## Physics: Principles with Applications (7th Edition)

(a) $T = 1.5\times 10^5~N$ (b) $T = 1.8\times 10^5~N$
(a) Let $M$ be the mass of the hull. Let $\rho_s$ be the density of steel. We can find the volume $V$ of the hull. $V = \frac{M}{\rho_s}$ $V = \frac{18,000~kg}{7.8\times 10^3~kg/m^3}$ $V = 2.31~m^3$ We can find the tension $T$ in the cable. The tension will be equal to the apparent weight of the hull, which is the actual weight minus the buoyancy force. Let $\rho_w$ be the density of water. $T = Mg - F_B$ $T = Mg - V~\rho_w~g$ $T = (18,000~kg)(9.80~m/s^2) - (2.31~m^3)(1.00\times 10^3~kg/m^3)(9.80~m/s^2)$ $T = 1.5\times 10^5~N$ (b) We can find the tension $T$ in the cable when the hull is completely out of the water. $T = Mg$ $T = (18,000~kg)(9.80~m/s^2)$ $T = 1.8\times 10^5~N$