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BUOYANCYArchimedes Principle11

Archimedes Principle Any body immersed in a fluid is acted upon by an upward force (buoyant force) which is equal to the weight of the displaced fluid2

3Procedure in solving BuoyancyFor homogeneous solid bodies of volume V floating in homogeneous fluid at rest.

If the body of height H has a constant horizontal cross-section are such as vertical cylinders, blocks, etc.:

Identify the forces acting and apply conditions of static equilibrium.

4Procedure in solving BuoyancyIf the body id of uniform vertical cross-sectional area A, the area submerged As is:

Identify the forces acting and apply conditions of static equilibrium.

5A spherical body has a diameter of 1.5 m, weighs 8.5 kN, and anchored to the sea floor with a cable as shown in the figure. Calculate the Tension of the cable when the body is completely immersed, Assume sea-water = 10.1 kN/m3

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CE Board Nov. 1977An iceberg having specific gravity of 0.92 is floating on a salt water of sp.gr. 1.03. If the volume of ice above the water surface is 1000 cu. m., what is the total volume of the ice.8

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A cylindrical buoy 0.6 m in diameter and 1.8 m high weighs 205 kg. It is moored in salt water to a 12m. Length of chain weighing 12 kg/m of length. At a high tide, the height of buoy protruding above water is 0.84 m. Density of steel and salt water are 7790 kg/cu.m. and 1030 kg/cu.m.What is the depth of water during high tide?What could be the height of protrusion of the buoy if the tide dropped 2.1 m.?How far is the bottom of the buoy from the ground at low tide?100.6 m0.84 m0.96 mhWc = 205 kgFB

110.6 m0.84 m0.96 mWc = 205 kgFB0.6 mWc = 205 kgFBWhen the tide drop to 2.1 m2.1 m

1.8 -

12When the tide drop to 2.1 m

0.6 mWc = 205 kgFB2.1 m1.8 -

13When the tide drop to 2.1 m

0.6 mWc = 205 kgFB2.1 m1.8 -

14The End15