Final answer:
In this case, the magnitude of the net torque on the system when it is first released is 2mgd, where m represents the mass of each block and d represents the distance between the pivot point and each block.
Step-by-step explanation:
When the system is first released, the net torque acting on the system can be calculated by considering the torques produced by the gravitational forces acting on each block.
Let's assume that the distance between the pivot point and each block is represented by d. Since the rod is massless, we can ignore its contribution to the net torque.
The gravitational force acting on each block can be calculated as F = mg, where m represents the mass of each block and g represents the acceleration due to gravity.
The torque produced by the gravitational force acting on each block can be calculated as τ = F * d = mgd.
Since there are two blocks, the total torque produced by the gravitational forces is given by the sum of the torques produced by each block:
Total torque = 2 * mgd = 2mgd.
Therefore, the magnitude of the net torque on the system when it is first released is 2mgd, where m represents the mass of each block and d represents the distance between the pivot point and each block.
Your question is incomplete, but most probably the full question was:
Two blocks, each of mass m, are attached to the ends of a massless rod which pivots as shown in the figure (Figure 1). Initially the rod is held in the horizontal position and then released. Calculate the magnitude of the net torque on this system when it is first released Express your answer in terms of the variables m, I₁. I₂. and appropriate constants.