Does Salt Water Affect the Weight of Displaced Water?

[archied]

hi all, I'm completely stuck on this question I've got no idea how to tackle it, I've looked back at previous examples and i don't know how this fits in with what I've see before.
heres the question

a boat floating in fresh water displaces water weighing 35.6kn

what is the weight of the water displaced if it were floating in salt water with a density of 1.10x10^3kg/m^3?

just something that came to mid could you work it out by going

(density of fresh water divided by water displaced)times(density of salt water)

 

[HallsofIvy]

Actually, it's much simpler than that! A boat (or other object) will sink down into water until it has displaced a weight of water equal to it's own weight (Archimede's principle). If the boat initially displaced 35.6 KN, then it weighs 35.6 KN and will displace that weight in ANY liquid.

(If the problem had given the VOLUME of fresh water displaced and then asked for the volume of saltwater displaced, then you would have to take the density into account.)

 

[thepatientmental]

Hello,

I understand that you are stuck on this density and buoyancy question. It can be a bit tricky, but I will do my best to explain it to you.

First, let's review the concept of buoyancy. When an object is placed in a fluid, it displaces some of the fluid's volume. The weight of the displaced fluid is equal to the weight of the object, which is known as the buoyant force.

Now, let's apply this concept to the question. We know that the boat is floating in fresh water and displacing water weighing 35.6kn. This means that the weight of the boat is equal to the weight of the water it displaces, which is 35.6kn.

Next, we need to find the weight of the water displaced in salt water. We can use the formula for density, which is mass divided by volume. In this case, the mass is the weight of the water displaced and the volume is the volume of water displaced.

So, the weight of the water displaced in salt water can be calculated as follows:

Weight of water displaced in salt water = (density of salt water) x (volume of water displaced)

= (1.10x10^3 kg/m^3) x (volume of water displaced)

But, we know that the volume of water displaced is the same in both fresh water and salt water (because the boat is the same size and shape). So, we can rewrite the equation as:

Weight of water displaced in salt water = (1.10x10^3 kg/m^3) x (volume of water displaced in fresh water)

Now, we can substitute the value of the volume of water displaced in fresh water, which is 35.6kn, into the equation:

Weight of water displaced in salt water = (1.10x10^3 kg/m^3) x (35.6kn)

= 38.4kn

Therefore, the weight of the water displaced in salt water is 38.4kn. I hope this helps you understand the question better and how to approach it. Keep practicing and you will become more confident in solving these types of problems. Good luck!