Driving on the ice In Northern Minnesota and likely other pl

Driving on the ice. In Northern Minnesota (and likely other places as well) people will drive their ice houses out onto frozen lakes (please do not try this - it is terrifying and surprisingly dangerous if not done at the right time of the year). This habit seems problematic, especially if the ice were to give way. The goal of this problem is to see if there are any cars that could float (assuming the doors and trunk were sealed properly). We shall begin with the Volkswagen Jetta. According to Wikipedia it is 4.64 m long. 1.78 m wide, and 1.45 m tall. Determine whether or not the Jetta sinks in freshwater if it weighs 14.4 kN. What about a Toyota RAV4? It weighs approximately 3.360 lbs., is 147 inches long. 67 inches wide, with a height of 64.8 inches. As a first approximation, you can assume that each car is effectively a box.

Solution

Volkswagen Jetta : Weight = 14.4 kN

Volume = L*W*H = 4.64*1.78*1.45 = 11.97 m³

Weight of water displaced = Weight of Car = 14.4 kN

Volume of water displaced = Weight of water/ specific weight of water = 14.4 kN / 9.807 kN/m³ = 1.47 m³

Since volume of water displaced is less than volume of car, it will remain afloat.

Submersion Height (portion of car sinking) = Volume of water displaced /(L*W) = 1.47/(4.64*1.78) = 0.178 m

TOYOTA RAV4

Weight = 3360 lbf

Volume = L*W*H = 147*67*64.8 = 638215.2 in³ or 369.3375 ft³

Weight of water displaced = Weight of Car = 3360 lbf

Volume of water displaced = Weight of water/ specific weight of water = 3360 lbf /62.43 lbf/ft³ = 53.82 ft³

Since volume of water displaced is less than volume of car, it will remain afloat.

Submersion Height (portion of car sinking) = Volume of water displaced /(L*W) = (53.82*144)/(67*64.8) = 1.78 ft or 21.36 in