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I recently observed that on a weighing machine(rectangular) a object weighed the same as the half portion of the same object with it's other half hanging out of the boundary of the machine (The object was halfway on the edge in the second case). In both the cases the weight was equal though only half the portion of the same object was lying inside the boundary of the weighing machine. Why did this happen?

It would be better if you explain with the help of a diagram.

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3 Answers 3

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Of course, the weight shown would be the same.

When the object is half hanging out of the machine but not touching anything other than the machine, the whole weight of the body is exerted on the machine, regardless of the fact that a part is outside the machine. This is because it is on contact only with the machine and it is putting its weight on it.

If you try the same experiment with the part outside the machine supported by something else, then you will see that the weight shown has decreased. This is because now the weight is divided into two parts, and the machine shows only half the total weight. If you use another machine of same dimensions, the you should see that the weight shown in both are equal.

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  • $\begingroup$ I accept your answer. This was what I thought firstly but, if the entire weight of the object is acting on the machine and I support the other half by another object with the same height as machine. Will the machine show half the weight it showed earlier? ( The other object is in the same plane) $\endgroup$
    – user160598
    Jan 18, 2018 at 4:10
  • $\begingroup$ It should, if you support it properly. $\endgroup$ Jan 18, 2018 at 10:36
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good weighing machines have not one but several sensors under the weighing pan or platform that measure force. by artfully summing the outputs of those sensors, the effects of off-center loads can be nulled out.

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  • $\begingroup$ But in the second case when the other half portion is not in the frame. So no sensors could "measure" the portion's weight. $\endgroup$
    – user160598
    Jan 18, 2018 at 4:07
  • $\begingroup$ take a scale apart and have a look. $\endgroup$ Jan 18, 2018 at 4:21
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As long as the object doesn't fall off the platform (see warning in Countto10's comment), your scale, assumed to be perfect in the sense that it measures any vertical force equally on all points of the platform, will measure the full weight of the object because the gravitational force on the center of gravity propagates down to the scale's platform.

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    $\begingroup$ @Countto10 - Hi, under normal circumstances the c.o.g. and c.o.m. should be the same anyway $\endgroup$
    – freecharly
    Jan 17, 2018 at 22:19
  • $\begingroup$ So do you mean the object in the second case is showing the same weight because it centre of gravity lies on the machine (infact on the edge)? $\endgroup$
    – user160598
    Jan 18, 2018 at 4:05

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