When the scale shows higher digits in the elevator, how do I calculate the real weight?

  • 2
    $\begingroup$ The scale doesn't show an increase/decrease of weight in an elevator unless it is accelerating. $\endgroup$ – CuriousOne May 26 '15 at 20:41
  • $\begingroup$ that s what I meant $\endgroup$ – EFmiza May 27 '15 at 7:14
  • $\begingroup$ Duplicate of Accelerating an elevator $\endgroup$ – John Rennie May 27 '15 at 11:06

What the scale in the elevator reads is the normal force. From Newton's second law, we know that $F_{\text{net}} = ma$ where $m$ is mass and $a$ is acceleration. There are only two forces on the person, the force of gravity down (equal to $mg$) and the normal force up (which I will call $F_N$). Newton's second law then yields

$$ ma = F_{N} - mg $$


$$F_{N} = m(g+a)$$

Remember $F_N$ is what the scale reads. If the elevator accelerates up ($a>0$), the reading of the scale ($F_N$) is higher than the person's weight. If the elevator accelerates down ($a<0$), the reading of the scale ($F_N$) is lower than the person's weight. If the elevator is at rest or moving at a constant velocity, the scale reads the same as the person's actual weight.


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