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 \( \begin{array}{ll}\text { acceleration. (Figure 1) } & \text { Part B } \\ & \text { Find } T_{1}, \text { the tension in the upper rope. } \\ M_{1}, M_{2} \text {, and } g .\end{array} \)

To find the tension \( T_{1} \) in the upper rope connecting the two masses \( M_{1} \) and \( M_{2} \), we need to account for both masses and the gravitational force acting on them. The system is in equilibrium if we consider that both blocks are either at rest or moving with constant velocity. The tension \( T_{1} \) exerted on the upper mass \( M_{1} \) must counteract the weight of both \( M_{1} \) and \( M_{2} \). Therefore, we can write the equation for the forces acting on the upper mass: \[ T_{1} = M_{1} g + M_{2} g \] This simplifies to: \[ T_{1} = (M_{1} + M_{2}) g \] This formula gives us the tension in the upper rope in terms of the masses and the gravitational acceleration. Now, if we think about this in real-world terms, consider that this setup resembles a simple pulley system, where understanding the forces in such a system can help us design elevators or cranes. By analyzing the tensions correctly, engineers ensure that structures have the strength needed to support these forces without failing! Additionally, a common mistake when solving for tensions in such problems is forgetting to consider both weights acting on the upper mass, or misinterpreting directionality. Remember that the tension is acting upward against the force of gravity pulling downward! This can lead to incorrect results if not addressed properly.