Figure 5.11(a) shows a flatboat hull that, when fully loaded, weighs 150 kN. Parts (b)–(d) show the top, front, and side views of the boat, respectively. Note the location of the center of gravity, cg. Determine whether the boat is stable in fresh water.

Question

Accepted Answer

First, find out whether the boat will float.

This is done by finding how far the boat will sink into the water, using the principles of buoyancy stated in Section 5.1. Complete that calculation before going to the next panel.

The depth of submergence or draft of the boat is 1.06 m, as shown in Fig. 5.12, found by the following method:

Equation of equilibrium:

[latex]\sum{F_v} = 0 =F_b - W[/latex]

[latex]W = F_b [/latex]

Submerged volume:

[latex]V_d = B imes L imes X[/latex]

Buoyant force:

[latex]F_b = \gamma_f V_d= \gamma_f imes B imes L imes X[/latex]

Then, we have

[latex]W = F_b = \gamma_f imes B imes L imes X[/latex]

[latex]X = \frac{W}{B imes L imes \gamma_f} =\frac{150kN}{\left(2.4m ight)\left(6.0m ight) } imes \frac{m^3}{\left(9.81kN ight) } = 1.06m[/latex]

It floats with 1.06 m submerged. Where is the center of buoyancy?

It is at the center of the displaced volume of water. In this case, as shown in Fig. 5.13, it is on the vertical axis of the boat at a distance of 0.53 m from the bottom. That is half of the draft, X. Then [latex]y_{cp}=0.53m.[/latex]

Because the center of gravity is above the center of buoyancy, we must locate the metacenter to determine whether the boat is stable. Using Eq. (5–5)

[latex]MB = {I}/{V_d}[/latex] (5–5)

, calculate the distance MB and show it on the sketch.

The result is MB = 0.45 m, as shown in Fig. 5.14. Here is how it is done:

[latex]MB = {I}/{V_d}[/latex]

[latex]V_d = L imes B imes X = \left(6.0m ight) \left(2.4m ight) \left(1.06m ight) = 15.26m^3 [/latex]

The moment of inertia I is determined about the axis X–X in Fig. 5.11(b) because this would yield the smallest value for I. See parts (c) and (d) of Fig. 5.11 for the dimensions L and B.

[latex]I = \frac{LB^3}{12} = \frac{\left(6.0m ight) \left(2.4m ight)^3 }{12} = 6.91 m^4[/latex]

Then, the distance from the center of buoyancy to the metacenter is

[latex]MB = {I}/{V_d} = {6.91m^4}/{15.26m^3} = 0.45m[/latex]

The position of the metacenter is found from

[latex]y_{mc}=y_{cb} + MB =0.53m + 0.45m = 0.98m[/latex]

Is the boat stable?

Result Yes, it is. Because the metacenter is above the center of gravity, as shown in Fig. 5.14, the boat is stable.

That is, for [latex]y_{cg}=0.80m[/latex] and [latex]y_{mc}=0.98m[/latex] , [latex]y_{mc}\gt y_{cg}[/latex]. Now, read the next panel for another problem.

Question 5.5: Figure 5.11(a) shows a flatboat hull that, when fully loaded...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions