197k views
2 votes
An elevator (mass 5000 kg ) is to be designed so that the maximum acceleration is 7.00×10⁻² g. What is the maximum force the motor should exert on the supporting cable?

User Iddober
by
8.3k points

1 Answer

4 votes

Final answer:

The maximum force the motor should exert on the supporting cable is 52430 N.

Step-by-step explanation:

To calculate the tension in the cable supporting the elevator, we need to consider the maximum acceleration and the mass of the elevator. The formula to calculate the tension in the cable is T = m * (g + a). Here, m is the mass of the elevator and a is the maximum acceleration. g is the acceleration due to gravity, which is approximately 9.8 m/s². Given the mass of the elevator as 5000 kg and the maximum acceleration as 7.00×10⁻² g, we can calculate the tension in the cable as follows:

T = 5000 kg * (9.8 m/s² + (7.00×10⁻² * 9.8 m/s²))

T = 5000 kg * (9.8 m/s² + 0.07 * 9.8 m/s²)

T = 5000 kg * (9.8 m/s² + 0.686 m/s²)

T = 5000 kg * 10.486 m/s²

T = 52430 N

The maximum force the motor should exert on the supporting cable is 52430 N.

User AreusAstarte
by
8.6k points