Mass A, 15 kg, is on a table, connected by a massless rope and frictionless pulley, to mass B, 8 kg, which is hanging off the side of the table. The kinetic friction coefficient is 0.14. Find the acceleration of the blocks.
The Attempt at a Solution
I've followed the problem all the way through, and I understand the gist of it, but I'm stuck on one tiny conceptual issue. Combining equations since the acceleration of block A equals that of block B, gives you a = (massB)(g) - (friction)(massA)(g) / (massA + massB). My question is that when trying to solve it on my own, I isolated block A. The forces on block A are the force of gravity, the normal force, the tension from the rope, and the force of friction. Since block A is not moving vertically, I look at only the tension and the friction. The tension is only due to the weight of block B, so that's (massB)(g). The opposing force of friction is (friction coeff)(massA)(g). So, in my mind, the net force on block A is F = (massA)(a) = (massB)(g) - (friction coeff)(massA)(g). Basically the same thing as the correct equation above, except I only divide by the massA, not (massA + massB). If I'm only looking at block A, why do I need to include the mass of block B?? Pls help!!