Solving Pulley on a Table Problem with Mass A & B

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SUMMARY

The problem involves two masses, A (15 kg) on a table and B (8 kg) hanging off the side, connected by a massless rope over a frictionless pulley. The kinetic friction coefficient is 0.14. To find the acceleration of the blocks, the correct approach is to set up two equations: one for block A and one for block B, accounting for tension and friction. The net force on block A is expressed as F = (massA)(a) = T - (friction coeff)(massA)(g), while block B's equation will include the weight of mass B and tension. Combining these equations yields the correct acceleration formula.

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Homework Statement



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!
 
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levi2613 said:
The tension is only due to the weight of block B, so that's (massB)(g).
This is an error: The tension does not simply equal the weight of block B. (If it did, the net force on Block B would be zero.)
The opposing force of friction is (friction coeff)(massA)(g).
Good.
So, in my mind, the net force on block A is F = (massA)(a) = (massB)(g) - (friction coeff)(massA)(g).
Since you don't know the tension (you'll have to solve for it), the best you can do is to call it "T" and write: F = (massA)(a) = T - (friction coeff)(massA)(g).

You'll need a second equation to combine with this to solve for the acceleration and the tension. Get that second equation by isolating Block B and writing a similar force equation.

(When you combine those two equations you'll get the equation you began with.)
 
Oh my goodness, THANK YOU SO MUCH! You have no idea how much this was hurting my brain! =P
 

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