Answer to Question #133574 in Physics for rose auburn

Question

A brick rests on a large piece of wood floating in a bucket of water. The brick slides off and sinks. Does the water level in the bucket go up, go down, or stay the same? Explain.

Answer 1

Assume that the volume of the brick is V, its density is $\rho_b$.

Initially, the brick was sitting on the wood with density and volume $\rho_w, V_w$, and they were floating together. Assume that initially, the brick was not even touching water. $V_s$ below is the volume of wood submerged under water. Since they are floating, the force of gravity is in balance with buoyancy force:

$$g\rho_bV+g\rho_wV_w=g\rho V_s.$$

The volume of wood submerged below the water surface is

$$V_s=\frac{\rho_bV}{\rho}+\frac{\rho_wV_w}{\rho}.$$

Initially, what makes water in bucket rise (compared to bucket with only water) was V_s only.

Finally, when the brick falls, the wood is submerged for

$$V_{s2}=\frac{\rho_wV_w}{\rho}.$$

Finally, what makes water in bucket rise is

$$V_{s2}+V=\frac{\rho_wV_w}{\rho}+V.$$

Sine density of brick is higher than density of water, in the equation for V_s above we have a coefficient $\rho_b/\rho>1$ before brick volume.

In the last equation for V_s2, we see that the coefficient for V is 1. That is why the water level in the bucket will go down:

$$V_s>V_{s2}+V.$$