A metal block 2 ft square and 10 in. deep is floated on a body of liquid which consists of an 8-in layer of water above a layer of mercury. The block weighs 120 lb/ft3. What is the position of the bottom of the block?If a downward vertical force of 500 lb is applied to the center of this block, what is the new position of the bottom of the block?Assume that the tank containing the fluid is of infinite dimensions.
A metal block 2 ft square and 10 in. deep is floated on a body of liquid which consists of an 8-in layer of water above a layer of mercury. The block weighs 120 lb/ft3. What is the position of the bottom of the block?
If a downward vertical force of 500 lb is applied to the center of this block, what is the new position of the bottom of the block?
Assume that the tank containing the fluid is of infinite dimensions.
Given:
Metal block is a square block having a dimension of 2 ft.
Depth of block = 10 in
Depth of water = 8 in
Weight per unit volume of the block = 120 lb/ft3
By the Archimedes principle
Weight of block = Weight of total volume of liquid displaced
Let the depth of the block under the mercury be 'x'.
Thus,
Total weight of block =
The density of water = 62.4 lb/ft3
The density of mercury =
Therefore,
The position of the block from the water surface = 8 in + 0.83
= 8.83 in
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