An elevator (mass 5000 kg) is to be designed so that the maximum acceleration is 0.0620 g. What is the maximum force the motor should exert on the supporting cable?
(in N)
What is the minimum force the motor should exert on the supporting cable?
(in N)
To find the maximum and minimum forces the motor should exert on the supporting cable, we need to consider the forces acting on the elevator.
Let's start by finding the weight of the elevator using the mass given and the acceleration due to gravity. The weight is the force exerted by the Earth on the elevator, given by the formula:
Weight = mass * acceleration due to gravity
Where the acceleration due to gravity is approximately 9.8 m/s^2.
Weight = 5000 kg * 9.8 m/s^2
Weight = 49000 N
Now, let's consider the maximum and minimum forces the motor should exert on the supporting cable.
1. Maximum force:
The maximum acceleration is given as 0.0620 g. To find the maximum force, we can use Newton's second law of motion:
Force = mass * acceleration
Since we want to find the force required to accelerate the elevator to the maximum acceleration, we can substitute the given values:
Force = 5000 kg * (0.0620 * 9.8 m/s^2)
Simplifying this equation:
Force = 3064 N
Therefore, the maximum force the motor should exert on the supporting cable is 3064 N.
2. Minimum force:
The minimum force occurs when the elevator is at rest or moving with a constant velocity. In these cases, the net force acting on the elevator is zero. So, the minimum force the motor should exert on the supporting cable is equal to the weight of the elevator, which is 49000 N.
Therefore, the minimum force the motor should exert on the supporting cable is 49000 N.