Final answer:
The maximum force the motor should exert on the supporting cable of an elevator with a mass of 4550 kg and a maximum acceleration of 0.0690 g is 47688.41 N, which includes both the force to provide the acceleration and the weight of the elevator.
Step-by-step explanation:
To determine the maximum force the motor should exert on the supporting cable for an elevator of mass 4550 kg with a maximum acceleration of 0.0690 g, we first convert the acceleration in terms of meters per second squared:
Maximum acceleration = 0.0690 x 9.8 m/s² = 0.6762 m/s²
Using Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration, we calculate the force (not including the force due to gravity, which would be the weight of the elevator):
Maximum force = 4550 kg x 0.6762 m/s² = 3078.41 N
However, to find the total force the motor needs to exert, we must also consider the weight of the elevator:
Weight of elevator = 4550 kg x 9.8 m/s² = 44610 N
Therefore, the maximum force the motor should exert on the cable is the sum of these two forces:
Maximum force = Force due to acceleration + Weight of the elevator
Maximum force = 3078.41 N + 44610 N = 47688.41 N
The maximum force the motor should exert on the supporting cable is thus 47688.41 N.