Question

An elevator (mass 4500 kg ) is to be designed so that the maximum acceleration is 0.0680 g . what is the maximum force the motor should exert on the supporting cable?

Answer 1

For this case what you should do is to start from the definition that force is equal to mass by acceleration.
We have then:
f = m * a
where,
m = 4500Kg
a = 0.0680g
g = 9.81 m / s ^ 2
Substituting the values:
f = (4500) * ((0.0680) * (9.81))
f = 3001.86 N
answer:
the maximum force the motor should exert on the supporting cable is
f = 3001.86 N