Buoyant force relates to the amount of force needed to displace some fluid like a gas or a liquid, by submerging it. It follows this equation:
Fb = rhoVg, where Fb is buoyant force, rho is the density of the fluid, V is the volume of the fluid displaced by the object, and g is the acceleration due to gravity, 9.8m/s^2.
***As a quick aside, I'm showing how this equation works. You can skip ahead if you're already familiar.
When you submerge an object in a fluid, you are pushing some of that fluid up against gravity. Recall that F = ma, or force = mass*acceleration. We have the acceleration, g. And volume and density (mass/volume) multiply to just equal mass. So really what this equation is asking us for is how much force is needed to lift the water moved by the object.
**** Aside done
The problem says the log is submerged, meaning that it's entire volume is being used to displace the water. So V = Volume of the log. = 11.0 cubic feet
The density of water, rho, is 1g/cm^3
Acceleration from gravity, g, is 9.8m/s^2
We can almost plug everything into the equation. But first we need some unit conversions so we can get the answer in pounds. I like to use dimensional analysis for this, where I use a series of unit conversions (ie 1ft = 12 in, so 1ft/12in = 1) to get to the units I want from the units I have. (If this is confusing, I highly recommend practicing with a few unit conversions you can already easily make. It is very useful for when you have to do multiple unit conversions.)
Force = rhoVg
= (1g/cm^3)*(1kg/1000g)*(100cm/1m)^3*(11.0 ft^3)*(1m/3.28ft)^3*(9.8m/s^2)*(1N/kgm/s^2)*(0.225lb/N)
= 687 (g*kg*cm^3*ft^3*m^3*m*s^2*N*lb)/(cm^3*g*m^3*ft^3*s^2*kgm*N)
The following units are on both the top and bottom and will cancel out:
g, kg, cm^3, ft^3, m^3, m, kg, s^2, N
This just leaves lbs.
So the answer is 687lb
