Parking on a lake

Classical Mechanics Level 2

Bob has made a wager and intends to win with a demonstration. The wager involves the ice surface that has formed on a large and deep lake. Bob is not aware of how thick the ice may be on the lake surface but knows that the ice forms very uniformly and will be of similar thickness everywhere on the surface of the lake. Firstly, Bob walks out onto the ice to a location where an ice fisher has drilled a hole through the ice. The holes happens to be hundreds of meters form the nearest shore. Bob carefully measures the vertical distance from the top of the ice surface to the water surface in the fisher’s hole. The top of the ice surface is found to be ~ 15mm above the surface of the water in the hole. Bob marks a circular ring on the ice surface 10 meters in radius around the hole. Bob promptly goes to the lake shore and returns with his pickup truck (2500 kg). Bob parks his truck on the frozen lake. The truck is carefully located so that the center of mass of the truck is directly above the hole that was drilled through the ice. Now to win the wager Bob takes a chain saw and cuts a vertical slit through the ice into the water below. He makes the cut, completely around the truck’s location on his mark at a 10-meter radius. The truck is now centered on a 20-meter diameter disk of ice that has been completely separated from the other ice on the lake.

What will the outcome of this demonstration?

Assumptions and givens

The lake water specific gravity is 1.00 kg / liter.

The truck mass is 2500 kg

The water is deep.

The original vertical distance from the ice upper surface to the water in the hole is ~15 mm.

The ice on the lake is ridged enough that the ice disk remains nearly flat with negligible distortion due to the truck’s mass sitting on the ice disk.

The ice on the lake is elastic and flexible enough to allow movement over large distance. i.e. the ice surface may move up or down at location away from the shore with no support contributed by the distant ground surfaces.

The truck would break through the ice before it arrived at the location of the hole. The vertical distance from the top of the ice to the water will remain at ~ 15mm The truck and the ice disk will sink to the bottom of the lake The vertical distance from the top of the ice to the water will be ~3 mm The vertical distance from the top of the ice to the water will be ~ 11mm There is insufficient information to determine the outcome The vertical distance from the top of the ice to the water will be ~7 mm

1 solution

Darryl Dennis
Dec 26, 2020

The mass of the pickup truck is 2500kg. In order to float a mass, the same mass of water must be displaced.

To float the disk with the truck sitting on it an additional mass 2500kg of water must be displaced.

$Water\quad \frac { kg }{ { m }^{ 3 } } =1000$

water volume displaced to float the mass of the truck

${ 2500kg }_{ w }=2500/1000=2.5{ m }^{ 3 }$

surface area of the ice disk 20 meters in diam.

$Area=\pi { r }^{ 2 }=314.{ m }^{ 2 }$

Vertical displacement of the ice disk to achieve 2.5 cubic meters

$\frac { 2.5 }{ 314 } =\quad .0079m$

The ice disk will be forced down by approximately 8 mm to displace 2500 kg of water.

$15-8=7mm$

the truck will remain sitting on the floating ice disk with approximately 7 mm of vertical distance from the top of the ice to the water

PS with a bit more information the ice thickness can be calculated.

ice has a density of approximately 92% of water

$thickness\quad =\quad \frac { { h }_{ above\_ water } }{ { portion }_{ above\_ water } } =\frac { .015m }{ .08 } =.188m$

The ice is about .19m thick

Parking on a lake

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