Floating on Ice: Calculating Volume Needed to Stay Afloat

Nanu Nana
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Homework Statement


A person (with mass 60.0 kg) is located on a volume of ice, floating on the water. Calculate the smallest volume of the ice so that the person would remain above the water ( ice density = 917kg / m3)[/B]

Homework Equations


F= m xg
Archimedes F = density x V xg [/B]

The Attempt at a Solution


If the object is floating then we know that F(g) = F(a)
therefore
m xg = d x V x g
g cancels out
m = d x v
60kg= d (ice) x V
V= 60kg / 917 kg/m3
V = 0.065 m 3
But the answer should be 0.72 m3 [/B]

Thank you :)
 
on Phys.org
Should the force of gravity include the weight of the ice as well as the person?

Also, are you sure you want to use the density of ice when finding the buoyant force?
 
Last edited:
TSny said:
Should the force of gravity include the weight of the ice as well as the man?

Also, are you sure you want to use the density of ice when finding the buoyant force?
You have to use density of water ?
 
Nanu Nana said:
You have to use density of water ?

Can you state Archimedes' principle in words?
 
TSny said:
Can you state Archimedes' principle in words?
Its a principle that states that a body immersed in a fluid is buoyed up by the force which is equal to the weight of displaced fluid
 
OK. What type of fluid is being displaced?
 
TSny said:
OK. What type of fluid is being displaced?
Water ?
 
Yes. But instead of typing "water ?", you should type "water!":smile:
 
Ok water!
 
  • #10
Good. So, how would you calculate the weight of the fluid displaced?
 
  • #11
W= rho x V x g
 
  • #12
OK. What density would you use here?
 
  • #13
1000 kg/ m3
 
  • #14
Yes (water). And what volume V should be used to find the weight of the fluid displaced?
 
  • #15
0.0654 m3
 
  • #16
I wasn't clear. When calculating the weight of fluid displaced, should you use the volume of the ice, the volume of the person, or the total volume of both?
 
Last edited:
  • #17
Total volume of both
 
  • #18
Nanu Nana said:
Total volume of both
We don't want the person to get wet.
 
  • #19
The volume of ice then
 
  • #20
Right. So you have that the buoyant force acting upward on the system is ρwater Vice g.

How does this force relate to the weights of the person and the ice?
 
Last edited:
  • #21
Aren't they equal?
 
  • #22
Could you state precisely what should be equal?
 
  • #23
Fz = F( A)
 
  • #24
What is the total downward force acting on the system ?
 
  • #25
F= mass x a
60.0 kg x 9.81 m/s2=588.6 N
 
  • #26
OK. That's the downward force of gravity on the person. Is there any other downward force acting on the system of person and ice?
 
  • #27
On ice aswell
 
  • #28
Exactly. How would you express the force of gravity on the ice in terms of the volume of ice and the density of ice? Just in symbols.
 
Last edited:
  • #29
TSny said:
Exactly. How would express the force of gravity on the ice in terms of the volume of ice and the density of ice? Just in symbols.
F=m xg = density xVx g
 
  • #30
Good. To be clear. let's write this as ρice Vice g.

You know that the total downward force on the system must equal the total upward force (since the system is floating). How would this look as an equation?
 

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