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Hi,

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

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.

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

Surface area of cylinder (A) = πr

Weight of cylinder (W

F

9310N=1030kg⋅m

9310N=5865kg⋅ms

rearranging for h

h=9310N/5865kg.ms

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 (W

F

9996N=1030kg⋅m

9996N=5865kg⋅ms

rearranging for h

h=9996N/5865kg.ms

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)?

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) = πr

^{2}=π(0.860/2)^{2}=0.581m^{2}Weight of cylinder (W

_{c}) = 950kg⋅9.8ms^{-2}=9310NF

_{b}=ρVg=ρAhg=W_{c}=9310N9310N=1030kg⋅m

^{-3}⋅9.8m⋅s^{-2}⋅0.581m^{2}⋅h9310N=5865kg⋅ms

^{-3}⋅hrearranging 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 (W

_{cm}) = (70kg + 950kg)⋅9.8ms^{-2}=9996NF

_{b}=ρVg=ρAhg=W_{cm}=9996N9996N=1030kg⋅m

^{-3}⋅9.8m⋅s^{-2}⋅0.581m^{2}⋅h9996N=5865kg⋅ms

^{-3}⋅hrearranging 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)?

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