Strategy We draw a free-body diagram for each mass separately, as shown above. Then we analyze each one to find the required unknowns. The forces on block 1 are the gravitational force, the contact force of the surface, and the tension in the string. Block 2 is subjected to the gravitational force and the string tension. Newton's second law applies to each, so we write two vector equations: For block 1: T + W1 + Ñ = mãi For block 2: T + W2 = m2a2. Notice that IT is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects Solution The component equations follow from the vector equations above. We see that block 1 has the vertical forces balanced, so we ignore them and write an equation relating the x-components. There are no horizontal forces on block 2, so only the y-equation is written. We obtain these results: Block 1 Block 2 E Fy = may IF2 = maz T= = m101; Ty - mag = m2 d2y. When block 1 moves to the right, block 2 travels an equal distance downward; thus, @1, = - @2y. Writing the common acceleration of the blocks as @ = @ir = - @2y, we now have T=m1a and T - m29 = - m2a. From these two equations, we can express a and T in terms of the masses m1 and m2, and g : a = m2 m1 + m2 g and T= mym2 m1 + m2 g. Significance Notice that the tension in the string is less than the weight of the block hanging from the end of it. A common error in problems like this is to set T = m2g. You can see from the free-body diagram of block 2 that cannot be correct if the block is accelerating.