The given problem can be solved using the following free-body diagram:
The diagram is the free-body diagram for the pulley that is holding the weight. Where:

Now we add the forces in the vertical direction:

Adding like terms:

Now, since the velocity is constant this means that the acceleration is zero and therefore the sum of forces is zero:

Now we solve for "T" by adding "mg" from both sides:

Now we divide both sides by 2:

Now we substitute the values and we get:
![T=\frac{(64\operatorname{kg})(9.8(m)/(s^2))}{2}]()
Solving the operations:

Now we use the free body diagram for the second pulley:
Now we add the forces in the vertical direction:

The forces add up to zero because the velocity is constant and the acceleration is zero:

Solving for the force:

Therefore, the pulling force is equal to the tension we determined previously and therefore is:
