Archimedes Principle: Calculating Relative Density for a Block of Oak and Lead

Nanu Nana
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. Homework Statement
A block oak weighs 90 N in air. A piece of lead weighs 130 N in water. The two weigh together 100N in water. Calculate the relative density of the wood.

Homework Equations


F=mxg
F(archimedes)= density x Volume x g

The Attempt at a Solution


F (g)oak = 90 N
F= mxg
m= 90N/9.81 m/s2
m= 9.17 gram
Oak in water = 130-100 N =30 N
Farchimedes oak = 90N-30N =60N
60N =1000kg/m3 x V x 9.81 m/s2
V=60N/9810
V=0.00611
Now that I have calculated V I can find density
rho = 9.17kg/ 0.00611 m3
density = 1499
Relative density is 1499 /1000 kg/m3 = 1.499
But the correct answer is 0.75
 
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Check your signs. In water, oak+lead=100N, while lead =130N.
 
How to do find N Force of oak in water ??
 
Nanu Nana said:
How to do find N Force of oak in water ??
Your method was ok but you made a mistake with the signs. Manipulate the two equations I wrote above.
 
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100+130?
 
Nanu Nana said:
100+130?
Please, don't make wild guesses.
If Ow+Lw=100 and Lw=130, what is Ow?
 
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Sorry /D its -30 so 90-(-30) = 120
 
Nanu Nana said:
Sorry /D its -30 so 90-(-30) = 120
Right. Does this get you to the right answer?
(How do you interpret the weight of -30N?)
 
haruspex said:
Right. Does this get you to the right answer?
(How do you interpret the weight of -30N?)
Yes I just calculated it. Negative force means opposite direction . But i don't know in this context
 
  • #10
Nanu Nana said:
Yes I just calculated it. Negative force means opposite direction . But i don't know in this context
If the wood were immersed in the water without being bound to the lead, what would happen?
 
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  • #11
It floats ofcourse
 
  • #12
Nanu Nana said:
It floats ofcourse
Right, hence the negative net force.
 
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  • #13
haruspex said:
Right, hence the negative net force.
Oh I see . thank you very much :)
 

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