Answer to Question #209717 in Mechanics | Relativity for Young

Question

A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 591N. As the elevator later stops , the reading is 391N. The magnitude of the acceleration is the same as during starting and stopping

Calculate weight of the person

Answer 1

Explanations & Calculations

• Weight of the person is the reading the scale reads $\small (R_0=mg)$ when it is stationary (or during uniform speeds, but not experienced in elevators more often).
• What the scale senses is that the force (mutual) between the person & the scale: the normal force.
• Then it is about assessing this normal force during any acceleration & that could be done by applying Newton's second law on the person along the moving direction.

$\qquad\qquad \begin{aligned} \small \uparrow F&=\small m a\\ \small R-mg&=\small m(\pm a) \end{aligned}$

• Then, as it goes up with acceleration

$\qquad\qquad \begin{aligned} \small 591-mg&=\small m(+a)=ma\cdots(1) \end{aligned}$

• Then, during the stopping period: where it continues to go up decelerating

$\qquad\qquad \begin{aligned} \small 391-mg&=\small m(-a)=-ma\cdots(2) \end{aligned}$

• By (1) + (2) we get the real weight

$\qquad\qquad \begin{aligned} \small 591-mg+391-mg&=\small ma-ma\\ \small 982&=\small 2mg\\ \small mg&=\small R_0=\bold{491\,N} \end{aligned}$