Pulling my hair out over this one

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The discussion revolves around calculating the buoyant force on a mass that displaces 352.8 ml of water. The key formula used is the buoyant force (Fb) equals the weight density of the fluid (water) multiplied by the volume of displaced fluid. Participants clarify that the density of water can be expressed as 1 g/cm³ in SI units, simplifying the conversion process. After some confusion over unit conversions, the final calculation yields a buoyant force of approximately 3.5 N. The conversation highlights the importance of understanding unit conversions in physics problems.
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1. An object with a mass of 56 g displaces 352.8 ml of water when it is completely immersed. What is the buoyant force on the mass?

Notes:

* Report your answer in Newtons and use N as the unit.
* Report your answer to one decimal place.
* Acceleration due to gravity = 9.8 N/kg




2. Fb = Dw(fluid) * V(displaced fluid)




3. Dw of water = 62.4lb/ft^3; V of displaced fluid is 352.8cm^3 . I can't figure out how to get them into the same unit. It has to be in pounds to convert to N for the answer.
 
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You shouldn't have to put anything *into* pounds to convert to N, since Newtons are SI derived.

I presume by Dw of water you mean its density? In which case, it's easy, since that's how the SI unit of mass is derived! 1 cm^3 of water = 1 g
 
e.bar.goum said:
You shouldn't have to put anything *into* pounds to convert to N, since Newtons are SI derived.

I presume by Dw of water you mean its density? In which case, it's easy, since that's how the SI unit of mass is derived! 1 cm^3 of water = 1 g

Dw is the weight density of water, straight out of my book and it is represented as 62.4ft^3.

What does 1 cm^3 = 1 g have to do with this? I seriously need some guidance before I lose my mind. I am taking an online class with a non existent instructor to answer questions while working full time in the military and 2 school age kids who need help and attention. I have been staring at my computer, books, notes for over 3 hours trying to dissect this and I am out of patience. I just need some help!
 
The problem statement is dealing with units of grams and milliliters. These are SI units. It's either malicious or silly to give you the density of water in imperial units. The density of water in SI is 1 g/cm3, or if you prefer, 1 g/ml, or 1 kg/L, or even 1000 kg/m3.
 
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I am trying to help you! By "1 cm^3 = 1 g" I meant that 1 cubic centermeter of water has the mass of one gram. That is, the density of water can be given by 1g/cm^3.

So, you have two options, you either use the canonical density of water, 1g/cm^3, or convert the density of water in imperial units, 62.4 lb/ft^3 (oh, you crazy Americans!) to density in metric units. To do this, you need to know the conversion factors between ft and cm and lb and kg. Frankly, you'll have to look that up yourself, as I don't keep those numbers in my head (Ok, I do, 2.54 cm = 1", 28.35 (ish) g = 1 oz).

Remembering the metric density of water is easy though, 1g/cm^3, none of this imperial nonsense!
 
Well, that makes a lot more sense now that I see what you mean by g. Not g as in gravity but g as in gram. I still don't see the Wd of water in g/cm^3 though. Maybe I am too tired or just so overwhelmed with this.
 
Sorry to confuse.
 
No worries, I appreciate your efforts.
 
Cheers.

Do you get it now?
 
  • #10
Got it! 352.8cm^3 = .000353m^3. Wd of water is 1000kg/m^3.
1000kg/m^3*.000353m^3=.353kg
1kg = 9.8 N so 9.8*.353=3.459

Thank you. You got me pointed in the right direction.
 
  • #11
Cool! Well done.
 
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