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?

Question

A cubical wooden block floats between oil and water, submerged 1.5 cm in oil, illustrating buoyancy concepts.

Accepted Answer

Hey everyone in this problem, we have a solid cylinder of diameter 10 cm and height cm placed into a container, Oil of density 850 kg/m Cube. And fresh water are placed in the container. The cylinder floats at the boundary of the two liquids with eight cm of its length in water were asked to determine the mass and the density of the cylinder. Alright, so we have a little diagram of what's going on here. Let's just label some of this. We're told that 8cm Of the length is in water. So this distance here is going to be 8cm. This has a total height of 25 cm or cylinder, Which tells us that the distance or the amount of this cylinder in the oil 25 -8 is going to be cm. All right now, let's go ahead and draw a free body diagram here. So we have our cylinder and this is like an Archimedes problem. We have this buoyant force acting upwards and the weight acting downwards. So Archimedes principle tells us that when an object is completely a partially immersed in a fluid, the fluid exerts an upward force on the object equal to the weight of the fluid displaced by the objects. Now, we know that the sum of the forces were in an equilibrium situation, it's going to be zero. In this case, we're looking at forces in the Y direction, we're gonna take up to be positive. And so we can look at the sum of the forces in the Y direction equal to zero. We have the buoyant force acting in the positive Y direction, and we have the weight acting in the negative Y direction. And so we have the F B minus W is equal to zero, which tells us, that tells us sorry that these are equal F B. Is equal to the weight W. Now, when we're talking about this buoyant force, what do we have to consider? We have some sort of force due to the oil and we have some sort of force due to the water and recall that this is going to be given by row G. V. Okay, so rho the density G the gravitational acceleration and the the volume of the cylinder that is in that particular liquid or fluid. Okay, so, when we're talking about F. B. We have to do again the oil and the water. So, we're gonna have row the density of the oil times G. The gravitational acceleration times V. The volume of the cylinder that is in oil plus row W. The density of water, G. The gravitational acceleration and V. W the volume of the cylinder in water. And all of this is going to equal the weight, which we know is m times G. All right, So, what do we have here? The density of oil and the density of water? Our values that we can look up in a table in our textbook the gravitational acceleration G is in every single term. So we can divide by G. That's gonna cancel out we have these two volumes which we can determine because we know the height that's in each fluid. We know the diameter of our cylinder. We can determine these volumes. B. The only thing we're left with is M. And we want to find the mass. And so this gives us a way to calculate that mass M that we're looking for. And so we have that the mass M is going to be the density of the oil times the volume of the cylinder in the oil, which is going to be pi r squared times a height that is in the oil plus the density of water road W times pi r squared H. And in this case the height that's in the water. All right. We are given the diameters and the heights in centimeters. We want to convert these into meters in order to do this calculation. So let's start with the radius of our cylinder. The radius R. That were given They were given the diameter of 10. So the radius is going to be 10 centimeters divided by two Which is five cm. And we want to convert this into meters. So we're gonna multiply by one m per centimeters. This is equivalent to dividing by 100. The unit of centimeter cancels we divide by 100 and we get a radius of 0.05 m. We can do the same for these two heights. The height of oil. The height of water. H oil we know is equal to 17cm. We divide by 100 or multiply by one m per 100 centimeters. The unit of centimeter divides out. And we're left with 0.17 m. And similarly for the height of the water or the height of the cylinder in the water. This is eight centimeters. We divide by a 100 And we're left with 0.08 m. Alright, so we have all these dimensions in their proper units. Let's get back to this equation that I've started blue for our mass M. And so we have that the mass M is equal to the density of oil were given in. The problem is kilogram per meters. Cute times pi The radius squared 0.05 m squared times the height of the cylinder in the oil. 0.17 m. Then we're gonna add the density of water. and recall that the density of water is a 1000 kilograms per meter cubed. And then we have pi r squared, it's the same radius 0.5 m squared. And the height in water is 0.8 m. And if we work all of this out we get a mass of 1.76 kg. So we found the mass of the cylinder. Now I need to figure out the density of the cylinder and recall that the density rope of the cylinder, it's going to be equal to the mass divided by the volume. We're talking about the volume here. We're talking about the volume of the entire cylinder, not just one part of it. And so this is going to be the mass divided by pi r squared H. Now, h here the entire cylinder is centimeters and doing the same conversion we did above one m per 100 centimeters. Okay to go from centimeters to meters. We divide by 100 and we get 0.25 m. And so our density becomes 1.7632 kg divided by pi Times 0.05 m 2.25 m. We get a density of 897. kilograms per meter. Cute. And there you have it. We have our mass. We have our density. So we're done with the problem. Let's go back up to our answer choices and select the correct one. We found that we had a mass of approximately 1.76 kg and the density of approximately 898 kg per meter cubed. And so we have answer choice. C Thanks everyone for watching. I hope this video helped see you in the next one

Verified Solution

Key Concepts

Buoyancy

Gauge Pressure

Density