Calculating Density/Specific Body Mass: Need Help?

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Poquinha
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If you have a body that weighs 200 N dipped in oil that has density equal to 0.8 g / cm³. This same body when immersed in water begins to weigh 60 N. How do I find the density / specific body mass?

Already tried using the formula of the apparent weight = actual weight - buoyant and not worked.

Where can I be missing someone help, please?
 
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Poquinha said:
If you have a body that weighs 200 N dipped in oil that has density equal to 0.8 g / cm³. This same body when immersed in water begins to weigh 60 N. How do I find the density / specific body mass?

Already tried using the formula of the apparent weight = actual weight - buoyant and not worked.

Where can I be missing someone help, please?
Is this a homework question? If so please follow the homework guidelines in the homework forum.

You seem to understand that the body displaces 140N worth of water. Next you need to find the volume that of that weight of water. That gives you the volume of the body. Once you have that, you can find the weight of the displaced oil, followed by the apparent weight in oil.
 
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Jeff Rosenbury said:
You seem to understand that the body displaces 140N worth of water. Next you need to find the volume that of that weight of water. That gives you the volume of the body. Once you have that, you can find the weight of the displaced oil, followed by the apparent weight in oil.
Jeff, as I read the question, 200 N is the apparent weight of the object when immersed in oil.

apparent weight (in oil) = actual weight - buoyant force (of oil)

And 60 N is the apparent weight in water

apparent weight (in water) = actual weight - buoyant force (of water).

By itself, that is not enough to yield a solution. (Two equations and three unknowns). But what if one could relate the buoyant force from the oil and the buoyant force from the water in some way?
 
jbriggs444 said:
Jeff, as I read the question, 200 N is the apparent weight of the object when immersed in oil.

apparent weight (in oil) = actual weight - buoyant force (of oil)

And 60 N is the apparent weight in water

apparent weight (in water) = actual weight - buoyant force (of water).

By itself, that is not enough to yield a solution. (Two equations and three unknowns). But what if one could relate the buoyant force from the oil and the buoyant force from the water in some way?
You may be correct. Or there may be a missing comma somewhere. I assumed the comma thing since it leads to a solution.
 
If you assume the comma after the 200 N, the mention of oil does not make sense anymore.
You don't need to assume it to have a solution.
 
Clearly I misunderstood the problem. I assumed the 200N was in air.

If the 200N is in oil, then the Volume (V) times the Density (ρ) = mass (m). (102g ≈ 1N on earth.)

So:
ρbdyV/102 - ρwtrV/102 = 60N, and
ρbdyV/102 - ρoilV/102 = 200N.

ρwtr = 1 g/cc. ρoil = 0.8 (given).
ρbdy is unknown; V is unknown.

That's 2 equations and 2 unknowns.
 
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Your equations seem to be dimensionally wrong. Assuming that you divide by 102 g. Density times volume is mass. And you divide by mass. So this cannot be the Newtons on the right hand side.
And they miss one parenthesis each.
 
nasu said:
Your equations seem to be dimensionally wrong. Assuming that you divide by 102 g. Density times volume is mass. And you divide by mass. So this cannot be the Newtons on the right hand side.
And they miss one parenthesis each.
That's 102 g/N, so dimensionally:
((g/cc)(cc))/(g/N) = N.

Thanks for the catch on the extra ")". I edited them out.
 
Oh, so it's just an original way to write 1/g where g is the gravitational acceleration. :)
Usually you write the weight as W=ρVg.
 
nasu said:
Oh, so it's just an original way to write 1/g where g is the gravitational acceleration. :)
Usually you write the weight as W=ρVg.
Sorry, g is grams. There are about 102 grams per Newton for g(gravitational acceleration)=9.8m/s2. Or at least that's what some random internet site said.