A boat pulls two barges down the river. The barge connected to the boat (barge A) has an inertia of 2x10^5 kg, and the other barge (barge B) has an inertia of 3x10^5 kg. the resistive force between barge A and the water is 7x10^3 N, and the resistive fore of barge B to the water is 9x10^3 N. The common acceleration of all three boats is .4 m/s^2. Even thought the ropes are huge, the gravitational exerted on them is much smaller than the pulling forces. (a). What is the tension in the rope that connects the boat to barge A? (b). What is the tension in the rope that connects the two barges? (C). Repeat steps a and b for the case in which the order of the barges is reversed. So for part (a) I took the contact force of the rope to barge A which is 2x10^5kg * .4m/s^2= 8x10^4 N and added the negative of the resistive force; 8x10^4N - 7x10^3N= 7.3x10^4N. I wasn't sure if I should also subtract the contact force of the second rope on barge A which would cancel out the two contact forces and the answer would be the resitive force of barge A and the water. (b). Contact force of the second rope on barge B; 3x10^5kg * .4m/s^2 = 1.2x10^5N; subtract the resistive force of barge B to the water; 1.2x10^5N - 9x10^3N = 1.11x10^5N. Again not sure if I should add the contact force of the first rope on barge B. (c). They would be the same unless I do subtract/add the parts I'm not sure about.