University Physics with Modern Physics (14th Edition)

(a) $a = 3.17~m/s^2$ (b) $T_A = 158~N$ $T_B = 95.1~N$
(a) Note that the mass of the system is 60.0 kg. $F = ma$ $a = \frac{F}{m} = \frac{190~N}{60.0~kg} = 3.17~m/s^2$ (b) To find the tension $T_A$ in rope A, let's consider the system of the second and third boats with a total mass of 50.0 kg. $T_A = ma = (50.0~kg)(3.17~m/s^2) = 158~N$ To find the tension $T_B$ in rope B, let's consider the system of the third boat with a mass of 30.0 kg. $T_B = ma = (30.0~kg)(3.17~m/s^2) = 95.1~N$