Question

Required information

You are designing a high-speed elevator for a new skyscraper. The elevator will have a mass limit of 2400 kg (including
passengers). For passenger comfort, you choose the maximum ascent speed to be 18.0 m/s, the maximum descent speed
to be 10.0 m/s, and the maximum acceleration magnitude to be 1.80 m/s2. Ignore friction.
What is the minimum upward force that the supporting cables exert on the elevator car?
KN

Answer 1

The minimum upward force that the supporting cables exert on the elevator car is 4320 N.

Step-by-step explanation:

The minimum upward force that the supporting cables exert on the elevator car can be calculated by considering the maximum acceleration and mass of the elevator. According to Newton's second law of motion, the net force on an object is equal to the product of its mass and acceleration.

So, the minimum upward force can be calculated as:

F = m * a

where F is the force, m is the mass, and a is the acceleration.

Given that the maximum acceleration magnitude is 1.80 m/s² and the mass limit is 2400 kg (including passengers),

F = 2400 kg * 1.80 m/s² = 4320 N

Therefore, the minimum upward force that the supporting cables exert on the elevator car is 4320 N.

Answer 2

19,224 N

Step-by-step explanation:

The given parameters are;

The mass limit of the elevator = 2,400 kg

The maximum ascent speed = 18.0 m/s

The maximum descent speed = 10.0 m/s

The maximum acceleration = 1.80 m/s²

Given that the acceleration due to gravity, g ≈ 9.81 m/s²

The minimum upward force that the elevator cable exert on the elevator car,
`F_(min)` , is given in the downward motion as follows;

`F_(min)` = m·g - m·a

∴
`F_(min)` = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N

The minimum upward force that the elevator cable exert on the elevator car,
`F_(min)` = 19,224 N