Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 115: 42

Answer

The scale reads $645N$ during the acceleration.

Work Step by Step

1) Find the acceleration $a$ of the elevator There are 2 forces acting on the elevator: the force of the hoisting cable and the total gravitational force acting on the elevator, scale and person from earth. These forces are in opposite direction. According to Newton's 2nd Law: $$F_{cab}-\sum mg=\sum ma$$ We already have $F_{cab}=9410N$ $\sum mg=(60+815)\times9.8=8575N$ Therefore, $$a=\frac{9410-8575}{(60+815)}=0.95m/s^2$$ 2) Find the apparent weight of the woman As the scale actually reads the normal force $F_N$ acting on the woman, let's consider Newton's 2nd Law on the woman as the elevator moves up: $$F_N-m_{woman}g=m_{woman}a$$ $$F_N=m_{woman}(g+a)=60\times(9.8+0.95)=645N$$
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