Answer:
a) [tex] W_B = F_B = 62.4 \frac{lb}{ft^3} (6ft*28ft*93ft)= 974937.6 lb[/tex]
b) [tex] W_g= 62.4 \frac{lb}{ft^3} * (9ft*28ft*93ft) -974937.6 lb =487468.8 lb[/tex]
Explanation:
Part a
For this case we have the situation illustrated on Figure 1. We will have two forces involved in equilibrium the weight [tex] W_B[/tex] and the Bouyance force[tex] F_B[/tex], and since the system is on equilibrium we have:
[tex] \sum F_{vertical}=0[/tex]
So then we have:
[tex] W_B = F_B = \gamma_{w} V_s[/tex]
Where [tex] V_s[/tex] represent the submerged volume. [tex]\gamma_w[/tex] represent the specific weight for the fluid. So we can replace and we have:
[tex] W_B = F_B = 62.4 \frac{lb}{ft^3} (6ft*28ft*93ft)= 974937.6 lb[/tex]
Part b
As we can see on figure 2 attached we have the illustration for this case. We add the weight for the grain and now the depth is 9ft.
W can do the balance of forces in the vertical and we got again:
[tex] W_B +W_g = F_B[/tex]
Where [tex] W_g[/tex] represent the weight for the grain.
And if we solve for [tex] W_g[/tex] we got:
[tex] W_g = F_B -W_B[/tex]
[tex] W_g =\gamma_w V_S -W_B[/tex]
Where [tex] \gamma_w[/tex] represent the specific weight of rthe water and [tex] V_s[/tex] the submerged volume. If we replace we got:
[tex] W_g= 62.4 \frac{lb}{ft^3} * (9ft*28ft*93ft) -974937.6 lb =487468.8 lb[/tex]
