Question 2.40: A piece of plastic with a specific gravity of 0.65 and a vol...

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

s_{W} : = 1                          s_{W} : = 0.65
\gamma _{W}: = 9810 \frac{N}{m^{3} }                   \gamma _{b}: = s_{b} \gamma _{W}=6.377 \times 10^{3} \frac{N}{m^{3} }
\rho _{W}:= 1000 \frac{kg}{m^{3} }                   \rho _{b}:= s_{b} \rho _{W} = 650 \frac{kg}{m^{3} }

Therefore, because the specific gravity (specific weight and density) of the plastic is less than the specific gravity (specific weight and density) of the water, the plastic will float.
(b) The free body diagram for floating plastic body in the water is illustrated in Figure EP 2.40.

(c) The proportion of the plastic body that floats above the water line is determinedby pplication of Newton’s second law of motion for fluids in static equilibrium given in Equation 2.254 \sum{F_{z} } = – W + F_{B} =0 W = F_{B} \gamma _{b} V_{b} = \gamma _{f} V_{ dispfluid} as follows:

\sum{F_{z}} = – W + F_{B}=0
V_{b} := 5 m^{3}                 L:= 2.5 m                 H:= 2m                 w:= 1 m

Guess value:            W:= 10 N                 F_{B}: = 10 N                 V_{d}: = 1 m^{3}
d:= 0.5 m                 P_{sub} := 0.5                 P_{float} := 0.5

Given

-W + F_{B} = 0                 W = \gamma _(b) . V_{b}                F_{B} = \gamma _(b). V_{d}                 p_{sub} = frac( V_{d} )(V_{d} )

P_{float}= 1 W- P_{sub}                    V_{d} = L.d.w                    V_{b}= L.H.w
\left ( \begin{matrix} W \\ F_{B} \\ V_{d} \\ d \\ p_{sub} \\ p_{float} \end{matrix} \right ) := Find (W, F_{B} , V_{d} , p_{sub} p_{float})
W = 3.188 \times 10^{4} N                    F_{B}= 3.188 \times 10^{4} N                    V_{d} = 3.25 m^{3}
d= 1.3 m                               p_{sub} = 0.65                                 p_{float}= 0.35

Furthermore, the buoyant force, F B is equal to the weight of the object, W.