How Does Placement of Lead Affect the Buoyancy of a Wood Block?

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Homework Help Overview

The discussion revolves around a physics problem involving buoyancy, specifically how the placement of lead affects the buoyancy of a wooden block. The block has a specified mass and density, and the goal is to determine the mass of lead required for the block to float with a certain volume immersed in water.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the weight of the wood block, its volume, and the buoyant force. There are attempts to calculate the volume of the wood and the corresponding buoyant force based on the volume submerged. Questions arise regarding the impact of lead placement on buoyancy and the calculations involved.

Discussion Status

Some participants have provided calculations related to the volume of the wood and the buoyant force, while others question the reasoning behind certain steps. There is an ongoing exploration of how the placement of lead affects the overall buoyancy and the resulting calculations.

Contextual Notes

Participants are working under the constraints of the problem statement, which specifies the masses and densities involved, as well as the requirement for the block to float with a certain volume submerged. There is also a discussion about the implications of lead placement on buoyancy, which remains unresolved.

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Homework Statement



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?

Homework Equations


Fb=MG
mg=.89Fb

The Attempt at a Solution


I don't really know where to begin, .89Fb=mg. but I am 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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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.
 
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?
 
tigers4 said:
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?
 
(.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?
 
tigers4 said:
(.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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