-2
$\begingroup$

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

$\endgroup$
3
  • 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
3
$\begingroup$

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

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

AKA

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.