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) For Parts \( A \) and \( B \) assume the blocks are at rest. Part A Part B Case 2: Accelerating blocks For Parts C and D the blocks are now accelerating upward (due to the tension in the strings) with acceleration of magnitude \( a \). Part C Part D Find \( T_{1} \), the tension in the upper rope. Express your answer in terms of some or all of the variables \( M_{1}, M_{2}, a \), and \( g \).

To find the tension \( T_{1} \) in the upper rope when the blocks are accelerating upward with an acceleration \( a \), we will analyze the forces acting on both blocks. ### Step 1: Analyze the forces on each block 1. **For Block \( M_{1} \)** (the upper block): - The forces acting on it are: - The tension \( T_{1} \) acting upward. - The weight \( M_{1}g \) acting downward. - According to Newton's second law, the net force \( F_{net} \) acting on block \( M_{1} \) is given by: \[ F_{net} = T_{1} - M_{1}g \] - Since the block is accelerating upward with acceleration \( a \), we can set up the equation: \[ T_{1} - M_{1}g = M_{1}a \] 2. **For Block \( M_{2} \)** (the lower block): - The forces acting on it are: - The tension \( T_{2} \) acting upward (which is equal to \( T_{1} \) since \( T_{1} \) is the tension in the upper rope). - The weight \( M_{2}g \) acting downward. - The net force \( F_{net} \) acting on block \( M_{2} \) is given by: \[ F_{net} = T_{2} - M_{2}g \] - Since block \( M_{2} \) is also accelerating upward with acceleration \( a \), we can set up the equation: \[ T_{2} - M_{2}g = M_{2}a \] - Since \( T_{2} = T_{1} \), we can substitute \( T_{1} \) into this equation: \[ T_{1} - M_{2}g = M_{2}a \] ### Step 2: Solve for \( T_{1} \) Now we have two equations: 1. \( T_{1} - M_{1}g = M_{1}a \) 2. \( T_{1} - M_{2}g = M_{2}a \) We can solve the first equation for \( T_{1} \): \[ T_{1} = M_{1}g + M_{1}a \] Now we can also solve the second equation for \( T_{1} \): \[ T_{1} = M_{2}g + M_{2}a \] ### Step 3: Combine the equations Since both expressions equal \( T_{1} \), we can set them equal to each other: \[ M_{1}g + M_{1}a = M_{2}g + M_{2}a \] ### Step 4: Isolate \( T_{1} \) To find \( T_{1} \) in terms of \( M_{1}, M_{2}, a, \) and \( g \), we can use either of the equations derived. Using the first equation: \[ T_{1} = M_{1}g + M_{1}a \] Thus, the final expression for the tension \( T_{1} \) in the upper rope is: \[ T_{1} = M_{1}(g + a) \] This expression shows that the tension in the upper rope depends on the mass of the upper block \( M_{1} \), the gravitational acceleration \( g \), and the upward acceleration \( a \).