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

Click For Summary
SUMMARY

The discussion centers on calculating the relative density of a block of oak and a piece of lead using Archimedes' Principle. The oak weighs 90 N in air and the lead weighs 130 N in water, with a combined weight of 100 N in water. The calculations reveal that the relative density of the oak is 1.499, but the correct relative density is 0.75, indicating a sign error in the calculations. The negative force interpretation is clarified, emphasizing the importance of understanding buoyancy and net forces in fluid mechanics.

PREREQUISITES
  • Understanding of Archimedes' Principle
  • Basic knowledge of weight and buoyancy
  • Familiarity with the formula F=mg
  • Ability to manipulate equations involving density and volume
NEXT STEPS
  • Study the concept of buoyant force in fluids
  • Learn about the calculation of density using the formula density = mass/volume
  • Explore the implications of negative forces in physics
  • Investigate the principles of floating and sinking in different fluids
USEFUL FOR

Students studying physics, particularly those focusing on fluid mechanics and density calculations, as well as educators seeking to clarify concepts related to buoyancy and Archimedes' Principle.

Nanu Nana
Messages
95
Reaction score
5
. 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
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: Nanu Nana
100+130?
 
Nanu Nana said:
100+130?
Please, don't make wild guesses.
If Ow+Lw=100 and Lw=130, what is Ow?
 
  • Like
Likes   Reactions: Nanu Nana
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?
 
  • Like
Likes   Reactions: Nanu Nana
  • #11
It floats ofcourse
 
  • #12
Nanu Nana said:
It floats ofcourse
Right, hence the negative net force.
 
  • Like
Likes   Reactions: Nanu Nana
  • #13
haruspex said:
Right, hence the negative net force.
Oh I see . thank you very much :)
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
695
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
11K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
6K