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A Spherical buoy has a diameter of 2m and weighs 3kN. It is designed to be used in a variety or circumstances by it being floated in water and anchored to the floor with a chain. Draw the free body diagram and determine the following given thta the volume of a sphere is,

$$Volume =\frac{\pi D^3}{6}$$

  1. The upthrust or buoyant force (Fb) in sea water.

  2. The tension in the chain (T)

Assume at 4Degrees the specific gravity of sea water is 1.042, g = 9.81m/s^2

1.

Equation

$$Pf Vd g$$

$$Fb = PfVdg = (1.042kg/m^3)(4.188m^3)(9.81m/s^2)$$ $$ = 42.8N$$

Is this correct? Trying to learn this. I think it is correct. But wanted to check so i understand how to do it correctly.

Then i am going to find question 2.

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  • $\begingroup$ Newton is a measure of force, providing one kilogram of mass with an acceleration of one meter per second per second. Three kiloNewtons would be three thousand times that force. It's not a weight. Do you mean 3 kilotonne? $\endgroup$
    – fred_dot_u
    May 3 '20 at 11:17
  • $\begingroup$ @fred_dot_u no it says 3kN. I Will Show a picture of the question. So i have gone wrong? $\endgroup$ May 3 '20 at 11:22
  • $\begingroup$ The creator of the problem has confused units of measure, force versus mass. I'm not qualified to suggest that you've gone wrong, especially if you are using mass figures to replace the incorrect force reference. $\endgroup$
    – fred_dot_u
    May 3 '20 at 11:25
  • $\begingroup$ I have just watched some YouTube videos, and learned from there to complete the question. I did not know their was a problem. So the 3kN is incorrect? $\endgroup$ May 3 '20 at 11:32
  • $\begingroup$ Do i need to work out the mass instead of weight and use the mass? $\endgroup$ May 3 '20 at 11:52
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A Buoy by definition has to float to be visible to the ships to alert the to the submerged obstacles or the depth datum.

Assuming the weight of the sphere and the tie, W the sphere sinks such that the buoyancy of the part under the water, called a sphere cap, is equal to the wight of the sphere plus approximately the weight of the tie.

Assuming the water $ \rho=1 $

In a calm sea the buoy will set at H, submersion height such that:

$ V=W= \frac{\pi H^2}{3}(3r-H) $

  • r the radius of the sphere

  • H is the height of the submersion (Cap)

they usually anchor the buoy to the floor of the sea with an anchor which is many times heavier the sphere buoyancy even if fully submerged so the maximum tension on the tie is the full buoyancy minus the weight of the sphere.

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buoy and displaced water

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