Simple Archimedes Principle

[djeitnstine]

This (textbook) question seems so simple yet I have been having the hardest time solving it :S I know there's something completely obvious I'm missing.

Homework Statement

A styrofoam slab has a thickness h and density $$\rho_{s}$$. When a swimmer of mass m is resting on it, the slab floats in fresh water with its tip at the same level as the water surface. Find the area of the slab.

Homework Equations

$$\Sigma F=F_{buoyant}-Mg=0$$
$$F_{buoyant}=Mg$$
$$F_{buoyant} = \rho_{f}ghA$$

(Archimedes Principal: Any object completely or partially submerged in a fluid experiences an upward buoyant force whose magnitude is equal to the weight of the fluid displaced by the object)

When totally submerged $$\Sigma F= (\rho_{f}-\rho_{o})V_{o}g$$

Where $$\rho_{o}$$ is the density of the object

The Attempt at a Solution

I don't even know. I wrote out a bunch of stuff. I know the density of water is $$1.00 (10^{3} \frac{kg}{m^{3}})$$. Of course I tried substituting and that gets me no where. I tried making a free body diagram and all that says is that the Buoyant force is equal to weight of the swimmer and the board (duh).

$$\Sigma F= F_{buoy} - F_{board} - F_{swimmer} = 0$$

Also a simple manipulation showed that $$(\rho_{f}-\rho_{s}) \Delta h A = m_{s}$$. I think is right?

Honestly I think some more numbers are missing :S

 

[djeitnstine]

Ok I read the solution and it was only a formula that they wanted... So I was correct.

 

[nlightner]

It seems like you have a good understanding of the concepts involved in this problem, but are struggling with the calculations. Let's break down the problem and see if we can find the missing information.

First, we know that the buoyant force (F_buoyant) is equal to the weight of the fluid displaced by the object. In this case, the object is the styrofoam slab and the fluid is the water. The weight of the water displaced by the slab is equal to the weight of the swimmer and the slab combined, so we can write the equation as F_buoyant = (m_swimmer + m_slab)g.

Next, we can use the equation for the buoyant force, F_buoyant = \rho_{f}ghA, to solve for the area of the slab. We know the density of water (\rho_{f}) is 1.00 (10^{3} \frac{kg}{m^{3}}) and the height of the slab (h) is given. We also know the mass of the swimmer (m_swimmer) and the density of the slab (\rho_{s}), but we need to find the mass of the slab (m_slab) in order to solve for the area.

To find the mass of the slab, we can use the equation you mentioned, (\rho_{f}-\rho_{s}) \Delta h A = m_{s}. This is the equation for the buoyant force when an object is completely submerged in a fluid, and it tells us that the difference between the density of the fluid and the density of the object is equal to the ratio of the mass of the object to the volume of the object. In this case, the object is the slab and the volume is the area of the slab multiplied by its thickness (A\Delta h).

So, by rearranging this equation, we can solve for the mass of the slab: m_slab = (\rho_{f}-\rho_{s})A\Delta h. Now that we have the mass of the slab, we can substitute it into our equation for the buoyant force and solve for the area (A).

I hope this helps! Remember to always carefully consider the given information and use the appropriate equations to solve the problem. Good luck!