specific gravity EK

SDN Members don't see this ad. (About Ads)

Attachment 18584

Hopefully, it's visible now. If not....

A brick with a density of 1400 kg/m^3 is placed on top of a piece of Styrofoam floating on water. If one half the volume of the Styrofoam sinks below the water, what is the ratio of the volume of the Styrofoam compared to the volume of the brick? (Assume the Styrofoam is massless.)

Can't see the image.

Let the brick has volume V. That means that it will weight 1400 * V. Half of the styrofoam displaces water which weights the same, in other words 1400 * V. Since water has density of 1000 kg/m^3, the volume of the water is 1400*V/1000=1.4 V. That's only half of the volume of the styrofoam, then the volume of the styrofoam is 2*1.4V. The ration in question is 2.8V/V = 2.8, answer C.

Wouldn't it be A?

Originally Posted by MedPR

Wouldn't it be A?

Why? Where's the error in my solution/what is the alternative solution? With the brick being denser than water you are guaranteed that you'll need the ration to be >1. And with only half the styrofoam submerged, it will have to be >2.

Originally Posted by milski

Why? Where's the error in my solution/what is the alternative solution? With the brick being denser than water you are guaranteed that you'll need the ration to be >1. And with only half the styrofoam submerged, it will have to be >2.

I don't want to do the math, but conceptually this is how I'm thinking of it.

If the styrofoam floats on the water without the brick on it, then you add the brick and the styrofoam now sinks below the water, doesn't that mean that the brick must be weigh more than the styrofoam? So isn't the brick more dense than the styrofoam?

Originally Posted by MedPR

I don't want to do the math, but conceptually this is how I'm thinking of it.

If the styrofoam floats on the water without the brick on it, then you add the brick and the styrofoam now sinks below the water, doesn't that mean that the brick must be weigh more than the styrofoam? So isn't the brick more dense than the styrofoam?

The styrofoam is massless. Anything weights more than it. But even if it has some minimal weight, how does that give you anything?

Unfortunately it is C. I'll bold the steps I don't understand. If someone could clarify, those I would be really grateful

Step-by-step solution:

The brick displace a volume of water equal to its weight.
The density of the break is 1.4 times of that of water, so it will displace 1.4 times of the volume displaced by water.
Since this is only 1/2 the volume of the styrofoam (why are the equating the brick with the foam??), the full volume must be 2.8 times larger than the volume of the brick.

I guess I can't type out what I'm thinking. It just seems to me that the ratio would be 1:2 considering the styrofoam wouldn't sink without the brick. So, it would seem that the brick must have a greater volume.

Originally Posted by milski

Let the brick has volume V. That means that it will weight 1400 * V. Half of the styrofoam displaces water which weights the same, in other words 1400 * V. Since water has density of 1000 kg/m^3, the volume of the water is 1400*V/1000=1.4 V. That's only half of the volume of the styrofoam, then the volume of the styrofoam is 2*1.4V. The ration in question is 2.8V/V = 2.8, answer C.

I didn't do any math, but I thought about the question intuitively. I came up with milski answer.

Dear aspiringdoc...As much as I appreciate your input, the fact that you got it...without an explanation, it ain't helping.

Originally Posted by SaintJude

Unfortunately it is C. I'll bold the steps I don't understand. If someone could clarify, those I would be really grateful

Step-by-step solution:

The brick displace a volume of water equal to its weight.
The density of the break is 1.4 times of that of water, so it will displace 1.4 times of the volume displaced by water.

The whole system is floating. The buoyancy force is the volume displaced * water density. The weight is volume of brick * brick density. The weight and the buoyancy force are equal:

volume displaced * water density = volume of brick * brick density
volume displaced/volume brick = brick density/water density=1400/1000=1.4

Since this is only 1/2 the volume of the styrofoam (why are the equating the brick with the foam??), the full volume must be 2.8 times larger than the volume of the brick.

We just found that half of the volume of the styrofoam is 1.4 the volume of the brick. Then the whole volume of the styrofoam is 2.8 the volume of the brick.

Originally Posted by MedPR

I guess I can't type out what I'm thinking. It just seems to me that the ratio would be 1:2 considering the styrofoam wouldn't sink without the brick. So, it would seem that the brick must have a greater volume.

Keep trying to write down what you think - that should help you more than me just telling you that it's not right. How much the styrofoam sinks is related only to the volume of the styrofoam (that gives you the buoyancy force, based on how much it is submerged) and the weight of the brick.

Note that the volume of the brick does not come into play - you can have all sorts of volumes on top of the styrofoam, as long as they weight the same, the styrofoam will sink the same.

Originally Posted by SaintJude

Dear aspiringdoc...As much as I appreciate your input, the fact that you got it...without an explanation, it ain't helping.

You're right, sorry. I didn't want to keep repeating what had been stated. Density of water 1000 kg/m^3 and the bricks density is 1400 kg/m^3. specific gravity = denisty of object/density of water =1400/1000 = 1.4. Therefore, if the brick is placed on the massless styrofoam and the styrofoam is submerged by 1/2 then it must have double the specific gravity (2.8) of the brick.

Originally Posted by aspiringdoc09

You're right, sorry. I didn't want to keep repeating what had been stated. Density of water 1000 kg/m^3 and the bricks density is 1400 kg/m^3. specific gravity = denisty of object/density of water =1400/1000 = 1.4. Therefore, if the brick is placed on the massless styrofoam and the styrofoam is submerged by 1/2 then it must have double the specific gravity (2.8) of the brick.

So milski is explaining/deriving how you get to the specific gravity equation. And you are going straight to it, but you are both saying exactly the same thing.

I'm not sure, since it's just 1 problem, but I think I might have found the huge hole in my understanding of archimedes. I kept forgetting that volume*density = mass. Though milski keeps saying weight... Same difference since gravity is all the same though.

Originally Posted by MedPR

I'm not sure, since it's just 1 problem, but I think I might have found the huge hole in my understanding of archimedes. I kept forgetting that volume*density = mass. Though milski keeps saying weight... Same difference since gravity is all the same though.

Yes, I am being a bit sloppy there. density * volume is mass, the forces that we are talking about are weight. You have an extra g in the equations which cancels at the end but that's no excuse.

So when an object floats, the buoyant force is equal to the weight of the object? Since the styrofoam is massless, the buoyant force is zero?

But, once you add on the brick, the buoyant force now equals the weight of the brick. Since volume*density = mass..

1000*(volume of H2O displaced) = 1400*(volume of brick)

1.4=(volume of brick)/(volume of H2O displaced)

From there I'm stuck. What do you know from the fact that half of the styrofoam is submerged?

Originally Posted by MedPR

So when an object floats, the buoyant force is equal to the weight of the object? Since the styrofoam is massless, the buoyant force is zero?

But, once you add on the brick, the buoyant force now equals the weight of the brick. Since volume*density = mass..

1000*(volume of H2O displaced) = 1400*(volume of brick)

1.4=(volume of brick)/(volume of H2O displaced)

From there I'm stuck. What do you know from the fact that half of the styrofoam is submerged?

You are almost there.

The volume of H2O displaced is half of the volume of styrofoam - remember it's submerged halfway. So you have:

1.4=(volume of brick)/(0.5*volume of styrofoam)

Originally Posted by milski

You are almost there.

The volume of H2O displaced is half of the volume of styrofoam - remember it's submerged halfway. So you have:

1.4=(volume of brick)/(0.5*volume of styrofoam)

Thank you. I'm having trouble understanding the concepts of the three situations (floating, partially submerged, fully submerged).

Gotta do EK 1001 at some point to get this into my head.

I would try to have the concepts nailed before moving to any problems. But that's me. Good luck!

Floating - equilibrium between weight and buoyancy force.

0 % submerged - 0 buoyancy force.
100 % submerged - max buoyancy force, fluid density*volume*g

Anything in between is just the corresponding fraction of the max buoyancy.

Originally Posted by milski

I would try to have the concepts nailed before moving to any problems. But that's me. Good luck!

Floating - equilibrium between weight and buoyancy force.

0 % submerged - 0 buoyancy force.
100 % submerged - max buoyancy force, fluid density*volume*g

Anything in between is just the corresponding fraction of the max buoyancy.

How can there be 0 buoyant force if the object is floating? What force is responsible for a piece of paper not sinking into a dish of water?

Originally Posted by MedPR

How can there be 0 buoyant force if the object is floating? What force is responsible for a piece of paper not sinking into a dish of water?

I never said that it would float if the buoyant force is 0. You vary the buoyancy force by changing how much the body is submerged in the fluid. Ideally, you'll be able to establish equilibrium between it and the weight and the body will float. If it's already floating, the equilibrium is already established.

The paper - most likely surface tension. A whole different story, so let's not go there.

Originally Posted by milski

I never said that it would float if the buoyant force is 0. You vary the buoyancy force by changing how much the body is submerged in the fluid. Ideally, you'll be able to establish equilibrium between it and the weight and the body will float. If it's already floating, the equilibrium is already established.

The paper - most likely surface tension. A whole different story, so let's not go there.

Huh? What's the difference between "0% submerged" and "floating"?

Originally Posted by MedPR

Huh? What's the difference between "0% submerged" and "floating"?

Floating means that the body is in equilibrium, it's not raising, it's not sinking. Based on the body and the fluid, that will happen at a specific amount of the body being submerged - 1%, 5%, 99%, whatever.

%0 submerged just means that it's not displacing any water.