a circular lid on top of a closed tank is 3 ft in diameter and wighs 250 lbs. an

Question

a circular lid on top of a closed tank is 3 ft in diameter and wighs 250 lbs. an open manometer tube is tapped into the side of the tank, and the water rises in the tube to a height of 26 ft, as shown. If a large block of concrete is suspended from the lid to hold it in place, what is the required volume of the block if the specific weight of concrete is 150 lb/ft^3 ?

(I apologize for the inverted image, but you get the idea.) I understand that certain equations involving bouyant force are involved but cant seem to grasp them off the top of my head. Thanks

Explanation / Answer

In order to hold the lid in its place, the net force acting on the lid should be zero.
=> Downward acting force = Upward acting force
=> Weight of lid + weight of concrete = Buoyant force + hydro-static force on the lid

=> 250 + V*150*32.2 = V*()*g + (h)**g*A

=> 250 + V*150*32.2 = V*(62.4)*32.2 + (26 - 6)*(62.4*32.2)*(3.14*1.5^2)

=> V*(4830 - 2009.28 ) = (20*62.4*32.2*3.14*1.5^2) - 250 = 283805.268

=> V = 283805.268 / 2820.72 = 100.61 ft^3.

Where, V is the required volume of the block.

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