If you are not interested in a solution that involves the changing angle, and only want to know how the tension in the tape changes with the number and placement of additional rollers, then I believe the problem reduces to a fairly simple one.

It would have helped if you included the distances from the fulcrum for each of the rollers, but no matter, I will just assume some distances in my solution. You can then apply your actual distances and solve your specific problem accordingly.

I am going to assume that 840 N⋅m of torque is required to raise the lever. (You use whatever the actual value is needed in your case, based on 70 N⋅L where L is the actual length from fulcrum to point e) That amount of torque is required anywhere along the length of the lever to raise it.

In my example, if the torque is applied at the end where you say 70 N of force is needed, the distance to the fulcrum is 12 meters, obviously. (70 N x 12 m = 840 N⋅m) I should mention this should be a cross product of vectors, [math] \vec\Gamma = \vec{r}\times\vec{F} [/math] but I suspect you don’t want to get into all that.

Taking all the forces you have on your diagram in descending order, they are 300 N, 175 N, 120 N, 90 N and 70 N. The respective distances from fulcrum, in my example, then are 2.8 m, 4.8 m,7 m, 9.33 m, and 12 m. Again, you use your actual lengths, not mine.

The total force, F, that is applied at the end is the tension in the tape, and it will be the same at all of the lifting points. As you say, it must be less than 70 N as the work is distributed among the five lift points.

But the total torque will always be 840 N⋅m regardless of how that gets distributed, in my example.

So, in this simplified case where we are ignoring vectors, we can just sum over each individual (force x distance) on the lever to get the net torque of 840 N⋅m

840 N⋅m = 2.8 F + 4.8 F + 7 F + 9.33 F + 12 F = 36 F

F = 23.3 N

That is a big force advantage from 70 N but the tape will need to be pulled through a longer distance to compensate.

It is always very possible I did not understand your question so let me know if this makes any sense to you. If you give me the actual lengths, I can work the actual problem for you instead of using this example. Or, you can use Gawd’s slip shod method.