# Archimedes's Principle Question

1. Nov 29, 2016

Hi,

I have an issue with a question I am trying to answer and I don't seem to be getting the correct answer...

Question
A 950kg cylindrical can buoy floats vertically in sea water, the diameter of the buoy is 0.860m. Calculate the additional distance the buoy will sink when a 70kg man stands on top of it.

Attempt
OK, so I know that for a floating object, the buoyant force is equal to the weight of the object, so calculating for just the cylinder on its own...

Density of sea water (ρ) = 1030kg⋅m-3
Surface area of cylinder (A) = πr2=π(0.860/2)2=0.581m2

Weight of cylinder (Wc) = 950kg⋅9.8ms-2=9310N
Fb=ρVg=ρAhg=Wc=9310N
9310N=1030kg⋅m-3⋅9.8m⋅s-2⋅0.581m2⋅h
9310N=5865kg⋅ms-3⋅h

rearranging for h
h=9310N/5865kg.ms-3
h=1.59m

So that should mean that 1.59m of the cylinder is submerged in water when it is just the cylinder.

Then I can do exactly the same again, with the additional 70kg on top of the cylinder...

Weight of cylinder and man (Wcm) = (70kg + 950kg)⋅9.8ms-2=9996N
Fb=ρVg=ρAhg=Wcm=9996N
9996N=1030kg⋅m-3⋅9.8m⋅s-2⋅0.581m2⋅h
9996N=5865kg⋅ms-3⋅h

rearranging for h
h=9996N/5865kg.ms-3
h=1.70m

So now I can just calculate Δh = 1.70m - 1.59m = 0.117m (calculated using non rounded figures)

I know I could have saved a lot of working out and just done the the calculations for 70kg only, neglecting the cylinder and got the same answer.

According to my text book, the answer to this question is 0.177m

Have I done something stupid here (which is entirely plausible)?

Last edited: Nov 29, 2016
2. Nov 29, 2016

### Staff: Mentor

What about the additional weight of the cylinder, if it sinks by $0.117$ cm?
(I've forgotten it, too, the first time I did the math and wondered, why the book's answer is an equivalent of $103$ kg.)

3. Nov 29, 2016

### anorlunda

$\frac{70 kg}{1030 kg*m^{-3}}= 0.068 m^3$

You do the rest. How far does the buoy have to sink to displace that volume of water?

The initial weight and depth of the buoy are not needed for this problem.

4. Nov 29, 2016

Thanks guys,

Anorlunda, if I calculate the displacement of the cylinder using the 0.068m3, I still get the answer of 0.117m.

(V/A=h)

I am sorry if you meant for me to look elsewhere here, but I still do not see why the answer I am getting is 0.117m and the text book says the answer is 0.177m.

Fresh 42, are you saying that the book is wrong or am I just missing something?

Thanks again.

5. Nov 29, 2016

### anorlunda

Then we do not understand what your question is. Try rephrasing it. What are you asking?

6. Nov 29, 2016

Sorry, what I am asking is do I have the right answer (0.117m) and the text book is wrong (0.177m) or am I making a mistake somewhere in my calculations?

7. Nov 29, 2016

### anorlunda

I believe that the book is wrong, probably a single character typo.

8. Nov 29, 2016

### Staff: Mentor

I calculated $11.7$ cm as well and first thought the mistake is, not to take the part of the cylinder's mass into account, that is newly under water. A desperate try to make any sense of the $950$ kg given, and a wrong one.
That's what I think, too.

9. Nov 29, 2016

Thank you both very much, I have never had an issue with the Sears and Zemansky University Physics book before but have lots of problems with their Mastering Physics website. Was hoping I could trust the book!

At least I know I had the right idea.

Thanks again!

10. Nov 29, 2016

### haruspex

Very likely. When a letter occurs twice consecutively in a word, typing the wrong letter twice is very common. Probably applies to digits too.