Given data;
* The mass of the elevator is 5000 kg.
* The acceleration of the elevator is,
[tex]a=3ms^{-2}[/tex]Solution:
The free body diagram of the elevator is,
The weight of the elevator is,
[tex]W=mg[/tex]where m is the mass of the elevator and g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} W=5000\times9.8 \\ W=49000\text{ N} \end{gathered}[/tex]The net force acting on the elevator is,
[tex]F_{\text{net}}=ma[/tex]where a is the acceleration of the elevator moving upwards,
Substituting the known values,
[tex]\begin{gathered} F_{\text{net}}=5000\times3 \\ F_{\text{net}}=15000\text{ N} \end{gathered}[/tex]From the free body diagram, the tension acting on the cable is,
[tex]\begin{gathered} T-W=F_{\text{net}} \\ T=F_{\text{net}}+W \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} T=15000+49000 \\ T=64000\text{ N} \\ T=64\text{ kN} \end{gathered}[/tex]Thus, the tension acting in the cable is 64 kN.
