## Tricky Buoyancy Problem Involving Water Levels

I'm trying to work through the below problem but it is a bit tricky and I would love any helpful tips:

A boat with a lead block on it is sitting in a pool. On the side of the boat is a red line right at the water level. On the side of the pool is a blue line right at the water level. The lead block is then tossed over the side of the boat but still connected to the boat by a string. Will the red line now be above/below or the same as the new water level? Will the blue line now be above/below or the same as the new water level?

So here is my current explanation:

The boat originally just had the force of the lead block's weight pushing directly downwards. Once the lead block is in the water, the is the force of the lead block's weight pushing downwards, but there is also some buoyant force pushing upwards. Therefore, the boat is not as heavy, resulting in the red line being above the new water level.

Because the lead block is now submerged underwater, it will cause the water levels to rise, meaning the blue line will now be below the new water level.

But I'm hesitant to put this as my answer as I feel like this could easily be a trick question. I believe this because I'm not quite sure how to relate the water displaced by the boat with the block on it compared to the water displaced by boat with the block in the water. Also, it seems contradictory for me to say that the boat will rise (resulting in a lower boat volume in the water) and yet the water level will rise as well.

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 Quote by haleyking11 The boat originally just had the force of the lead block's weight pushing directly downwards.
What about the weight of the boat itself?
 Because the lead block is now submerged underwater, it will cause the water levels to rise, meaning the blue line will now be below the new water level.
But you say the boat is displacing less water.

I think it is safe to assume that the lead is denser than the boat.
 Yes, I did take into account the weight of the boat itself, just forgot to put that in my explanation, and to answer the second part of your question where I say the boat is displacing less water, that is where I know I am contradicting myself, but I'm not sure which part is wrong. So in the beginning, there is the weight of the boat+the weight of the block pushing downwards on the boat. Then, when the block is dropped over the side, there is still the weight of the boat pushing downwards, and a slightly smaller value of the original block's weight also pushing downwards on the boat (I say slightly smaller due to the buoyancy that the block will have while submerged in the water). Therefore, the boat will we higher up in the water (the red line will be above the boat) and displace less water. That sounds about right. I guess where I'm getting tripped up is the fact that before, there was no block in the water, and now that there is, that makes me think the added volume of the block will increase the total water level. But is that wrong to say? Especially since the boat will be displacing less water? I agree that the lead will be denser than the boat. Since the boat (most likely) has a larger surface area in the water, does that mean that it will have the greater effect on the overall water level, making the overall water level drop? Thanks for responding

Recognitions:
Homework Help

## Tricky Buoyancy Problem Involving Water Levels

You are right, the volume of the block rises the water level in the pool.
Write up the equations for the depth the boat is submerged into water and the height of water level in the pool, and see the difference between both situations - block in and out of boat.

ehild
 Recognitions: Homework Help Say the volume of the lead block is Vp, its mass is Mp. The mass of the boat is Mb. The submerged volume of the boat with the block in is V1, that with the block out is V2. The volume of water in the pool is Vw. The cross-section of the pool is A, the height of the water level is H1 when the block is in the boat and H2 when the block is in the water. ρ is the density of water. Force of buoyancy =weight. gρV1=g(Mp+Mb), ρg(V2+Vp)=g(Mp+Mb) => V2+Vp=V1. The water level in the pool: When the lead block is in the boat: AH1=V1+Vw, when the block is in the water: AH2=V2+Vp+Vw. A(H2-H1)=V2+Vp-V1. Compare the equations in bold. ehild

 Tags buoyancy, tricky, water level