Problem

An ore sample weighs 17.50 $\mathrm{N}$ in air. W…

03:02

Problem 31 Medium Difficulty

When an open-faced boat has a mass of 5750 $\mathrm{kg}$ , including its cargo and passengers, it floats with the water just up to the top of its gunwales (sides) on a freshwater lake. (a) What is the volume of this boat? (b) The captain decides that it is too dangerous to float with his boat on the verge of sinking, so he decides to throw some cargo overboard so that 20$\%$ of the boat's volume will be above water. How much mass should he throw out?

Answer

a) 5.75 $\mathrm{m}^{3}$
b) 1150 $\mathrm{kg}$

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Physics 101 Mechanics

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Video Transcript

Okay, so in this problem, we have a boat floating underwater. Let's put here on boats and he's faulting on the water. So in the first item of this problem, we want to calculate the volume off the boat. And how are we going to do this? Let's remember off the arguments principle. So we have a force here, keeps you both floating and we have the weight off the boat so we can make this force is equal using the arguments principle. So we have that density off the water times the gravity acceleration times the volume off the water that is going to be called the volume off the boat. It's equal the mass off the boat times the gravity's acceleration. Therefore, we can cross. The gravity isn't here, so to simplify, we have the volume off. The boat is going to be equal the mass of the boat divided by density off the water. So let's calculate this. The mass of the boat is just 5000 750 kilograms divided by did institute off the water, which we know that it's one times 10 to the tree. So we have the volume off the boat equals five point 75 meters cubits. So that's the volume off the boat. Now that we have the volume of the boat, the second thing we need to calculate is a mess off the cargo, knowing that Heathrow 20% off the mess own underwater. So we're going to use this same equation here but to calculate the mess. So we know that the mess is going to be equal volume. The final volume multiply by the density off the water. But who is the final volume is just 80% of the total volume because he throw away 20% so we can see that the mess of the cargo it's going to be 0.8 times the volume that we already calculate. Riches, uh, 5.75 5.75 times. Do you density off the water, which is one times 10 to the tree. So if recalculate this, we have a total mess that is still on the boat, both 4000 and 600 kilograms, and that's to find a lesser charge us a problem. So actually, that's see, he wants to know how much you throw away. How much must throw out. Okay. Okay. So this is the mess. The left on the boat. So to calculate what the problem actually wants, we just need to subtract 5000 750 miners, 4000 and 60. And we know the mess he threw out is going to be equal 1000 150 kilograms. That's the actual answer of this problem, That's all. Thanks for watching.

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