Logs sometimes float vertically in a lake because one end has become waterlogged and denser than the other. What is the average density of a uniform-diameter log that floats with 20 percent of its length above water? Use a value of 1000 kilograms per meter cubed for the density of water.
As we start, let’s highlight some of the critical given information, and also name any assumptions we’ll make in solving this problem. First, we’re told that the log has a uniform diameter, meaning the diameter is constant all the way up and down the log. We’re also told that when it floats, 20 percent of its length along a vertical line is above water, meaning 80 percent is below. And we’ll also use a value of 1000 kilograms per meter cubed for the density of water. Now we’ll make the assumption that this density value is an exact number.
From here, let’s draw a diagram of what’s going on in this problem. So here we have our log where 20 percent is above water, and 80 percent of the log is submerged beneath the surface of the lake. We’re told the density of the water in the lake 1000 kilograms per meter cubed, and we want to solve for the density of the log, realizing that it will have a constant diameter from top to bottom. Now let’s consider the forces that are acting on this log. We’ll draw those forces in on our diagram.
First, we have the force of gravity that’s acting down on the log. We’ll label that 𝐹 sub 𝑔. And then we also have another force. It’s the buoyant force that acts on the log, which we’ll label 𝐹 sub 𝑏. Now these two forces, we know, in magnitude, are equal to one another. And the reason we know that, is because we’re assuming that the log is stationary in the water. It’s moving neither up nor down. We can express that mathematically by saying that 𝐹 sub 𝑏 minus 𝐹 sub 𝑔 is equal to zero. Let’s start to figure out what these two forces are and see if in finding them, we’re able to solve for our ultimate intended answer of the density of the log symbolized by the Greek letter 𝜚 sub log.
First, let’s work at 𝐹 sub 𝑏, the buoyant force. In order to do that, let’s begin recollecting a definition about buoyant force. The buoyant force that acts on an object is equal to the volume of fluid that object displaces multiplied by the density of that fluid displaced, all multiplied by 𝑔, the acceleration due to gravity. As a side note, you may notice that when we multiply a volume times a density, what we end up with is a mass in kilograms. So when we see that, we realize that the equation for buoyant force is very much like the equation for weight force, mass times gravity. Let’s use this equation for buoyant force to insert for our 𝐹 sub 𝑏 in our own application.
So in our case, 𝐹 sub 𝑏 is equal to the volume that the log displaces, we’ll call that 𝑣 sub disp as in this equation, multiplied by the density of the fluid, which in this case is 𝜚 sub 𝑤, the density of the water in the lake. Now all of that, according to our buoyant force equation, is multiplied by 𝑔, the acceleration due to gravity.
Now let’s consider the gravitational force that acts on our log. Well, as usual, we know that that force is equal to the mass of the object, in this case our log, times 𝑔, the acceleration due to gravity. Here, we’ve written the mass of log as 𝑚 sub log. And then when we multiply that by 𝑔, we have our gravitational force. And we know that altogether, these two terms equal zero.
Now let’s take a step here and rewrite one of our terms. If we look at the volume displaced by the log, we know, because of what the problem statement told us, how much fractionally of the total logs volume that is. We’re told in the problem statement that 80 percent of the log’s volume is below the level of the water. What that means is that 𝑣 disp, our displaced volume of water due to the log, is equal to 80 percent of the volume of the log. We can write that out as an equation off to the side. Here, where we’ve written in our equation 0.8, that’s an exact number. It’s exactly 80 percent of the log’s volume that is submerged and therefore displacing water in the lake.
Let’s plug 0.8 𝑣 log into our equation, so that now we have the volume and the mass of the log in one equation. Now we’re making great progress because as we look at this equation, we see that if we divide both sides by 𝑔, that term completely cancels out which simplifies our expression, so that we now have just three terms. Drawing our cleaned-up version of this equation, those three terms are: the volume of the log, the density of the water, and the mass of the log.
Now at this point, let’s recall an equation relating density, mass, and volume. The density of an object, the Greek letter 𝜚, is defined as the mass of that object divided by its volume. So as we search for 𝜚 sub log, the density of our log, we don’t see that in our equation, but we do see the mass and the volume of the log as terms there. So let’s make it our aim to rearrange this equation to get the mass of the log divided by its volume. As a first step, we’ll add the mass of the log, 𝑚 sub log, to both sides. Having done that, we see that the mass of the log cancels out on the left-hand side of our equation. Now let’s go and divide both sides of our equation by 𝑣 sub log. Doing that, 𝑣 sub log cancels out of the left-hand side of the equation. And now let’s look at what we have left.
We can rewrite a cleaned-up version. But at a first glance, what we’ve got is the mass of the log divided by its volume being equal to 0.8 times the density of the lake water. We recall by our definition that the mass of the log divided by its volume is equal to the density of the log, 𝜚 sub log. So this is great. We’re very close to solving our problem. Let’s rewrite this equation one last time to really home in on our solution. 𝜚 sub log, the density of our log, is equal to 0.8, where that is an exact number, multiplied by the density of the water of the lake. Now we’ve been given that density of the lake water, 1000 kilograms per meter cubed. So let’s plug that in to our equation now.
And when we do this and multiply through, we’ll find that the density of the log, 𝜚 sub log, is equal to 800 kilograms per meter cubed. So the log is 80 percent as dense as the water, and 80 percent of the log is submerged below the water.