AAMC 4..buoyancy question

[shaqkillz451]

I don't understand the explanation to question 18 on the physical science.

An object with 15 grams mass is immersed in benzene and suffers an apparent loss of mass of 5 grams. What is the approximate specific gravity of the object? (Data: Specific gravity of benzene = 0.7).

I know that if i could find the density of the object, I could do this. But i'm not sure how to find the density.

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[slyloxy]

I don't understand the explanation to question 18 on the physical science.

An object with 15 grams mass is immersed in benzene and suffers an apparent loss of mass of 5 grams. What is the approximate specific gravity of the object? (Data: Specific gravity of benzene = 0.7).

I know that if i could find the density of the object, I could do this. But i'm not sure how to find the density.

Draw force diagram easier to see
mass of object pointing down buoyancy force pointing up

Find the buoyancy force
B-Wo=5
B=5+15= 20g
Let V volume of object and whole object is immersed in benzene
V* 0.7= mass of benzene displaced=20g
V=20/0.7=200/7 cm3
density of object =m/v=15/(200/7)=0.525=specific gravity

 

[axp107]

I didn't know how to do this problem either.. what does "apparent loss of mass" mean

 

[RPedigo]

I didn't know how to do this problem either.. what does "apparent loss of mass" mean

It means that if you put it on a scale dry, it weighs more than if it's immersed in benzene. The benzene supplies a buoyant force that reduces the necessary normal force on the object (if it's sitting on a scale), thus making it's "apparent mass" smaller.

 

[axp107]

I think slyloxy did it wrong.. thats not the answer

If the Buoyant force is = to mg... isn't it just the "normal" force of the liquid? For most problems I've done so far on buotancy, you just set mg = to the Buoyant force and solve for unknown variables..

I guess since I don't understand the concept, I'm having a harder time doing this problem

 

[xlr8]

I think slyloxy did it wrong.. thats not the answer

If the Buoyant force is = to mg... isn't it just the "normal" force of the liquid? For most problems I've done so far on buotancy, you just set mg = to the Buoyant force and solve for unknown variables..

I guess since I don't understand the concept, I'm having a harder time doing this problem

that is true when there no net acceleration, but an apparent weight means the buoyant force isn't large enough to keep the object from sinking further (mg is larger than Fb), so you are solving for another force that is applied upwards so there is no net acceleration: A + Fb = mg

they say the apparent mass suffers a 5 gram loss, so im assuming they are indirectly telling you the buoyant force

(.7)Vg=5g, V = 50/7
mg = (Pobject)Vg 15 = (x)(50/7)
solving from there gets 2.1 i think

 

[turquoise_water]

since the apparent weight loss is (5 ) * 10 = 50 N

The weight of the object is (15) * 10 = 150 N , which is 3 times the apparent weight loss, therefore, its specific gravity, and density will be 3 times that of Benzene.

3 * 0.7 =2.1

i hope this helped, this is how EK does it .

 

[slyloxy]

since the apparent weight loss is (5 ) * 10 = 50 N

The weight of the object is (15) * 10 = 150 N , which is 3 times the apparent weight loss, therefore, its specific gravity, and density will be 3 times that of Benzene.

3 * 0.7 =2.1

i hope this helped, this is how EK does it .

yeah u guys are right, I miscalcalated forgot the 5g is buoyant force and not the net force on the object when immersed. thanks guys

 

[BrokenGlass]

I don't understand the explanation to question 18 on the physical science.

An object with 15 grams mass is immersed in benzene and suffers an apparent loss of mass of 5 grams. What is the approximate specific gravity of the object? (Data: Specific gravity of benzene = 0.7).

I know that if i could find the density of the object, I could do this. But i'm not sure how to find the density.

I was only gone for a day and I no longer feel in control of this board. Everybody seems to be posting whatever they wish. That's just unacceptable! Anyhow, here is my solution:


Actual weight of the oject is 0.015 N

Apparent weight of the object is 0.010 N (in benzene)

So the buoyant force (Fb) = 0.015 N - 0.010 N = 0.005 N (in benzene).

Obviosly this object sinks in benzene (if it floated, apparent weight would be zero).


When object sinks, p object / p fluid = Weight / Fb (Courtesy of Berkeley Review)


p object / p benzene = 0.015 N / 0.005 N = 3


=> p object = 3 * p benzene


We are given that p benzene / p water = 0.7

=> p water = p benzene / 0.7


We want to know p object / p water.


p object / p water = (3 * p benzene) / (p benzene / 0.7) = 2.1

 

[RPedigo]

I was only gone for a day and I no longer feel in control of this board. Everybody seems to be posting whatever they wish. That's just unacceptable! Anyhow, here is my solution:


Actual weight of the object is 0.015 N

Apparent weight of the object is 0.010 N (in benzene)

Might want to check your decimal places... it's 0.015kg, not 0.015N.


But other than that, BrokenGlass has the correct solution, since the ratios are going to be the same even with the decimal mixup.

 

[BrokenGlass]

Might want to check your decimal places... it's 0.015kg, not 0.015N.


But other than that, BrokenGlass has the correct solution, since the ratios are going to be the same even with the decimal mixup.

True, true. As is always the case with the mistakes I make, they LITERALLY tend to cancel each other out.

 

[axp107]

I get the rest of the problem, but how do you figure out that the 5g force is the buoyant force? And not 10g? Can someone explain this without math calculations


What forces are acting on the block?

The force of gravity, the buoyant force.. anything else?

so Fg = (0.015) * 10 pointing down

 

[RPedigo]

I get the rest of the problem, but how do you figure out that the 5g force is the buoyant force? And not 10g? Can someone explain this without math calculations


What forces are acting on the block?

The force of gravity, the buoyant force.. anything else?

so Fg = (0.015) * 10 pointing down

So before it weighed 15g (or 0.15N). In benzene, it weighed 5g less, or (0.05N less). The only way that it can weigh less in this case is if there's an additional force pointing up that accounts for that 0.05N difference. The only force that can point up in this case is the buoyant force.

 

[axp107]

So if we were to draw a force diagram of the mass "in" benzene.. would it look like this:

0.15 N down, 0.05 N up... so does this mean the object is accelerating downwards

 

[BrokenGlass]

So if we were to draw a force diagram of the mass "in" benzene.. would it look like this:

0.15 N down, 0.05 N up... so does this mean the object is accelerating downwards

The object is sitting at the bottom of the container. When we talk about "apparent weight, " we usually mean the "normal force." The buoyant force "helps" normal force to counteract the weight, so the normal force doesn't have to be as large as it would have been had there been no buoyant force. So your "apparent" weight is less than your "real weight."

When you step on a scale every morning (ok, only girls do that, but bear with me here AND assume you are not a girl), the scale is displaying the normal force, which happens to equal your weight because the only 2 forces acting on you in that scenario are your weight and the normal force, and they are equal in magnitude and opposite in direction.

I don't know if this made any sense...

 

[slyloxy]

The object is sitting at the bottom of the container. When we talk about "apparent weight, " we usually mean the "normal force." The buoyant force "helps" normal force to counteract the weight, so the normal force doesn't have to be as large as it would have been had there been no buoyant force. So your "apparent" weight is less than your "real weight."

When you step on a scale every morning (ok, only girls do that, but bear with me here AND assume you are not a girl), the scale is displaying the normal force, which happens to equal your weight because the only 2 forces acting on you in that scenario are your weight and the normal force, and they are equal in magnitude and opposite in direction.

I don't know if this made any sense...

it makes perfect sense. good job broken glass