Chapter 4, Problem 16P

To determine

The magnitude and direction of acceleration of a moving elevator.

Answer to Problem 16P

Solution: Elevator is moving downward with acceleration of magnitude $-0.25{\text{ m/s}}^{\text{2}}$

Explanation of Solution

Given:

The normal force or the weight experienced in the moving elevator, $F_N=0.75F_g$

Actual weight of the woman, $F_g=mg$

Here, m is the mass of the woman and g is the gravitational acceleration.

Formula Used:

From Newton’s second law of motion:

  ${\displaystyle \sum F}=ma$

Here, m is the mass and a is the acceleration.

Calculation:

Free body diagram:

  

Substitute the values and solve for acceleration ( a ).

  $\begin{array}{l}{\displaystyle \sum F}=ma\\ F_N-F_g=ma\\ 0.75mg–mg=ma\\ -0.25=a \\ a=-0.25{\text{ m/s}}^{\text{2}}\end{array}$

Negative value of acceleration indicates that elevator is moving downwards.

Conclusion: Using Newton’s second law of motion,the acceleration is $-0.25{\text{ m/s}}^{\text{2}}$ . The negative value of acceleration indicates that elevator is moving downwards.