Question

An elevator, mass 4750kg, is designed so that the maximum acceleration is 0.50m/s2. What are themaximum and minimum forces the motor exerts on the cable?

Answer 1

The free body diagram of the elevator can be shown as,

According to free body diagram, the net force acting on the elevator is,

$F_n=T_{\max }-mg$

According to Newton's law,

$F_n=ma$

Plug in the known expression,

$\begin{gathered} ma=T_{\max }-mg \\ T_{\max }=ma+mg \\ =m(a+g) \end{gathered}$

Substitute the known values,

$\begin{gathered} T_{\max }=(4750kg)(0.50m/s^2+9.8m/s^2) \\ =(4750\text{ kg)(}10.3m/s^2)(\frac{1\text{ N}}{1kgm/s^2}) \\ =48925\text{ N} \end{gathered}$

Thus, the maximum force exerted on the cable is 48925 N.

For minimum, tension the net force can be expressed as,

$F_n=mg-T_{\min }$

Plug in the known expression,

$\begin{gathered} ma=mg-T_{\min } \\ T_{\min }=mg-ma \\ =m(g-a) \end{gathered}$

Substitute the known values,

$\begin{gathered} T_{\min }=(4750kg)(9.8m/s^2-0.5m/s^2) \\ =(4750\text{ kg)(}9.3m/s^2)(\frac{1\text{ N}}{1kgm/s^2}) \\ =44175\text{ N} \end{gathered}$

Thus, the minimum force exerted on the cable is 44175 N.