# Buoyancy help needed

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bobbricks

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#1

A block of timber 130 mm thick is floating in water with 62 mm submerged below the water surface. Assuming the density of water to be 1000 kg/m3 and the acceleration due to gravity to be 9.81m/s2, calculate the density of the timber block (in kg/m3)

I'm stuck on this question.

I've worked out that Upthrust=pvg (where p is density, v is volume and g is acceleration due to gravity). Therefore the weight of displaced fluid= weight of object since it is floating:

p(object) x V(object)=p(fluid) x V(fluid) [the g constants on both sides cancel out]

p(fluid) is 1000

so p(object)xV(object)=1000 x V(fluid)

I'm not sure what to do from here, since we don't know all 3 side lengths of the object to work out the volume of the fluid..?

I'm stuck on this question.

I've worked out that Upthrust=pvg (where p is density, v is volume and g is acceleration due to gravity). Therefore the weight of displaced fluid= weight of object since it is floating:

p(object) x V(object)=p(fluid) x V(fluid) [the g constants on both sides cancel out]

p(fluid) is 1000

so p(object)xV(object)=1000 x V(fluid)

I'm not sure what to do from here, since we don't know all 3 side lengths of the object to work out the volume of the fluid..?

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bobbricks

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#2

Maybe I could make up the side lengths? So the block has dimensions 1m x 1m x 0.13m (with 0.062m of 0.13m submerged below the water). Therefore p(object) x 0.13 =1000 x 0.62 so p(object) x 0.13 = 62 so p(object) =62/13 so the density is about 477 kgm^(3) ?

Is this correct?

Is this correct?

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noodledoodle

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#3

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#3

(Original post by

Maybe I could make up the side lengths? So the block has dimensions 1m x 1m x 0.13m (with 0.062m of 0.13m submerged below the water). Therefore p(object) x 0.13 =1000 x 0.62 so p(object) x 0.13 = 62 so p(object) =62/13 so the density is about 477 kgm^(3) ?

Is this correct?

**bobbricks**)Maybe I could make up the side lengths? So the block has dimensions 1m x 1m x 0.13m (with 0.062m of 0.13m submerged below the water). Therefore p(object) x 0.13 =1000 x 0.62 so p(object) x 0.13 = 62 so p(object) =62/13 so the density is about 477 kgm^(3) ?

Is this correct?

In the case of this question the bottom area of the block does not matter because the bottom area of the block of timber is equal to the bottom area of the ''water block'' that is displaced. Thus when u form the equation u will be able to cancel out the area on both sides.

After subbing all the values and cancelling the relevant values you will end up with a proportion of the density of block to density of water

After getting the ratio you will get 477kgm^-3. So yes your answer is correct but your method is wrong.

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bobbricks

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#4

(Original post by

You are never supposed to make up your own dimensions.

In the case of this question the bottom area of the block does not matter because the bottom area of the block of timber is equal to the bottom area of the ''water block'' that is displaced. Thus when u form the equation u will be able to cancel out the area on both sides.

After subbing all the values and cancelling the relevant values you will end up with a proportion of the density of block to density of water

After getting the ratio you will get 477kgm^-3. So yes your answer is correct but your method is wrong.

**noodledoodle**)You are never supposed to make up your own dimensions.

In the case of this question the bottom area of the block does not matter because the bottom area of the block of timber is equal to the bottom area of the ''water block'' that is displaced. Thus when u form the equation u will be able to cancel out the area on both sides.

After subbing all the values and cancelling the relevant values you will end up with a proportion of the density of block to density of water

After getting the ratio you will get 477kgm^-3. So yes your answer is correct but your method is wrong.

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Stonebridge

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#5

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#5

(Original post by

A block of timber 130 mm thick is floating in water with 62 mm submerged below the water surface. Assuming the density of water to be 1000 kg/m3 and the acceleration due to gravity to be 9.81m/s2, calculate the density of the timber block (in kg/m3)

I'm stuck on this question.

I've worked out that Upthrust=pvg (where p is density, v is volume and g is acceleration due to gravity). Therefore the weight of displaced fluid= weight of object since it is floating:

p(object) x V(object)=p(fluid) x V(fluid) [the g constants on both sides cancel out]

p(fluid) is 1000

so p(object)xV(object)=1000 x V(fluid)

I'm not sure what to do from here, since we don't know all 3 side lengths of the object to work out the volume of the fluid..?

**bobbricks**)A block of timber 130 mm thick is floating in water with 62 mm submerged below the water surface. Assuming the density of water to be 1000 kg/m3 and the acceleration due to gravity to be 9.81m/s2, calculate the density of the timber block (in kg/m3)

I'm stuck on this question.

I've worked out that Upthrust=pvg (where p is density, v is volume and g is acceleration due to gravity). Therefore the weight of displaced fluid= weight of object since it is floating:

p(object) x V(object)=p(fluid) x V(fluid) [the g constants on both sides cancel out]

p(fluid) is 1000

so p(object)xV(object)=1000 x V(fluid)

I'm not sure what to do from here, since we don't know all 3 side lengths of the object to work out the volume of the fluid..?

Volume of water displaced = hA where h is the amount submerged (given)

Volume of block is xA where x is the given thickness.

The A cancels to give you the answer.

This is the mathematically correct way of solving these problems.

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noodledoodle

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#6

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#6

(Original post by

But surely if the lengths I've used are kept constant throughout it, it doesn't matter if I do this (as they would eventually cancel out)? So are you saying this method won't work for other questions like this (because a wrong method can't produce a right answer, can it)?

**bobbricks**)But surely if the lengths I've used are kept constant throughout it, it doesn't matter if I do this (as they would eventually cancel out)? So are you saying this method won't work for other questions like this (because a wrong method can't produce a right answer, can it)?

It is an incorrect method that produces the right answer. But to most teachers its less about the answer and more about how u got there. Hence in structured questions more marks are allocated to workings and methkds than to tbe answer itself.

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