How can you use pressure to rescue a friend who has fallen through ice?

[Zfool]

Hey guys my first homework question I'll be asking here... as simple as it may sound this question is kinda stupid and pissed me off so here goes...

Your friend has fallen through the ice and you decide to attempt a rescue. Having just come from a physics class on pressure, how will you do this?

Now, the knowledge I have about pressure in liquids regarding hydrostatic pressure is that pressure obviously depeds on the depth of fluid involved, and that pressure is a scalar quantity. and I know that you would feel a greater hydrostatic pressure at the BOTTOM of the pool than we do near the SURFACE. In this unit we were only taught these equations: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/PressureGifs/Pressure06.gif
where P(gauge) must be rho*gd (pgd), therefore leading to P(gauge) = P(abs)-P(atm)

I'd appreciate any help/clues to effectively write the solution... I do not know how detailed I should be explaining this. Again, let me ask you, doesn't this qusetion sound stupid as hell? :P yeah I thought so too...

 

[Bystander]

"... friend has fallen through the ice ..." Think about how your friend might have fallen through the ice, then think about the definition of pressure.

 

[Zfool]

"... friend has fallen through the ice ..." Think about how your friend might have fallen through the ice, then think about the definition of pressure.

hmm I do know the definition of pressure... but man the ice... is the answer a lot simpler than what I'm thinking through?... I mean haha the only reason why I think he fell through the ice is because his weight is greater and the buoyant force couldn't hold him up or something, or that he was just a fatass

...hmmm more clues maybe? I'm almost done thinkiing through this question...

 

[Bystander]

"hmm I do know the definition of pressure"

State it, please. At least to yourself.

 

[Zfool]

...pressure: is the magnitude of the normal force per unit surface area, (mg/v), where m/v = p(roh)g and Pressure difference involving depth being = pgd

 

[Zfool]

I think I got it, correct me if I'm wrong...

I would lay flat on the ice surface and slide over to him to rescue or something... because to minimize the normal force acting on the surface, equalizing the pressure. The word 'normal force' was what you're wanting me to refer to right?

 

[Bystander]

Pretty close. Your friend went through walking, presumably --- his mass times g equals the force he applied to the ice distributed over the area of his feet. You, to stay dry, flatten yourself on the ice, grab a sheet of plywood to spread the force over a larger area, wear skis or snowshoes, something to reduce the pressure you apply to the ice.