A pirates treasure chest has a rectangular top with a dimension of 0.5 m and 1 m

Question

A pirates treasure chest has a rectangular top with a dimension of 0.5 m and 1 m. It lies 10 m below the surface of the ocean. What is the magnitude of the force that acts on the rectangular top? (Assume that the atmospheric pressure is 1.0x105 Pa and the density of the sea water is 1000kg/m3). A pirates treasure chest has a rectangular top with a dimension of 0.5 m and 1 m. It lies 10 m below the surface of the ocean. What is the magnitude of the force that acts on the rectangular top? (Assume that the atmospheric pressure is 1.0x105 Pa and the density of the sea water is 1000kg/m3). A pirates treasure chest has a rectangular top with a dimension of 0.5 m and 1 m. It lies 10 m below the surface of the ocean. What is the magnitude of the force that acts on the rectangular top? (Assume that the atmospheric pressure is 1.0x105 Pa and the density of the sea water is 1000kg/m3).

Explanation / Answer

(F)Force acting will be sum of (F1)force due to atmospheric pressure and (F2)that due to water above it

F1=P*A=(1*10^5)*(0.5*1)=5*10^4 N=50000N

And

F2= density of seawater*volume of seawater above*g

Or F2=1000*(10*0.5*1)*10=50000N

F1+F2=50000+50000=F=100000N

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