You are out fishing, when you go to reach for the cooler, when you clumsily knoc

Question

You are out fishing, when you go to reach for the cooler, when you clumsily knock it into the water. The cooler is in the shape of a cube with side-length 0.2 meters. The cooler plus sandwiches inside weight 6 pounds (26.7 Newtons). How deep below the water's surface is the bottom of the cooler (how big is L in the figure; L is not the cooler's side-length), assuming it floats stationary with the top and bottom level with the water's surface (not tilted)? When the cooler was just sitting on the boat, there was actually a buoyant force on it from air (there is a buoyant force from air on any object in our atmosphere). How big is the buoyant force of air on the cooler? How many times smaller is that force than the weight of the cooler? Assume you are at sea level where the density of air is 1.225 kg/m^3

Explanation / Answer

here,

side of cooler , a = 0.2 m

weight of cooler ,W =26.7 N

a)

for the length L

Using archimedes principle

Buyoant force = weight of water displaced

weight of cooler = weight of water displaced

26.7 = 9.8 * 0.2^2 * L * 1000

L = 0.0681 m

the length L is 0.0681 m

b)

Buyoant force due to atmosphere , FB = Volume of cooler * pair * g

Buyoant force due to atmosphere , FB = 0.2^3 * 1.225 * 9.8

Buyoant force due to atmosphere , FB = 0.096 N

the Buyoant force due to atmosphere , FB is 0.096 N

W/FB = 26.7/.096

W/FB = 278

this force is 278 times smaller than the weight of cooler

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