# Pulleys / Newton's Laws

by martyk
Tags: laws, newton, pulleys
 P: 4 1. The problem statement, all variables and given/known data The figure shows two blocks connected by a cord (of negligible mass) that passes over a frictionless pulley (also of negligible mass). The arrangement is known as Atwood's machine. Block 1 has mass m1 = 0.900 kg; block 2 has mass m2 = 2.70 kg. What is the tension in the cord? Assume a y axis has its positive direction upward. 2. Relevant equations F = ma 3. The attempt at a solution Ok can someone tell me what is wrong with my approach here. I drew two free-body diagrams, and both basically the same. My free body diagrams have only 2 forces acting on the boxes. Upwards is the tension, and downwards is the weight of the boxes. So let T1 and W1 be the forces on the first box, and T2 and W2 be the forces on the second box. And based on this scenario, I know that T1 and T2 are equal, so is acceleration I summed up the forces acting on the first box, which leaves T1 - W1 = ma1, since the problem is asking for tension of the cord, I rearranged the equation to this T1 = ma1 + W1. I did the exact same with the second box. T2 - W2 = ma2, hence T2 = ma2 + W2. Since T1 = T2, I set these two equations equal to find A (note that a1 and a2 should be equal also). After plugging in the numbers in the parameters I got acceleration as -9.8m/s. Plugged it back to the tension equation.
HW Helper
PF Gold
P: 6,040
 Quote by martyk 1. The problem statement, all variables and given/known data The figure shows two blocks connected by a cord (of negligible mass) that passes over a frictionless pulley (also of negligible mass). The arrangement is known as Atwood's machine. Block 1 has mass m1 = 0.900 kg; block 2 has mass m2 = 2.70 kg. What is the tension in the cord? Assume a y axis has its positive direction upward. 2. Relevant equations F = ma 3. The attempt at a solution Ok can someone tell me what is wrong with my approach here. I drew two free-body diagrams, and both basically the same. My free body diagrams have only 2 forces acting on the boxes. Upwards is the tension, and downwards is the weight of the boxes. So let T1 and W1 be the forces on the first box, and T2 and W2 be the forces on the second box. And based on this scenario, I know that T1 and T2 are equal
very good so far
 , so is acceleration
their magnitudes are equal, but what about their directions?
 I summed up the forces acting on the first box, which leaves T1 - W1 = ma1, since the problem is asking for tension of the cord, I rearranged the equation to this T1 = ma1 + W1.
good
 I did the exact same with the second box. T2 - W2 = ma2, hence T2 = ma2 + W2.
Here's your error. It's all in the plus or minus sign.

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