If a lever fixed at one end is laid across a scale, and you stand on the other end of the lever, what will the scale read? Let's imagine we have a long lever fixed to the ground at one end by a hinge. We lay this lever across a scale with the scale placed a distance d1 away from the hinge, then tear it. Then a person stands on the end of the lever opposite from the hinge, a distance d2 away from the scale.
                                    o
                                   -|-
                                   / \
  o-----------------------------------
  ^       d1        ^       d2
hinge             Scale

Let's say that the person's weight is normally x (as measured by the scale when they stand directly on it). What will the scale read when they stand d1 distance away from the scale?
As a follow up, I'd like to imagine a similar scenario:
We take our setup from above, but now we weld the lever to the scale and cut off the hinge and the rest of the lever on that side. So now we just have:
                  o
                 -|-
                 / \
  ------------------
  ^       d
Scale

In this scenario, the person is standing a distance d away from the scale. What will their weight read in this scenario?
EDIT
I think we can ignore the second example.
Upon further thinking, the second example doesn't make much sense. A scale measures normal force, and in the second example, there doesn't appear to be any normal force, only a rotational force (torque). I could imagine that the second example might actually cause the scale to read 0 (half the sensors would be lifted and the other half would be pressed down).
 A: In the first case you have a class 2 lever which has a pivot (fulcrum) at one end, a force at the other end, and the load in between. The leverage applied to the load is a ratio of the distances of the load and the force from the pivot. If the scale is in the center, as it  appears to be in your diagram, the scale will read twice the man's weight as he is twice as far from the piivot as the scale is. The second case does not have a pivot so it cannot apply leverage. You should read this article, it should have all the info you need; https://en.wikipedia.org/wiki/Lever
A: In the second case: if the scale is a foot wide and d is 100ft, it will break. If d is 5ft you have a setup similar to the first case, which you should diagram.
A: These can be solved using equations for rotational and translational equilibrium of the rod.(considering rod to be massless)
Case 1
Torque equations about hinge for rotational equilibrium. N is normal force applied by scale.
$$mg(d_1+d_2)=N.d_1$$
$$N=mg\frac{d_1+d_2}{d_1}$$
Case 2
The net force in vertical direction must be zero.
$\implies N_{net}=mg$
This solution is for an ideal case , but it must hold true for a real case too as the conditions for equilibrium will be the same.
