A brass cube 6 in on a side weighs 67 lb. We want to hold this cube in equilibrium under water by attaching a light foam buoy to it. If the foam weighs 4.5 $lb/ft^3 ,$ what is the minimum required volume of the buoy?

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Question

A brass cube 6 in on a side weighs 67 lb. We want to hold this cube in equilibrium under water by attaching a light foam buoy to it. If the foam weighs 4.5 $lb/ft^3 ,$ what is the minimum required volume of the buoy?

Complete the solution before looking at the next panel.

Accepted Answer

Calculate the minimum volume of foam to hold the brass cube in equilibrium.

Notice that the foam and brass in Fig. 5.8 are considered as parts of a single system and that there is a buoyant force on each. The subscript F refers to the foam and the subscript B refers to the brass.

No external force is required.

The equilibrium equation is

[latex]\sum{F_v} =0[/latex]

[latex]0= F_{bB}+F_{bf}-W_B - W_F[/latex] (5–4)

[latex]W_B = 67lb[/latex] (given)

[latex]F_{bB}= \gamma_f V_{dB}=\left(\frac{62.4lb}{ft^3} \right) \left(6in\right)^3 \left(\frac{ft^3}{1728in^3} \right) =7.8lb[/latex]

[latex]W_F = \gamma_F V_F[/latex]

[latex]F_{bF}=\gamma_f V_F[/latex]

Substitute these quantities into Eq. (5–4):

[latex] F_{bB} +F_{bf}-W_B-W_F = 0 [/latex]

[latex]7.8lb+ \gamma_f V_F -67lb-\gamma_F V_F =0[/latex]

Solve for [latex]V_F[/latex], using [latex]\gamma_f = 62.4 {lb}/{ft^3}[/latex] and [latex]\gamma_F = 5.4 {lb}/{ft^3}[/latex] :

[latex]\gamma_f V_F - \gamma_F V_F = 67lb -7.8lb = 59.2lb[/latex]

[latex]V_F\left(\gamma_f - \gamma_F \right) =59.2lb[/latex]

[latex]V_F =\frac{59.2lb}{\gamma_f - \gamma_F} =\frac{59.2lb ft^3}{\left(62.4 - 4.5 \right)lb } [/latex]

[latex]V_F = 1.02ft^3[/latex]

Result This means that if [latex] 1.02ft^3[/latex] of foam were attached to the brass cube, the combination would be in equilibrium in water without any external force. It would be neutrally buoyant.

This completes the programmed example problems

Question 5.4: A brass cube 6 in on a side weighs 67 lb. We want to hold th...

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