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3 years ago

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A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show

that buoy does not float with its axis vertical. (II). What minimum pull should be applied to a chain attached to the center of the base to keep the buoy vertical?

1 answer:

aleksandrvk [35]3 years ago

4 0

Answer:

$GM$

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter $d=2m$

Length $l=2.5m$

Weight $W=22kN$

Specific weight of sea water $\mu= 10.25kN/m^3$

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

$W=(pwg)Vd$

Where

$V_d=\pi/4(d)^2y$

Therefore

$22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m$

Therefore

Center of Bouyance B

$B=\frac{y}{2}=0.26m\\\\B=0.75$

Center of Gravity

$G=\frac{I.B}{2}=2.6m$

Generally the equation for\BM is mathematically given by

$BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\$

Therefore

$BG=2.6-0.476\\\\BG=0.64m$

Therefore

$GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\$

Therefore

$GM$

So the bouy does not float with its axis vertical

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3 years ago

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$C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(x_{2}+x_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(x_{3}+x_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(x_{4}+x_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(x_{5}+x_{6})(x_{5} y_{6} -x_{6} y_{5})]]$

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5 0

3 years ago