Question

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?

Answer 1

33705 N

Step-by-step 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=
`(v-u)/(t)=(2.7)/(3)=0.9m/s^2`

Net force=
`F_(net)=T-mg`

`ma=T-mg`

`T=ma+mg=m(a+g)`

Where
`g=9.8m/s^2`

Using the formula

`T=3150(0.9+9.8)`

`T=33705 N`