Floating Ice Block: Height & Penguin Capacity

[mattmannmf]

A rectangular block of ice 5 m on each side and 0.5 m thick floats in seawater. The density of the seawater is 1025 kg/m3. The density of ice is 917 kg/m3.

a) How high does the top of the ice block float above the water level?


b) How many penguins of mass 19 kg each can stand on the ice block before they get their feet wet?

So what i did first was 917/1025= fraction of total volume of ice submerged..than subtracted it by 1 to find fraction of ice above water... I am just not sure where to go from there..i can take the fraction of the volume...but they are still just looking for height...

 

[mgb_phys]

For part A you know the fraction below the water - so what fraction of the total height is above water.

For B - a floating object displaces water equal to it's weight.
So if the block was totally submerged how much water would it displace. How much more mass is this than the mass of the ice - the difference is the penguins

 

[mattmannmf]

I tried that for A and got it wrong...

917/1025= .8946*5= 4.47 (under water)-5= .52 above water right? i checked it and it was wrong

 

[mgb_phys]

Erm if it's only .5 high it's difficult to see how you can have .52m above water.
What about (917/1025) * 0.5m below water or 0.5 - (917/1025) * 0.5m above.

Quick check 917 is roughly 90% of 1025 so you expect to have roughly 10% above water so the answer must be about 0.05m

 

[mattmannmf]

ahh yep...thats correct..idk why i was getting a decimal place off...

Im not really understanding B.. for what you wrote

 

[mgb_phys]

The volume of the berg is 5m * 5m * 0.5m
It weighs 917kg/m^3 so you can work out how many kg this is.
When it is underwater it is displacing the same volume (5m * 5m * 0.5m) of water that weighs 1025kg/m^3
So work out how much the ice weighs and how much the water weighs. The extra weight of the water is the weight needed to force the ice down = the penguins