Atwood Machine A classic problem in physics, similar to the one we just solved, is that of the Atwood machine, which consists of a rope running over a pulley, with two objects of different mass attached. It is particularly useful in understanding the connection between force and motion. In the figure below, m1 = 2.00 kg and m2 = 4.00 kg. Consider the pulley to be frictionless. (a) If m2 is released, what will its acceleration be? (b) What is the tension in the string? Block 1 Block 2 ? ? m2 m. W. m2 An Atwood machine and free-body diagrams for each of the two blocks. Strategy We draw a free-body diagram for each mass separately, as shown in the figure. Then we analyze each diagram to find the required unknowns. This may involve the solution of simultaneous equations. It is also important to note the similarity with the previous example. As block 2 accelerates with acceleration @2 in the downward direction, block 1 accelerates upward with acceleration @1. Thus, a = @1 = - @2. Solution 1. We have Forma, SF =T - mg = ma. For m2, Fy =T -m2g = - m2a. (The negative sign in front of m2@ indicates that m2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is (m2-m1)g=(m1+m2) a. Solving for a: a = m1 + m2 m2 - m1 9 = 4kg + 2 kg 4kg-2kg (9.8m/s2) = 3.27 m/s2. 2. Observing the first block, we see that T-mig =mia T = m1 (g +a) = (2kg) (9.8m/s2 + 3.27 m/s2) = 26.1 N. Significance The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system, (m2 - m1) g, to the total mass of the system, m1 + 2. We can also use the Atwood machine to measure local gravitational field strength.