Question 2.39: An object with aspecific gravity of 0.86 and avolume of 2 m^...

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

\gamma _{w} := 9810 \frac{N}{m^{3} }                             \rho _{w} := 1000 \frac{kg}{m^{3} }
s_{o} := 0.86                                       s_{b} := 0.86
\gamma _{w} := s_{o}.\gamma _{w}= 8.437 \times 10^{3} \frac{N}{m^{3} }                 \gamma _{b} := s_{b}.\gamma _{w}= 8.437 \times 10^{3} \frac{N}{m^{3} }
\rho _{o} := s_{o}.\rho _{w}= 860 \frac{kg}{m^{3} }                       \rho _{b} := s_{b}.\rho _{w}= 860 \frac{kg}{m^{3} }

Therefore, because the specificgravity (specificweight and density) of the object is equal to the specific gravity (specific weight and density) of the crude oil, the object will become suspended or neutrally buoyant.
(b) The free body diagram for the neutrally buoyant body in the crude oil isillustrated in Figure EP 2.39.
(c) The weight of the object (body) is determined by application of Newton’s second law of motion for fluids in static equilibrium given in Equation 2.253 \sum{F_{z} } = – W + F_{B} = 0 W = F_{B} \gamma _{b} V_{b}= \gamma _{f} V_{dispfluid} as follows:

where the volume of the displaced fluid, V_{d} is equal to the volume of the body, V_{b} because the body is completely submerged.

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

Given

– W + F_{B} = 0                W = \gamma _{b} V_{b}                F_{B} =\gamma _{o} V_{d}                V_{d}= V_{b}
\left ( \begin{matrix} V_{d} \\ W \\ F_{B} \end{matrix} \right ) := Find (V_{d},W, F_{B})
V_{d}:= 2 m^{3}                W:= 1.687 \times 10^{4} N                F_{B}: = 1.687 \times 10^{4} N

Furthermore, the buoyant force, F_{B} is equal to the weight of the object, W.