Question 2.38: An object with a specific gravity of 1.78 is completely subm...

(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 water as follows:

s_{w} := 1                                          s_{b} := 1.78
\gamma _{w} := 9810 \frac{N}{m^{2} }                       \gamma _{b} := s_{b}.\gamma _{w} = 1.746 \times 10^{4} \frac{N}{m^{2} }
\rho _{w} := 1000 \frac{Kg}{m^{3} }                     \rho _{b} := s_{b}.\rho _{w} = 1.78 \times 10^{3} \frac{Kg}{m^{3} }

Therefore, because the specificgravity (specificweight and density) of the object is greater than the specific gravity (specific weight and density) of the water, the object will sink to the bottom unless it is manually suspended in the water.
(b) The free body diagram for the manually suspended body in the water is illustrated in Figure EP 2.38.

(c) The volume of the object (body) is determined by application of Newton’s second law of motion for fluids in static equilibrium given in Equation 2.252 \sum{F_{z} } = W_{s}- W + F_{B} = 0 W_{s} = W – F_{B} W_{s} = \gamma _{b} V_{b}- \gamma _{f}V_{dispfluid}= ( \gamma _{b}- \gamma _{f}) V_{b}f 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_{b}: 1 m^{3}                V_{d}: 1 m^{3}                W: = 10 N                F_{B}:= 5 N

Given

W_{s} – W + F_{B}= 0                  W= \gamma _{b}. V_{b}
F_{B} = \gamma _{w}. V_{d}                  V_{d}= V_{b}
\left ( \begin{matrix} V_{b} \\ V_{d} \\ W \\ F_{B} \end{matrix} \right ) := Find (V_{b},V_{d},W, F_{B})
V_{b}= 0.105 m^{3}                            V_{d}: 0.105 m^{3}
W = 1.826 \times 10^{3} N                  F_{B}:= 1.026 \times 10^{3} N

Furthermore, the buoyant force, F B reduces the weight of the object by 1026 N as follows: