Need Help regarding some problems

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The discussion revolves around a user seeking assistance with physics problems related to buoyancy and hydraulic systems for extra credit. Key points include the calculation of weight density of seawater using buoyant force and volume, as well as determining the new floating depth of a boat after adding cargo. Participants emphasize the importance of understanding concepts rather than simply obtaining answers, encouraging the user to show their work for better guidance. The conversation highlights the necessity of proper unit management in calculations and the value of grasping underlying principles for problem-solving. Ultimately, the user expresses gratitude for the help and indicates a clearer understanding of the problems.
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Need Help regarding some problems...

Hi,
I have been trying to get these problems for some time now however at no prevail. If you guys can answer the questions below I promise I won't ever borther you guys again. Thanks a lot for those who do choose to help :shy:

1. The boat in the figure below has a weight (W) of 1400 N and floats at a depth of 8 m. The area of the hull is 60 m2.



(a) What is the weight density of the sea water in which it is floating ?

HINT : Consider the flotation principle
Rearranging the equation BF = WD x Volume,
WD = BF / volume ; volume = area x depth

(b) At what depth will the boat float if a 600 N cargo is added ?

HINT: Consider the flotation principle
Rearranging the equation BF = WD x Volume,
volume = BF / WD; and volume = area x depth
rearranging equation : depth = volume / area



2. Consider the hydraulic system below :


area in = 5 m2 ; area out = 80 m2

If a weight of 1600 N (Fout) must be lifted, how much force (Fin) must be applied to the input end ?

HINT - First determine what pressure must be provided at the output end to lift the weight of 1600 N.


3. A hot air balloon, with a mass of 14 kg., hoovers at an altitude of 8 km. where the weight density of the air is 2.8 N/m3. If a suitcase is dropped from the balloon, the balloon rises to an altitude where the weight density of the air is 2.1 N/m3. What is the weight of the suitcase ?

HINT: First determine the volume of the balloon; i.e., volume = BF / WD ; 'BF' is not 'mass'
Then using BF = WD x Volume, determine the new BF which can be used to determine the weight of the suitcase (i.e., this new BF is not equal to the weight of the suitcase)
 
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What have you tried? Where are you getting stuck? Those hints look like they should help a lot. Remember, we don't answer homework for you, we'll only help you if you truly make an effort to solve them first and show all your work so we can point out where any mistakes are.

Couple of hints:

1. they gave you enough already

2. Use mechanical advantage ratios.

3. To hover in a spot the balloon must ahve the same density as the air.
 
Look at the hints. Do you know what all the terms mean? Once you know that you should be able to handle the problems.
BF=bouyant force
WD=water density


PhysicsNoob88 said:
If you guys can answer the questions below I promise I won't ever borther you guys again.
If you were bothering us, we wouldn't be looking at your posts in the first place. We don't mind helping when your stuck.
 
Just to give you guys more information these are not HW problems, they are extra credit, to get an A in this class I have to get 102 on the next test. Each problem is with 3 extra credits so I need it badly. I would try and do it myself but I still have to read 4 chapters worth of material (100+ pages!) I really hate to ask for the answers but they are extra credit and I do need them badly. This is the last test which is why I mentioned I wouldn't bother you guys again. Thanks!
 
PhysicsNoob88 said:
Just to give you guys more information these are not HW problems, they are extra credit, to get an A in this class I have to get 102 on the next test. Each problem is with 3 extra credits so I need it badly. I would try and do it myself but I still have to read 4 chapters worth of material (100+ pages!) I really hate to ask for the answers but they are extra credit and I do need them badly. This is the last test which is why I mentioned I wouldn't bother you guys again. Thanks!
Well I for one will not be doing them for you. How do you have 4 chapters to read? You should be reading the material as you cover it in class. This helps you understand it better and you have less work to do when test time rolls around.
Having other people do your work for you is cheating and that is not the proper way to get an A. You are the one who earns your grades, not someone else.
If you were stuck on the problems and had actually tried them, I would help you, but I refuse to simply give you the answers. I'm sorry, but you're on your own for this test. best of luck. If you encounter any problems understanding any of your reading, feel free to post here and we'd be glad to help you. If you can tell us where you're getting confused we can help you understand it.
 
If you can't solve the problems, you don't deserve to get an A
 
You guys are right its wrong for me to ask for the answers, I have attemped to do problem 1(a) please tell me what you think of the answer I got below:

I used the hint given: WD = BF / volume ; volume = area x depth

I multiplyed the area (60 m2) by the depth (8m) to get 480 as the volume

Then I Got 1400N (I considered this the BF) and divided it by the volume 480.

The final answer I got the weight density of the sea water to be roughly 2.916..

Please tell me if I did it correctly, thanks! :approve:
 
The volume is in cubed meters. But your force is Newtons. So the result you're getting has WD defined in N per m3. But conventionally density is defined in terms of kilograms per m3. See this link.

P.S. What I am saying is, make sure your results have units, whether it is kg/m3 or N/m3.
 
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EnumaElish said:
The volume is in cubed meters. But your force is Newtons. So the result you're getting has WD defined in N per m3. But conventionally density is defined in terms of kilograms per m3. See this link.

P.S. What I am saying is, make sure your results have units, whether it is kg/m3 or N/m3.

So is the answer right? If i add kg/m3??
 
  • #10
I don't think so, never thought of any kind of missing acceleration?
 
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  • #11
PhysicsNoob88 said:
So is the answer right? If i add kg/m3??
Maxos is right, your answer is incomplete. A Newton is equivalent to kg m/s2, so your result is in N/m3 = (kg m/s2)/m3 = kg/(s2m2). But density is measured in kg/m3. So it looks like you are one step away from the answer. (Hint: how do you convert force into mass, or mass into force?)

At all times you should write numbers with their associated units. Instead of writing "1400", write "1400 N", even when (especially when) it is an input to a calculation. And always express your results in appropriate units, even in intermediate calculations. Keeping track of units is a great way to check for logical errors (or missing steps). You can multiply and divide units as you multiply and divide the associated numbers. For example: (1kg + 5kg)/(12m x 2m2) = 6kg/24m3 = 0.25 kg/m3.
 
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  • #12
Can someone PLEASE tell me how to start 1.(b)

I know I am supposed to use the forumulas but I can't seem to make sense of it all.

I know depth = volume / area

Is the depth going to be the same as 1(a)? that would make it 8m. However how do I find out the volume or the area, I tried to follow the hints but I can't seem to make sense of it, I am not asking for the answer but can someone tell me how to at least start the problem please?
 
  • #13
PhysicsNoob88 said:
I used the hint given: WD = BF / volume ; volume = area x depth

I multiplyed the area (60 m2) by the depth (8m) to get 480 as the volume

Then I Got 1400N (I considered this the BF) and divided it by the volume 480.

The final answer I got the weight density of the sea water to be roughly 2.916[N/m3, which was then converted to ... kg/m3 by converting force (N) into mass (kg) through this formula: ______ ]
Now you have a heavier boat, how much should you increase the volume so that you'll find the same WD (N/m3) that you found in (a)?
 
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  • #14
EnumaElish said:
Now you have a heavier boat, how much should you increase the volume so that you'll find the same WD (N/m3) that you found in (a)?

Well since 1400 N / 600 N would make it an 2.33 n/m^3 more. So would the depth then be simply 2.33 n/m^3 + my previous answer for (a)? :confused:
 
  • #15
What is the new weight for the boat, in Newtons?
 
  • #16
EnumaElish said:
What is the new weight for the boat, in Newtons?

Yes, so would the answer be 5.25 n/m^3?

If its not the right answer I have no idea what I am doing :cry:
 
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  • #17
Take a step back (and take deep breaths). Just fill in the blank:

After adding 600 N to its original weight of 1400 N the boat now weighs ______ Newtons.

(Hint: you need to add 1400 N and 600 N.)
 
  • #18
As far as I can understand, you are trying to solve part (a) again in part (b). In (b), you need to take WD (in N) that you calculated in (a) as a given and solve for the correct volume that would give you the same WD as in (a) using the new weight as the BF.
 
  • #19
EnumaElish said:
As far as I can understand, you are trying to solve part (a) again in part (b). In (b), you need to take WD (in N) that you calculated in (a) as a given and solve for the correct volume that would give you the same WD as in (a) using the new weight as the BF.

oh! I can't believe I missed doing that! I know how to solve it now! Thanks for all your help!
 
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