Question 2.6: A cube measuring 1 ft on a side is submerged so that its top...
A cube measuring 1 ft on a side is submerged so that its top face is 10 ft below the free surface of water. Determine the magnitude and direction of the applied force necessary to hold the cube in this position if it is made of
(a) cork \left(\rho=10 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}\right)
(b) steel \left(\rho=490 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}\right).
The pressure forces on all lateral surfaces of the cube cancel. Those on the top and bottom do not, as they are at different depths.
Summing forces on the vertical direction, we obtain
\Sigma F_{y}=-W+\left.P(1)\right|_{\text {bottom }}-\left.P(1)\right|_{\text {top }}+F_{y}=0where F_{y} is the additional force required to hold the cube in position.
Expressing each of the pressures as P_{\mathrm{atm}}+\rho_{w} g h, and W as \rho_{c} g V, we obtain, for our force balance
-\rho_{c} g V+\rho_{w} g(11 \mathrm{ft})\left(1 \mathrm{ft}^{2}\right)-\rho_{w} g(10 \mathrm{ft})\left(1 \mathrm{ft}^{2}\right)+F_{y}=0Solving for F_{y}, we have
F_{y}=-\rho_{w} g[(11)(1)-10(1)]+\rho_{c} g V=-\rho_{w} g V+\rho_{c} g VThe first term is seen to be a buoyant force, equal to the weight of displaced water. Finally, solving for F_{y}, we obtain
(a) \rho_{c}=10 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}
F_{y}=-\frac{\left(62.4 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}\right)\left(32.2 \mathrm{ft} / \mathrm{s}^{2}\right)\left(1 \mathrm{ft}^{3}\right)}{32.2 \mathrm{lb}_{\mathrm{m}} \mathrm{ft} / \mathrm{s}^{2} \mathrm{lb}_{\mathrm{f}}}+\frac{\left(10 \mathrm{lb}_{\mathrm{m}} \mathrm{ft}^{3}\right)\left(32.2 \mathrm{ft} / \mathrm{s}^{2}\right)\left(1 \mathrm{ft}^{3}\right)}{32.2 \mathrm{lb}_{\mathrm{m}} \mathrm{ft} / \mathrm{s}^{2} \mathrm{lb}_{\mathrm{f}}}
=-52.4 \mathrm{lb}_{\mathrm{f}}(\text { downward })(-233 \mathrm{~N})
(b) \rho_{c}=490 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}
F_{y}=-\frac{\left(62.4 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}\right)\left(32.2 \mathrm{ft}^{2}\right)\left(1 \mathrm{ft}^{3}\right)}{32.2 \mathrm{lb}_{\mathrm{m}} \mathrm{ft} / \mathrm{s}^{2} \mathrm{lb}_{\mathrm{f}}}+\frac{\left(490 \mathrm{lb}_{\mathrm{m}} \mathrm{ft}^{3}\right)\left(32.2 \mathrm{ft} / \mathrm{s}^{2}\right)\left(1 \mathrm{ft}{ }^{3}\right)}{32.2 \mathrm{lb}_{\mathrm{m}} \mathrm{ft} / \mathrm{s}^{2} / \mathrm{lb}_{\mathrm{f}}}
=+427.6 \mathrm{lb}_{\mathrm{f}}(\text { upward })(1902 \mathrm{~N})
In case (a), the buoyant force exceeded the weight of the cube, thus to keep it submerged 10 ft below the surface, a downward force of over 52 lb was required. In the second case, the weight exceeded the buoyant force, and an upward force was required.