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?

Respuesta :

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
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