Passage VII TBR Fluids and Solids Ch7

[bdc142]

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Experiment: hanging mass in a liquid (liquid's density is variable). Mass is "balanced in such a way that it was flush with the top the fluid in a graduated cylinder". It is balanced by a string connected to a 10g pan with a counterweight.

For the fluid medium of carbon tetrachloride, the counterweight needed is 24.7 g, fluid density 1.49 g/mL

#45. What is the approximate density of the hanging mass?
A. 1.25 g/mL
B. 2.00 g/mL
C. 3.00 g/mL
D. 5.00 g/mL

So TBR solves it by doing F weight = F buoyant + F apparent weight (aka tension). What I don't get is for F apparent, they don't include the mass of the pan, but only 24.7. Can anyone explain why?

 

[hdub]

I was actually just doing this problem and was wondering the same thing as well as how they got the value of .05 g N for the F weight... ??

 

[cloak25]

For an object that sinks, W/B = density of object/density of fluid.
the mass has a weight, W=50g. In CCl4, its apparent weight is 24.7 so the bouyant force = 50-24.7 = 25.3.

W=50
B=25.3
density of fluid = 1.49

50/25.3 = density of object/1.49
2/1.5 = density of object = 3

 

[bdc142]

For an object that sinks, W/B = density of object/density of fluid.
the mass has a weight, W=50g. In CCl4, its apparent weight is 24.7 so the bouyant force = 50-24.7 = 25.3.

W=50
B=25.3
density of fluid = 1.49

50/25.3 = density of object/1.49
2/1.5 = density of object = 3

But shouldn'it its apparent weight be 24.7 (Counterweight) + 10 (mass of pan)?

 

[bdc142]

I was actually just doing this problem and was wondering the same thing as well as how they got the value of .05 g N for the F weight... ??

They mean .05 kg. N is just there to represent 10 m/s^2, but won't matter since it will get cancelled out

 

[milski]

But shouldn'it its apparent weight be 24.7 (Counterweight) + 10 (mass of pan)?

Yes, it should. Seems like a typo...
50 g seems like an arbitrary weight for the object that they chose. It will get cancelled, so it does not really matter. Using just a letter, like m, would have been cleaner.

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[ilovemedi]

Sorry to bring this up again, but why did they use the counterweight/fluid density of Carbon Tetrachloride? I get that the object sinks under all those other fluids (including Carbon Tetrachloride), and the Carbon Tetrachloride has the highest density and we should input that... but then shouldn't our answer be GREATER than 3.00g/mL (the answer we get when we use the Carbon Tetrachloride's density value)?

Also, I agree with @bdc142, shouldn't we add the weight of the pan (10g) to the counterweight, which equals total weight on the right side to calculate the Tension, which = Normal force?

This is all Passage VII on TBR Fluids/Solids lecture...

 

[inasensegone]

Sorry to bring this up again, but why did they use the counterweight/fluid density of Carbon Tetrachloride? I get that the object sinks under all those other fluids (including Carbon Tetrachloride), and the Carbon Tetrachloride has the highest density and we should input that... but then shouldn't our answer be GREATER than 3.00g/mL (the answer we get when we use the Carbon Tetrachloride's density value)?

Also, I agree with @bdc142, shouldn't we add the weight of the pan (10g) to the counterweight, which equals total weight on the right side to calculate the Tension, which = Normal force?

This is all Passage VII on TBR Fluids/Solids lecture...

First of all, the mass in the fluid was given as 50 grams. The 1st line of the data table says that the counterweight necessary when the fluid is air, is 50 grams. Therefore from this you should conclude that the counterweights as listed in the data table include the 10 grams mass of the pan.

Now, since you know that the counterweights as listed are inclusive of all the relevant weights, you can determine the density of the hanging mass. Again, the hanging mass is 50 grams. The reason they used Carbon tetrachloride, is because for that data value, the counterweight is 24.7 grams which is close to half the mass of the hanging mass (50). This makes the math simple. If you only need roughly half of the mass of the object as a counterweight, then the carbon tetrachloride must be providing a buoyancy force to effectively cancel out roughly half off the mass of the object (the object's apparent weight in the carbon tetrachloride is roughly half of its weight in air). This means that the object is roughly twice as dense as carbon tetrachloride ~ 3 g/mL

(Remember, the object displaces an amount of liquid equal to its volume once it is submerged. This means the fluid provides a buoyancy force as given: B = Volume fluid displaced * density of fluid * gravity.... which again is roughly equal to half the weight of the object, therefore the object is ~ twice as dense)

 

[ilovemedi]

First of all, the mass in the fluid was given as 50 grams. The 1st line of the data table says that the counterweight necessary when the fluid is air, is 50 grams. Therefore from this you should conclude that the counterweights as listed in the data table include the 10 grams mass of the pan.

Now, since you know that the counterweights as listed are inclusive of all the relevant weights, you can determine the density of the hanging mass. Again, the hanging mass is 50 grams. The reason they used Carbon tetrachloride, is because for that data value, the counterweight is 24.7 grams which is close to half the mass of the hanging mass (50). This makes the math simple. If you only need roughly half of the mass of the object as a counterweight, then the carbon tetrachloride must be providing a buoyancy force to effectively cancel out roughly half off the mass of the object (the object's apparent weight in the carbon tetrachloride is roughly half of its weight in air). This means that the object is roughly twice as dense as carbon tetrachloride ~ 3 g/mL

(Remember, the object displaces an amount of liquid equal to its volume once it is submerged. This means the fluid provides a buoyancy force as given: B = Volume fluid displaced * density of fluid * gravity.... which again is roughly equal to half the weight of the object, therefore the object is ~ twice as dense)

So we're just suppose to automatically assume that the counterweight listed includes the 10g of the pan? What was the point of even stating that the pan was 10g, then? Should we always assume this in these situations?

Also, just to make sure - Tension-mg=0... And Tension is equal to Normal Force only? Not Normal + Buoyant force?

THANK YOU so much for your help!

 

[inasensegone]

So we're just suppose to automatically assume that the counterweight listed includes the 10g of the pan? What was the point of even stating that the pan was 10g, then? Should we always assume this in these situations?

Also, just to make sure - Tension-mg=0... And Tension is equal to Normal Force only? Not Normal + Buoyant force?

THANK YOU so much for your help!

I wouldn't say that you should always assume that... but in this case they gave the mass of the counterweight necessary when the object was suspended in air, as 50 grams. From this you can conclude that it has to include the 10 grams, otherwise it would actually equal 60 grams, and then you wouldn't have a balance.

Well you don't necessarily have to think of tension in this problem... but I suppose you can. If you do, you have to remember that something is providing the tension... in this case, the counterweight is providing the tension. So what you would actually get is:

mg - tension - buoyancy force = 0

The mg is the weight of the 50 gram object.

The tension is essentially equal to the weight of the counterweight.

The buoyancy force is variable depending on the density of the fluid in which the object is submerged in. The denser the fluid, the greater the buoyancy force, and the smaller counterweight (tension) that is necessary.

You're welcome! 🙂 Hope this additional explanation helps

 

[ilovemedi]

Thanks

I wouldn't say that you should always assume that... but in this case they gave the mass of the counterweight necessary when the object was suspended in air, as 50 grams. From this you can conclude that it has to include the 10 grams, otherwise it would actually equal 60 grams, and then you wouldn't have a balance.

Well you don't necessarily have to think of tension in this problem... but I suppose you can. If you do, you have to remember that something is providing the tension... in this case, the counterweight is providing the tension. So what you would actually get is:

mg - tension - buoyancy force = 0

The mg is the weight of the 50 gram object.

The tension is essentially equal to the weight of the counterweight.

The buoyancy force is variable depending on the density of the fluid in which the object is submerged in. The denser the fluid, the greater the buoyancy force, and the smaller counterweight (tension) that is necessary.

You're welcome! 🙂 Hope this additional explanation helps

Gahhh I'm so stupid. It specifically says in the passage that the recorded values were from BOTH counterweight and mass of pan. Ack! Thanks! The problem also states that "Fb=Fweight - Fapparent weight" and "apparent weight" is usually normal force... so we can assume that Tension = normal force? I guess I got too detailed and started worrying about the little details. I should just go w/ more simple equations lol. Thank you!

 

[sillyjoe]

Would you have been able to solve this question by solving for the volume using the scenario in water? For example, the passage gives the counterweight needed for water as 33.1g. Couldn't you use that to figure out the buoyant force in the problem by solving for volume since you know the density of water is 1? (equation F = pVg). Then you can divide 50 grams (mass of object) by the volume it displaces in water? When I solve this out I get 2.95 as the answer.

It seems to me like this is a simpler way of getting to the answer. Is this a correct method though?

 

[Mortyblurb]

They mean .05 kg. N is just there to represent 10 m/s^2, but won't matter since it will get cancelled out

they mean .05g N. g is not supposed to be grams, but rather indicate gravity. The weight of the object is F=mg=(50grams)g= .05g Newtons. Hope that clears things up