×
×

# (II) An elevator (mass 4850 kg) is to be designed so that ISBN: 9780130606204 3

## Solution for problem 13P Chapter 4

Physics: Principles with Applications | 6th Edition

• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants Physics: Principles with Applications | 6th Edition

4 5 1 372 Reviews
16
3
Problem 13P

Problem

(II) An elevator (mass 4850 kg) is to be designed so that the maximum acceleration is 0.0680g. What are the maximum and minimum forces the motor should exert on the supporting cable?

Step-by-Step Solution:

Problem 13P

Solution 13P:        Pre-calculation of minimum and maximum forces that the motor should be capable of delivering is an important part in elevator designing in terms of safety of passenger.                                                                                Step 1 of 4</p>

Concept:    Newton’s second law: The force F acts on mass m produces an acceleration a in the object. Mathematically,                                        F=ma

Step 2 of 4</p>

Taking upward direction to be positive and downward direction to be negative.Mass of elevator = m = 4850 kg

Acceleration of elevator (a) = 0.0680g

Free Body Diagram of the elevator in upward direction and downward direction: Step 3 of 4</p>

Minimum force would be exerted by the elevator, when the elevator is moving in the downward direction because the tension in the string would be minimum. When the elevator is coming down, the minimum tension will be exerted by the motor as shown in figure 1. Considering the forces along the vertical direction, we get,     Thus, minimum force that should be exerted by the motor on the supporting cable of the elevator is, Step 3 of 3

##### ISBN: 9780130606204

Unlock Textbook Solution