An elevator (mass 4175 kg) is to be designed so that the maximum acceleration is 0.0700g.

What is the maximum force the motor should exert on the supporting cable?
What is the minimum?

M*(g+a) = minimum

M*(g-a) = minimum

M should include the maximum passenger load in the maximum-force case.

M should be elevator mass only in the minimum case.

Whoever assigned the problem should realize that. What school is teaching this course?

I am curious. Teaching standards and student effort have suffered since online education has become the rage.

To find the maximum and minimum forces exerted by the motor on the supporting cable, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration:

F = m * a

Where:
F = Force applied (in Newtons)
m = Mass of the object (in kilograms)
a = Acceleration of the object (in meters per second squared)

In this case, the maximum acceleration is given as 0.0700g. To convert this to meters per second squared (m/s^2), we need to multiply it by the acceleration due to gravity (g), which is approximately 9.8 m/s^2.

So, the maximum acceleration is:
a_max = 0.0700 * 9.8 = 0.686 m/s^2

Now, we can calculate the maximum and minimum forces exerted by the motor.

Maximum Force:
To find the maximum force, we multiply the mass of the elevator by the maximum acceleration:

F_max = m * a_max

Substituting the given values:
F_max = 4175 kg * 0.686 m/s^2
F_max = 2861.75 N

Therefore, the maximum force the motor should exert on the supporting cable is 2861.75 Newtons.

Minimum Force:
To find the minimum force, we assume that the elevator is at rest. In this case, the acceleration is zero, and the force exerted by the motor is also zero.

Therefore, the minimum force the motor should exert on the supporting cable is 0 Newtons.