Place each foot on a scale: can you add the two to find your weight? I frequent a blog from a British psychologist, and every Friday he likes to pose an interesting puzzle or riddle. The Monday after that he posts the answer. They're good fun, and IANAP but this week's answer made my it-might-not-be-quite-as-simple-as-that detector go off.
My question boils down to this: let's say I have two identical scales, and I stand on the scales with one foot on each scale. The scales read W1 and W2. Does my weight equal W1 + W2?
 A: Yes, this is not the hardest problem ever, but here's the mechanics calculation that leads to the yes answer.
Draw a free body diagram of your body as you are standing still with one foot on each scale.  You experience three forces (I will label their magnitudes): (1) The force due to gravity pulling you down, $W$ (aka your weight), (2) the normal force $N_1$ of scale 1 pushing up on one foot, and (3) the normal force $N_2$ of scale 2 pushing up on your other foot.  Since your body is not accelerating, these forces balance by Newton's Second Law;
$$
  W = N_1+N_2
$$
Now the question is, how are these forces related to what the scales read?  Well, each scale reads the force of the corresponding leg that pushes down on it.  Let's call the magnitude of these forces $W_1$ and $W_2$.  As it turns out, Newton's Third Law tells us that the magnitude of the force that each scale exerts on each foot (the normal force) equals the corresponding magnitude of the force that each foot exerts on the scale;
$$
  W_1 = N_1, \qquad W_2 = N_2
$$
It follows that
$$
  W = W_1 + W_2
$$
as desired.
A: Well, It would work, but only with the condition that your weight doesn't shift around between the time you look at the first reading and the next. But either way, the sum of the readings on the scales always equals "more than twenty stone" however much that specific weight may be, as long as the body in question is in equilibrium, that is to say, not moving. Most scales operate by spring, and Hooke's law states that the force applied to a spring is directly proportional with the extension or compression of the spring by this force. The force used need not be gravity alone, but could be momentum or great amounts of heat, but the less of these there are, the more accurate your weight reading is. In a real life scenario, I would stand still, then take a snapshot of both scales together so that my weight doesn't shift.
