How to Calculate Foam Volume for Buoyancy Support of Cargo Block

[ACE_99]

Homework Statement

Determine the volume of foam needed to support one cargo block with the following conditions. The bottom of the block is 2 cm above the water line and when viewed from the top there is 1 cm of foam around all edges of the cargo block.

The dimension of the block are 3.5 cm high x 3.5 cm x 7.5 cm. It weighs 70 g.

The Attempt at a Solution

I realize this is probably a very simple problem but I just can't manage to wrap my head around it. I know that in order for the block to float the buoyant force exerted by the water must be equal to the weight of water displaced by the block but other than that I'm completely stumped any help would be great.

 

[LowlyPion]

Aren't you going to need the density of the foam?

If it is going to be doing the displacement of the water you will need to know its weight as well as the weight of the block that is needed.

You know at least that the area of the foam needs to be (3.5 + 2)cm wide and (7.5 + 2)cm long, so all you need is the depth.

 

[ACE_99]

Aren't you going to need the density of the foam?

If it is going to be doing the displacement of the water you will need to know its weight as well as the weight of the block that is needed.

This question is part of a prelab and the density of the foam wasn't given. The only other piece of information is that the foam block provided for the experiment is 60 cm x 30 cm x 2.5 cm and it weighs 120g.

 

[LowlyPion]

What do you mean they didn't give you the density?

They gave you the dimensions and the weight.

So how much water needs to be displaced to support the 70g block and the 5.5x9.5x2 cm of foam block above the water line? What is the net buoyant force per volume you have available and then translate that into the depth that you need below the waterline.

 

[ACE_99]

Can't believe that I completely missed that. Ok so given the volume and mass I found the density of the foam to be 26.6 kg/m^3.

So the buoyancy force would be

F = g*((($$\rho$$foam))(V) + 70g)
= 0.71 N

so knowing that the buoyant force = $$\rho$$gV I would then use the density of foam and the force I found to calculate the volume.

F = $$\rho$$gV
0.71= (26.6)(9.81)(V) --> solve for V

V = 0.0027 m^3

Does that sound right?

 

[LowlyPion]

I'd just work in grams / cc since those are what the dimensions are given in. If your answer needs to be in m³ then convert at the end.
(Note they gave you the weight in grams so I wouldn't use g to figure it as it cancels out everywhere.)

First calculate the foam density:
(60*30**2.5) cc = 120 g

ρ = .026667 g/cc

Then above the waterline:

70 g + 9.5*5.5*2*(.026667) = 72.787g

That just leaves the figuring of the below the waterline volume which we know is 9.5*5.5*X

72.787 + (.026667*52.25*X) = 52.25*X*(1)

So just solve for X.

X = 72.787/(52.25*.97333) = 1.43

I make it then the volume of foam needed including the 2 cm superstructure at (2+1.43)*9.5*5.5 = 179.22 cc of foam

 

[ACE_99]

Thanks LowlyPion, that helps a lot.