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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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5 0
3 years ago
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mars1129 [50]

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3 years ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete
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Answer:

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Y=\frac {K}{\sigma_c \sqrt {a\pi}}

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6 0
3 years ago
Six housing subdivisions within a city area are target for emergency service by a centralized fire station. Where should the new
Marina86 [1]

Answer:

Explanation:

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To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.

C_{x}=\frac{1}{6A} sum(({x_{i} +x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))     where i=1 to N-1 and N=6

C_{y}=\frac{1}{6A} sum(({y_{i} +y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))

A area of the polygon can be found by the following formulaA=\frac{1}{2} sum(x_{i} y_{i+1} -x_{i+1} y_{i}) where i=1 to N-1

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

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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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putting the values of x's and y's you will get

C_{y} =22.55

So coordinates for the fire station should be (15.36,22.55)

5 0
3 years ago
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