A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 65.0-kg woman to be able to stand on it without getting her feet wet?

Video transcript

Hey everyone. So this problem is working with buoyancy force. Let's see what it's asking from us. So a block of ice forms in a pond during the winter. We need to determine the least volume of the block required to keep a 68 kg Fisher holding a 50, sorry, a five kg load of their fishing gear without wetting their feet. So this is a little bit tricky, but we're gonna go through everything we know and then figure out how we can calculate the volume of the block. So recall that the buoyancy force is given by row G. V. Where rho is the density of the specific liquid that we're dealing with gravity is is just so jesus gravity and then V. Is the volume. So we're going to be solving for b the givens, we know from the problem we're dealing with water and ice. So the density of water Is 1000 kg per meter cubed. The density of ice is kg per meter cube. You just need to over call those those values. And then from here we're gonna do a free body diagram. So we know the buoyancy force is is acting in the positive Y direction and then in the um negative, why we have the weight of we're just gonna call this the weight of the load and then the weight of the ice. So the total load, it's gonna be the mass of the total load times gravity. Right, recall weight is just mass times gravity and the mass of the total load. We know we have a 68 kg fisher and a five kg fishing gear. So the mass of this total load is kg from our free body diagram, what's recall? Newton's second law where the sum of the forces equals mass times acceleration. This is a zero term, it's not moving, it's not accelerating, so that just goes to zero. So we have the buoyancy force is equal to the weight of the total load plus the weight of the ice block. Okay, so from here, we're just going to kind of write out what we know and then see if we have enough to solve for the volume. So this volume of the ice is what we're trying to get to. So we have the density of the water times gravity times the volume of the ice equals the mass of the load, times gravity plus the mass of the ice times gravity. All right, so the good news is we've gravitated each turn that cancels, we simplify it a little bit. Something else that we need to recall is that density is just mass over volume. So, if we don't know the mass of the ice, which we don't, we can actually rewrite this term in terms of volume of ice. So the mass of the ice is equal to the density of the ice, times the volume of the ice. So we're gonna substitute that in here and then let's see what we have density of the water times the volume of the ice equals the mass, the load plus the density of the ice, times the mass of the ice. Okay, so now we actually have the mass of the ice or the volume of the ice. We can put those terms together, we know everything else and we can solve for this volume of the ice. So we're gonna move this term over to the other side. It's gonna look like this total mass equals the mass of the load. We're gonna pull out the volume of the ice, oops, the volume of the ice times the density of the water minus the density of the ice, pulls the mass of the load. We're going to divide out the densities and we are left with the volume of the ice equals the mass of the total load over the density of the water minus the density of the ice. We already know that 73 kg Over 1000 kg per meter cubed -920 kg per meter cubed Plug that into our calculators. And we are left with the volume of the ice. 0.91 motors, cute. All right, let's go back up to our possible answers. And that is one. So, the answer to this question is C 0. m cubed. Alright, that's all for this problem. We'll see you in the next video