Question

A 64kg person stands in an elevator. Find the weight of the person and the normal force between the person and the elevator for each of the following situations.
a. When the elevator is traveling downward at a constant speed of 12.4m/s.
Weight =
Normal Force =
b. When the elevator is accelerating upward at 2.4m/s2
Weight =
Normal Force =
c. The elevator cable is cut, and the elevator goes into freefall.
Weight =
Normal Force =

Answer 1

a) Weight: $W=mg=64\ kg\cdot 9.8\ \dfrac{m}{s^2}=627.2\ N.$

Let's write the Newtons Second Law of Motion:

$$\sum F_y=ma_y=0,$$ $$N-W=0,$$ $$N=W=627.2\ N.$$

b) Weight: $W=mg=64\ kg\cdot 9.8\ \dfrac{m}{s^2}=627.2\ N.$

Let's write the Newtons Second Law of Motion:

$$\sum F_y=ma_y=ma,$$ $$N-mg=ma,$$ $$N=m(g+a)=64\ kg\cdot(9.8\ \dfrac{m}{s^2}+2.4\ \dfrac{m}{s^2})=780.8\ N.$$

c) Weight: $W=mg=64\ kg\cdot 9.8\ \dfrac{m}{s^2}=627.2\ N.$

In free fall the acceleration will be directed downward and we get:

$$N=m(g-a)=m(g-g)=0 N.$$

Answer:

a) $W=627.2\ N, N=627.2\ N.$

b) $W=627.2\ N, N=780.8\ N.$

c) $W=627.2\ N, N=0\ N.$