At the top of the motion, however, the weight is almost perpendicular to its torque arm; thus, the weight feels a lot “heavier” here; your arm must provide a force to counteract this increased torque.
Okay, so you may be looking at the diagram here and thinking, “hold up! What about the torque arm for the tricep effort! Isn’t that almost parallel to the effort?” Yes, very true. I’m going to go out on a limb (see what I did there?) and claim that even though the torque arm and force may be very close to parallel, they will never actually BE parallel due to the insertion point of the tricep on the bone. So for this particular case, the torque will never go to zero. Also, if you were to calculate the force needed to counteract the force of the weight, you would find:
Effort = [Weight * rsin(theta_W)] / [R * sin(theta_E)]
As theta_E goes to 0, the denominator will blow up and the effort needed to make the arm perfectly straight will go to infinity!
But I digress…