A cubical block of wood, 10.0 cm on a side, floats at the interfa...

Question

A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.50 cm below the interface (Fig. E12.33). The density of the oil is 790 kg/m^3. (a) What is the gauge pressure at the upper face of the block? (b) What is the gauge pressure at the lower face of the block? (c) What are the mass and density of the block?

Accepted Answer

Hey everyone. So this problem is working with hydrostatic pressure. We are given the problem and this diagram. So let's work through and see what they're given us and see what they're asking for. So a container is filled with water and oil of density 800 kg per meter. A cylindrical wooden block of diameter 12 centimeters in height, 15 centimeters is floating at the interface of the two liquids. If four cm of the block is in the water, determine the gauge pressure at the top and lower faces of the block. Okay, so Let's first recall that the density of water is 1000 kg per meter cubed. They've given us the density of the oil in the problem that's 800 kg per meter cubed. They also tell us that four cm of the block is in the water. So This distance here is four cm. And we are given in this diagram that the total height of the oil is 18 cm. If the total height of the block is 15 cm, we know that four cm is in the oil, Sorry, is in the water. That means 11 cm is in the oil and therefore the height of the oil above the block is cm -11 cm Or seven cm. Which we can rewrite as 0.07 m. Okay, Okay, so the next thing we need to do is recall that our hydrostatic pressure equation is given by p equals rho G. H where rho is the density of the liquid G is your gravity? That's another constant. Remember that's 9.8 m per second squared. And h is the height of the fluid acting on the block? So they're asking us it's a two part question. They're asking us to determine the pressure at the top and lower faces of the block. So let's take the top first. There's only one fluid acting on the top of the block. It's just the oil. So we are already given the density of the oil, that's 800 kg per meter cubed gravity. We know 9.8 m per second squared. And then the height of the oil acting on this block we figured out was 0.7 meters. We're gonna plug that indoor calculator and get 549 Pascal's. So that is the pressure acting on the top of the block. Let's take a look at our possible solutions and we can already eliminate a couple of these. So we're left with D. And E. And we're going to now solve for the pressure at the bottom of the block. So looking at this diagram helps us to see that there are two uh fluids acting um as for pressure on the bottom of this block. So we need to add both of those pressures together. So we're gonna have pressure from the oil plus the hydrostatic pressure from the water total of that is going to be the pressure of the bottom of the block. So for this one, the height of the oil acting on the bottom of the block is the total amount of the oil or total height of the oil. So that's 18 cm or .18 m. The height of the water Was given to us in the problem. There's four cm of the block submerged underwater. So that's 0.04 m. Okay from there, we can just plug and chug in our equation. So we have the total pressure from the oil is going to be density of the oil times gravity times the height of the oil plus the pressure from the water is going to be the density of the water times gravity times the height of the water. So let's plug in everything we know here. Renegade 800 kg per meter cubed Times 9.8 m/s squared Times 0.18 m plus For the water term 1000 kg per meter cubed gravity doesn't change 9.8 m per second squared. And then the height of the water, 0.04 m plug that in and we get p pressure at the bottom equals 1803 pascal's. So now, looking back at the answers that we have left, we see that E 1800 for p lower is the closest to our calculated answer. And that means that the answer for this problem is E That's all we have for this video. We'll see you in the next one

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