A spherical buoy (float) with a diameter of 2 m and a specific gravity of 0.88 is anchored to the bottom of the ocean floor (sseawater = 1.024) with a cord as illustrated in Figure EP 2.41b. (a) Determine if the unanchored buoy will sink, float, or become suspended (i.e., neutrally buoyant).

Question

A spherical buoy (float) with a diameter of 2 m and a specific gravity of 0.88 is anchored to the bottom of the ocean floor ([latex]s_( seawater) = 1.024 [/latex]) with a cord as illustrated in Figure EP 2.41b. (a) Determine if the unanchored buoy will sink, float, or become suspended (i.e., neutrally buoyant). (b) Draw the free body diagram for the unanchored buoy in the seawater. (c) Draw the free body diagram for the anchored buoy in the seawater. (d) Determine the tension in the chord if the anchored buoy is completely submerged in the seawater.

Accepted Answer

(a) In order to determine if the unanchored buoy will sink, float, or become suspended (i.e., neutrally buoyant), the specific gravity (specific weight, or density) of the buoy is compared to the specific gravity (specific weight, or density) of the seawater as follows:

[latex] s_{seaw} := 1.024 [/latex] [latex]s_{b} := 0.88 [/latex] [latex]\gamma_{w} : = 9810 \frac{N}{m^{3}} [/latex] [latex] ho _{w} : = 1000 \frac{kg}{m^{3}} [/latex]
[latex]\gamma_{b}:= s_{b} . \gamma_{w}= 8.633 imes 10^{3} \frac{N}{m^{3}} [/latex] [latex]\gamma_{seaw}:= s_{seaw} . \gamma_{w}= 1.005 imes 10^{4} \frac{N}{m^{3}} [/latex]
[latex] ho _{b}:= s_{b}. ho _{w}=880 \frac{kg}{m^{3}} [/latex] [latex] ho _{seaw}:= s_{seaw}. ho _{w}= 1.024 imes 10^{3} \frac{kg}{m^{3}} [/latex]

Therefore, because the specific gravity (specific weight and density) of the buoy is less than the specific gravity (specific weight and density) of the seawater, the unanchored buoy will float.
(b) The free body diagram for the floating unanchored buoy in the seawater is illustrated in Figure EP 2.41a.
(c) The free body diagram for the anchored buoy in the seawater is illustrated in Figure EP 2.41b.
(d) The tension in the cord if the anchored buoy is completely submerged in the sea water is determined by application of Newton’s second law of motion for fluids in static equilibrium as follows:

[latex]\sum{F_{z} } = - T - W + F_{B} = 0 [/latex]
[latex]D_{b} : = 2 m [/latex]

Guess value: [latex]W:= 100 N [/latex] [latex] F_{B}: = 100 N [/latex] [latex] T:= 100 N [/latex] [latex]V_{b}:= 10 m^{3} [/latex]

Given

[latex]-T - W + F_{B} = 0 [/latex] [latex]V_{b} = \frac{\pi . x^{3}_{b} }{6} [/latex]
[latex]W = \gamma_{b}. V_{b} [/latex] [latex] F_{B} = \gamma_{seaw}. V_{b} [/latex]
[latex] \left ( \begin{matrix} W \ F_{B} \ T \ V_{b} \end{matrix} ight ) := Find (W, F_{B}, T, V_{b}) [/latex]
[latex] W = 3.616 imes 10^{4}N [/latex] [latex] F_{B} = 4.208 imes 10^{4}N [/latex] [latex] T = 5.917 imes 10^{3}N [/latex]
[latex] V_{b}= 4.189 m^{3} [/latex]

Question 2.41: A spherical buoy (float) with a diameter of 2 m and a specif...

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