To find the tensions in the cords T1 and T2, we can analyze the forces acting on each of the objects and use Newton's laws of motion. Let's break down the problem step by step:
1. Start by drawing a free-body diagram for each object:
- Mass 2m: This mass is attached to the left end of cord T1. The forces acting on it are the tension T1 pulling to the left and the force of gravity (mg) pulling downward.
- Mass m: This mass is attached to the right end of cord T1 and also the left end of cord T2. The forces acting on it are the tension T1 pulling to the right, the tension T2 pulling upward, and the force of gravity (mg) pulling downward.
- Mass M: This mass is attached to the right end of cord T2, hanging freely. The only force acting on it is the force of gravity (Mg) pulling downward.
2. Consider the vertical equilibrium of mass M:
Since mass M is in equilibrium, the sum of the forces in the vertical direction must be zero. The only force acting on mass M is its weight (Mg) pulling downward. Therefore, Mg = T2.
3. Solve for the tensions in cord T2:
From step 2, we found that T2 = Mg.
4. Consider the horizontal equilibrium of mass m:
Since mass m is in equilibrium, the sum of the forces in the horizontal direction must be zero. The forces acting on it are T1 pulling to the right and T2 pulling to the left. Therefore, T1 = T2.
5. Solve for the tensions in cord T1:
From step 4, we found that T1 = T2 = Mg.
So, the tensions in the cords T1 and T2 are both equal to Mg.
Note: In this solution, we assume there is no friction between the objects and the surfaces, and the cords and the pulley are massless and frictionless.