoyancy force=(Mwood+Mlead)g?
...and you were doing so well...

In your final line, why are you multiplying the buoyancy force by 0.89? You've already calculated the buoyancy given that 0.89 of the wood's volume is submersed.

Time to put some numbers to the calculations.

What is the weight of the block (Newtons)?
What is the force due to buoyancy (Newtons)?

Is there a difference? If so, what is it and what's to be done about it?
tigers4
#5
Apr1-11, 12:25 PM
P: 24
(.89*.00642m^3)*(1000kg/m^3)=(3.84+m)
m=1.874kg

Why would it be different if the lead was on the bottom of the wood?
gneill
#6
Apr1-11, 12:35 PM
Mentor
P: 11,625
Quote Quote by tigers4 View Post
(.89*.00642m^3)*(1000kg/m^3)=(3.84+m)
m=1.874kg

Why would it be different if the lead was on the bottom of the wood?
Because when it's below the wood the lead will also displace some water -- essentially it weighs less underwater.


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