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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?

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Yes ........... – John Rennie Mar 18 '13 at 16:19
Yes but for one source of error. If you stand on a pair of scales, the force on each may have equal (but opposite) horizontal forces. This will likely mess up the reading for some designs of scale where the assumption was that the force would be vertical. – Dan Piponi Mar 18 '13 at 17:47
So basically what you're saying is, that if I "did a split", then there is a significant component of my weight that points perpendicular to the normal of the scale's surface, that isn't being measured; but if that's the case, wouldn't that make the scale move horizontally? Doesn't the fact that the scales stand still (assuming they do) imply that no such components exists? – toon81 Mar 18 '13 at 17:52
@toon81 No; friction could hold the scales in place against the horizontal component of the force. – Nathan Reed Mar 18 '13 at 19:30
up vote 3 down vote accepted

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.

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I needed those Wikipedia links to refresh my memory, but with help from what I remember from high school physics I understand perfectly, even though I'm a layman. Excellent, thanks a lot! – toon81 Mar 18 '13 at 17:48
@toon81 I figured the details and links would help. Cheers! – joshphysics Mar 18 '13 at 17:50

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.

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Are you saying that there are other sorts of scale for which this doesn't work (assuming my body is at an equilibrium of course)? – toon81 Mar 18 '13 at 16:33
Other scales will have the same result, just that instead of using a physical spring, they use a virtual one, or computer algorithms that also yield the correct weight. – Moving Massive Mar 18 '13 at 16:46

protected by Qmechanic May 30 '15 at 6:13

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