An elevator (cabin mass 550 kg) is designed for a maximum load of 2600 kg, and to reach a velocity of 2.7 m/s in 3 s. For this scenario, what is the tension the elevator rope has to withstand?

Respuesta :

Answer:

33705 N

Explanation:

We are given that

Mass of cabin=550 kg

Maximum load=2600 kg

Total mass,m=550+2600=3150 kg

Initial velocity=u=0

Final velocity=v=2.7 m/s

Time=3 s

We have to fine the tension of the elevator rope to withstand.

Acceleration=[tex]\frac{v-u}{t}=\frac{2.7}{3}=0.9m/s^2[/tex]

Net force=[tex]F_{net}=T-mg[/tex]

[tex]ma=T-mg[/tex]

[tex]T=ma+mg=m(a+g)[/tex]

Where [tex]g=9.8m/s^2[/tex]

Using the formula

[tex]T=3150(0.9+9.8)[/tex]

[tex]T=33705 N[/tex]