Two blocks with masses \( M_{1} \) and \( M_{2} \) hang one under the other. For this problem, take the positive direction to be upward, and use \( g \) for the magnitude of the free-fall acceleration. (Figure 1) Find \( T_{2} \), the tension in the lower rope. Express your answer in terms of some or at rest B assume the blocks are at rest. \( M_{1}, M_{2} \), and \( g \).

To find the tension \( T_{2} \) in the lower rope, we can apply Newton's second law to the lower block \( M_{2} \). Since the blocks are at rest, the forces acting on \( M_{2} \) must balance. The weight of \( M_{2} \) acting downward is \( M_{2} g \), and the tension \( T_{2} \) acting upward must equal this weight. Thus, we set up the equation: \[ T_{2} = M_{2} g \] This shows that the tension in the lower rope is simply the weight of the lower block \( M_{2} \). Now, if we want to find the tension \( T_{1} \) in the upper rope, we can analyze the forces acting on the entire system of both blocks, but in this case, we focused solely on \( M_{2} \) for \( T_{2} \). When the system is at rest, it's purely a matter of comparing the loading on the ropes with the forces due to gravity! Isn’t physics fun when you see it all balancing out?