• Support PF! Buy your school textbooks, materials and every day products Here!

Bouyant force and apparent weight

  • #1

Homework Statement




This question is continued from a problem that I have already solved. This is what I know from the previous problem that will be helpful:

r= 0.20 m
roh of solid sphere= 850 kg/m^3
mass sphere= 3.94 x 10 ^-5 kg

I also have 2 apparent weights that I solved for when the tension in the system is
a. 150 N ----> Normal force (apparent weight) Fn = -1.49 x 10^2
b. 0 N -----> Fn = -3.86 x 10^-4 N

(Hopefully those are correct)


Now- the problem I came for help on:

A room is partially filled with water with a sphere in it and the sphere is on a scale. What percentage of the sphere must be immersed in the water in order to achieve the same apparent weights as above (a and b).
And- What percentage of the sphere must be immersed to achieve an apparent weight of zero?


Homework Equations



% immersed = V immersed/ V total = roh obj/roh fluid
Bouyant force (Fb) = roh x V x g
Fn = mg + ma (a = O)

And from a full body diagram I have this:

Fn + Fb - mg = ma (whch is zero)



The Attempt at a Solution



Well....I really am not sure what to do. One fact that I know and I think will help me is that if an object is say 90% underwater, then the roh of the object is 90% of the fluid....

I need a kick in the right direction please.
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
Fn + Fb - mg = ma (whch is zero)
Start with this, the 3 forces acting on the sphere. Fill in the detailed formula for each force.
 
  • #3
Ok- here's what I got, but Im not sure what this implies...

mg + roh x V x g -mg = 0

mg cancels and I end up with roh x V x g = 0
 

Related Threads on Bouyant force and apparent weight

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
10K
  • Last Post
Replies
3
Views
9K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
14
Views
21K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
Top