A piece of plastic with a specific gravity of 0.65 and a volume of 5 m^3 (L= 2.5 m, H= 2m, w = 1m) is partially submerged in water. (a) Determine if the plastic will sink, float, or become suspended (i.e., neutrally buoyant). (b) Draw the free body diagram for the plastic body in the water.

Question

A piece of plastic with a specific gravity of 0.65 and a volume of 5 [latex] m^{3} [/latex] (L= 2.5 m, H= 2m, w = 1m) is partially submerged in water. (a) Determine if the plastic will sink, float, or become suspended (i.e., neutrally buoyant). (b) Draw the free body diagram for the plastic body in the water. (c) Determine the proportion of the plastic body that floats above the water line.

Accepted Answer

(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:

[latex] s_{W} : = 1 [/latex] [latex]s_{W} : = 0.65[/latex]
[latex]\gamma _{W}: = 9810 \frac{N}{m^{3} } [/latex] [latex]\gamma _{b}: = s_{b} \gamma _{W}=6.377 imes 10^{3} \frac{N}{m^{3} } [/latex]
[latex] ho _{W}:= 1000 \frac{kg}{m^{3} } [/latex] [latex] ho _{b}:= s_{b} ho _{W} = 650 \frac{kg}{m^{3} } [/latex]

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

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

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

Given

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

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

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

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

This Question has been Answered.

Already have an account?

Login

Related Answered Questions