A block of wood with uniform density weighs 400 N and floats in water. Its dimensions are H= 40cm by h= 40cm and b = 80 cm along the proposed axis of rotation, y as illustrated in Figure EP 2.42. (a) Determine if the floating block of wood is top heavy or bottom heavy. (b) Determine if the

Question

A block of wood with uniform density weighs 400 N and floats in water. Its dimensions are H= 40cm by h= 40cm and b = 80 cm along the proposed axis of rotation, y as illustrated in Figure EP 2.42. (a) Determine if the floating block of wood is top heavy or bottom heavy. (b) Determine if the floating block will be in a position of rotational stable equilibrium about its longitudinal y-axis if the block is rotated by [latex]10^{\circ }[/latex] (assumed to be a small rotational disturbance) clockwise about the y-axis. (c) Determine the magnitude and the type of moment that result from the rotational disturbance.

Accepted Answer

(a) In order to determine if the block is top heavy or bottom heavy, one must first determine the location of the center of gravity, G and the location of the center of buoyancy, B. A free body diagram of the floating wood is illustrated in Figure EP 2.42b. Because the block of wood has a uniform density, the location of the center of gravity, G is at the centroid of the vertical rectangular cross section, which is located at G = H/2 = 0.2 m from the bottom of the block and at h/2 = 0.2 m from either side of the block as shown in Figure EP 2.42b. In order to determine the location of B, one must first determine the unknown submerged depth, d. Application of Newton’s second law of motion for fluids in static equilibrium to the floating wood (body) in the vertical direction yields the following:

[latex]\sum{F_{z}= - W + F_{B}=0} [/latex]

The submerged depth, d and the buoyant force, [latex]F_{B} = \gamma _{w} V_{dispfluid} [/latex] dispfluid are evaluated as follows:

[latex]W:= 400 N [/latex] [latex]\gamma _{w} := 9810 \frac{N}{m^{3}} [/latex] [latex]h:= 0.4 m [/latex] [latex]b:=0.8 m [/latex]

Guess value: [latex]d:=0.1 m [/latex] [latex]F_{B} : = 100 N [/latex]

Given

[latex]- W + F_{B} = 0 [/latex] [latex]F_{B} = \gamma _{w}hdb [/latex]
[latex]\left ( \begin{matrix} d \ F_{B} \end{matrix} ight ) := Find (d, F_{B}) [/latex]
[latex]d = 1.027 m [/latex] [latex]F_{B} = 400 N [/latex]

and its location, B is at the centroid of the rectangular submerged volume, which is located at B = d/2= 0.064 m from the bottom of the block and at h/2 = 0.2 m from either side of the block, as shown in Figure EP 2.42b. Therefore, because G is located above B, the floating block is top heavy.
(b) In order to determine if the floating block will be in a position of rotational stable equilibrium about its longitudinal y-axis if the block is rotated by [latex]10^{\circ }[/latex] clockwise about the y-axis, one must determine the metacentric height, GM for a top-heavy floating body as follows:

[latex]GM = BM - GB [/latex]

The distance [latex] BM = (I_{y}/ V_{ dispfluid} ) [/latex] ( [latex] I_{y} [/latex] for a rectangular cross section is given in Table B.1 in Appendix B) is computed as follows:

[latex]I_{y} : = \frac{b. h^{3}}{12}= 4.267 imes 10^{-3} m^{4} [/latex] [latex]V_{ dispfluid} = h.d.b=0.041 m^{3} [/latex]
[latex]BM:= \frac{I_{y} }{V_{ dispfluid} } = 0.105 m [/latex]

The distance GB is computed as follows:

[latex]H:= 0.4 m [/latex] [latex]G:= \frac{H}{2} = 0.2 m [/latex] [latex]B:= G:= \frac{d}{2} = 0.064 m [/latex] [latex]GB:= G - B = 0.136 m [/latex]

And, thus, the metacentric height, GM for the top-heavy floating block of wood is computed as follows:

[latex]GM: = BM - GM = - 0.032 m [/latex]

Therefore, because the metacentric height, GM is negative, the floating block is not in a position of rotational stable equilibrium about its longitudinal y-axis if the block is rotated by [latex] 10^{\circ }[/latex] clockwise about the y-axis. Relative to the distance GB = 0.136m, a smaller distance [latex] BM = (I_{y}/ V_{ dispfluid} ) [/latex] = 0.105 m indicates that for a small rotational disturbance of θ= [latex] 10^{\circ }[/latex], the floating top-heavy unstreamlined block may be considered to be a “tall and slender body.” One may recall that [latex] I_{y} [/latex] represents the geometric design of the body, and the larger the value for [latex] I_{y} [/latex] , the more rotationally stable the body is about the y-axis. Furthermore, because of the shape of the block of wood and thus the actual design angle of rotation, θ may not be a good indicator of the stability of the top-heavy floating block of wood; instead, the metacentric height, GM is used as an indicator of the stability of the floating block of wood.
(c) The magnitude of the moment, [latex]M_{couple}[/latex] that results from the rotational disturbance is computed as follows:

[latex] heta := 10^{\circ } [/latex] [latex] M_{couple}: = W GM \sin ( heta )= - 2.198 Nm [/latex]

And, finally, because the metacentric height, GM is negative, the floating block is not in a position of rotational stable equilibrium, so the resulting moment, [latex]M_{couple}[/latex] is an over turning moment that is in the clockwise direction (as opposed to restoring moment that would be in the counterclockwise direction).

Table B.1
The Area, A, the Center of Area (Centroid), [latex]y_{ca} [/latex] and [latex]x_{ca} [/latex], the Centroidal Moment of Inertia, [latex]I_{x-ca} [/latex] , and the Moment of Inertia about the y-Axis, [latex]I_{y} [/latex] for Some Common Geometric Shapes
Shape	A	[latex]y_{ca} [/latex]	[latex]x_{ca} [/latex]	[latex]I_{x-ca} [/latex]	[latex]I_{y} [/latex]
	[latex]\frac{\pi BL}{4} [/latex]	[latex]\frac{L}{2} [/latex]		[latex]\frac{\pi BL^{3} }{64}[/latex]	[latex]\frac{\pi B^{3}L }{16}[/latex]
	[latex]\frac{\pi BL}{4} [/latex]	[latex]\frac{4L}{3\pi } [/latex]		[latex]\frac{\pi BL^{3} }{16}-\frac{4 BL^{3}}{9\pi }[/latex]
	[latex]\frac{2 BL}{3} [/latex]	[latex]\frac{3L}{5}[/latex]	[latex]\frac{3B}{8} [/latex]	[latex]\frac{\pi BL^{3} }{7}-\frac{18 BL^{3}}{75 }[/latex]
	[latex](\frac{2 BL}{3} )[/latex]	[latex]\frac{3L}{5}[/latex]		[latex]2(\frac{\pi BL^{3} }{7}-\frac{18 BL^{3}}{75 })[/latex]

Question 2.42: A block of wood with uniform density weighs 400 N and floats...

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