Fluid Dynamics Buoynacy Question

Question

I am actually studying maths but have to do a fluid dynamics module. I'm revising for my exams but can't get my head around this question even though I know it should be straight forward.

The densities of wood, copper and water are:

$\rho_{\text{wood}} = 600 \ \mathrm{kg/m^3}$

$\rho_{\text{copper}} = 8900\ \mathrm{kg/m^3}$

$\rho_{\text{water}} = 1000 \ \mathrm{kg/m^3}$

what mass of copper in kg must be attached to a piece of wood of mass $0.5$ kg so that the copper and wood, together, will just submerge under water?

I have the solution which is apparently $0.375$ kg. But I have no idea how we get to this.

Answer 1

Total mass $$m_t = m_1+ m_2 = 0.5 \text{kg} + m_2$$

Total volume $$v_t = v_1 + v_2 = 0.5 \text{kg}/ 600 \text {kg}/\text{m}^3 + m_2 / 8900 \text{kg}/\text{m}^3$$

Now, what's the expression for total density, and what does it need to equal?

Answer 2

Just solve $1000 = m_{t}/v_{t}=\rho_{t}$ and find $m_{2}$. This is saying total density of copper and wood must be equal to total density of water.