Chapter 4, Problem 14P

To determine

The maximum and minimum forces amotor can exert on supportingcable.

Answer to Problem 14P

Solution:The maximum and minimum forces possible on the cable are 50810 N and 44297 N respectively.

Explanation of Solution

Given:

Mass of the elevator, $m=4850\text{ kg}$

Maximum possible acceleration of the elevator, $a=0.0680g{\text{ m/s}}^{\text{2}}$

Formula Used:

According to Newton’s second law of motion:

  ${\displaystyle \sum F} =ma$

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

Calculation:

(I) For maximum tension, the elevator should move up, upward acceleration is a positive acceleration.

  

  $\begin{array}{l}{\displaystyle \sum F=ma}\\ T_{\max }-mg=ma \\ T_{\max }-\left(4850\right)\left(9.81\right)\\ T_{\max }=\left(4850\right)(0.0680g)\\ T_{\max }=50810\text{ N}\end{array}$

(II) For minimum acceleration, the elevator should move downward with a negative acceleration.

  

  $\begin{array}{l}{\displaystyle \sum F=ma}\\ T_{\min }-mg=ma \\ T_{\min }-\left(4850\right)\left(9.81\right)=\left(4850\right)(-0.0680g)\\ T_{\min }=44297\text{ N}\end{array}$

The maximum force on supporting cable is 50810 Nwhenthe elevator moves up.Similarly,the minimumforce on cable 44297 N is observed when the elevator moves down.