6.1 Solving problems with newton’s laws  (Page 6/12)

From the free-body diagram, we see that ${\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_s-\overrightarrow{w},$ so we have

Solving for $F_s$ gives us an equation with only one unknown:

or, because $w=mg,$ simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes $F_s-w=\text{−}ma.$ )

Solution

1. We have $a=1.20{\text{m/s}}^2,$ so that
   
   $F_{\text{s}}=(75.0\text{kg})(9.80{\text{m/s}}^2)+(75.0\text{kg})(1.20{\text{m/s}}^2)$
   
   yielding
   
   $F_{\text{s}}=825\text{N}\text{.}$
2. Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because $a=\frac{\text{Δ}v}{\text{Δ}t}$ and $\text{Δ}v=0.$ Thus,
   
   $F_{\text{s}}=ma+mg=0+mg$
   
   or
   
   $F_{\text{s}}=(75.0\text{kg})(9.80{\text{m/s}}^2),$
   
   which gives
   
   $F_{\text{s}}=735\text{N}\text{.}$

Significance

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In [link] (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Two attached blocks

Strategy