Aluminium weight buoyancy problem

[Aleisha]

An aluminium weight (B) is placed on top of a 0.472 kg wooden block (A) that is floating in water as shown below. The wooden block has a density of 533 kg.m-3 (3 s.f.).
Calculate the required mass of the Aluminium weight to just fully submerge the wooden block as shown in the diagram.
The density of water is 1.00 x 103 kg.m-3.
Given:
m(Wood block)=0.472kg
p (Density wood block)= 533 kg.m-3
p (Density water) = 1000 kg.m-3
Unknown:
m (Aluminium block weight)=?
Attempt:
V (Wood block)= FB/ p(object) x g FB= 0.472 x 9.80 = 4.6256
4.6256/(5.33 x 9.80)
= 8.86e-4 m3
m (Aluminum) = p(Water) x V (wood block)
= 1000 x 8.86e-4
= 0.886kg

I just need verification if I've done this problem correctly and if not could someone please explain where I've gone wrong? Thank you.

 

[kuruman]

There is no picture "shown below" or a diagram. Is the aluminum attached to the top or to the bottom of the wooden block?

On Edit: Oops, the problem mentions that the block is on top of the wood.

m (Aluminum) = p(Water) x V (wood block)

This equation is valid only if the wood is massless which it cannot be. The right side of the equation is the mass of the displaced water which must be equal to the mass of the aluminum plus the mass of the wood.

 

[Aleisha]

won't the water displaced just be the mass of the wood? I am still a bit confused?