Question 3.SP.9: The pontoon shown in Fig. S3.9 is 15ft long, 9ft9ft wide, ......

(a) Floating level, let d= the depth of submergence. Then

W=F_B ; \quad 15(9) 4(45)=15(9) d(62.4) ; \quad d=2.885 \mathrm{ft}

(b) At 12^{\circ}  tilt, let A D  be the water line (see Fig. S3.9).

Divide the buoyancy force into two components B_1  and B_2 , due to the rectangular block A E H K  and the triangular prism A D E  of displaced water, respectively.

D E=2 e=b \tan 12^{\circ}=9 \tan 12^{\circ}=1.913 \mathrm{ft} ; N I=e=0.957 \mathrm{ft}

As there is no net force, M N=d=2.885 \mathrm{ft} . Therefore

c=I M=M N-N I=2.885-0.957=1.928 \mathrm{ft}

 B_1  is at the centroid of the block A E H K , so

\begin{aligned}& G B_1=\frac{1}{2}(h-c)=\frac{1}{2}(4-1.928)=1.036 \mathrm{ft} ; \quad a_1=G B_1 \sin 12^{\circ}=0.215 \mathrm{ft} \\& F_1=\gamma L b c=45(15) 9(1.928)=11,710 \mathrm{lb} \\&\end{aligned}  

 B_2  is at the centroid of the triangle A D E , so

J E=b / 3, \quad I J=b / 6=1.5 \mathrm{ft}, \quad B_2 J=\frac{2}{3} e=0.638 \mathrm{ft}

 G  is at the centroid of the major rectangle, so M G=h / 2=2 \mathrm{ft} ,

\begin{gathered}G I=M G-M I=M G-c=2-1.928=0.0719 \mathrm{ft} \\a_2=I J \cos 12^{\circ}+\left(B_2 J-G I\right) \sin 12^{\circ}=1.585 \mathrm{ft} \\F_2=\gamma L b e=45(15) 9(0.957)=5810 \mathrm{lb}\end{gathered}

Counterclockwise moments about G  :

\begin{aligned}\text { Righting moment } & =F_2 a_2-F_1 a_1=5810(1.585)-11,710(0.215) \\& =6690 \mathrm{lb} \cdot \mathrm{ft} \end{aligned}