An object with a specific gravity of 1.78 is completely submerged in water. The objectis attached to a scale and weighed in the water and registers 800N. (a) Determine if the body of the object will sink, float, or become suspended (i.e., neutrally buoyant). (b) Draw the free body diagram for the

Question

An object with a specific gravity of 1.78 is completely submerged in water. The objectis attached to a scale and weighed in the water and registers 800N. (a) Determine if the body of the object will sink, float, or become suspended (i.e., neutrally buoyant). (b) Draw the free body diagram for the manually suspended body in the water. (c) Determine the volume of the object (body).

Accepted Answer

(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:

[latex] s_{w} := 1[/latex] [latex] s_{b} := 1.78[/latex]
[latex] \gamma _{w} := 9810 \frac{N}{m^{2} } [/latex] [latex] \gamma _{b} := s_{b}.\gamma _{w} = 1.746 imes 10^{4} \frac{N}{m^{2} } [/latex]
[latex] ho _{w} := 1000 \frac{Kg}{m^{3} } [/latex] [latex] ho _{b} := s_{b}. ho _{w} = 1.78 imes 10^{3} \frac{Kg}{m^{3} } [/latex]

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 [latex]\sum{F_{z} } = W_{s}- W + F_{B} = 0 [/latex] [latex] W_{s} = W - F_{B} [/latex] [latex] W_{s} = \gamma _{b} V_{b}- \gamma _{f}V_{dispfluid}= ( \gamma _{b}- \gamma _{f}) V_{b}f [/latex] as follows:

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

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

[latex] W_{s}:= 800 N[/latex]

Guess value: [latex] V_{b}: 1 m^{3}[/latex] [latex] V_{d}: 1 m^{3}[/latex] [latex] W: = 10 N[/latex] [latex] F_{B}:= 5 N[/latex]

Given

[latex] W_{s} - W + F_{B}= 0[/latex] [latex] W= \gamma _{b}. V_{b}[/latex]
[latex] F_{B} = \gamma _{w}. V_{d}[/latex] [latex] V_{d}= V_{b}[/latex]
[latex] \left ( \begin{matrix} V_{b} \ V_{d} \ W \ F_{B} \end{matrix} ight ) := Find (V_{b},V_{d},W, F_{B})[/latex]
[latex] V_{b}= 0.105 m^{3}[/latex] [latex] V_{d}: 0.105 m^{3}[/latex]
[latex] W = 1.826 imes 10^{3} N[/latex] [latex] F_{B}:= 1.026 imes 10^{3} N[/latex]

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

[latex] F_{B}:= W- W_{s} = 1.026 imes 10^{3} N [/latex]

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

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