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kobusy [5.1K]
3 years ago
10

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?
Engineering
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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