ab elevator (mass 4850kg) is to be designed so that the maximum acceleration is .0680g. what are the maximum and minimum forces the motor should exert on the supporting cable? i know the answers but don't know how to come up with the work! (5.08*10^4N and 4.43*10^4N)

Forceoncable= masselevator(g +- .068g)

+- means plus or minus.

thanks i just figured it out two minutes ago. thank you, though

You're welcome! I'm glad you were able to figure it out. However, I can still explain how to derive the solution for future reference.

To find the maximum and minimum forces the motor should exert on the supporting cable, we need to consider the maximum acceleration and the mass of the elevator.

The formula to calculate force is given by Newton's second law of motion:

Force = mass x acceleration

In this case, the mass of the elevator is given as 4850 kg and the maximum acceleration is 0.068g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

To calculate the maximum force, we substitute these values into the equation:

Maximum force = mass x maximum acceleration
= 4850 kg x (0.068 x 9.8 m/s^2)

Calculating this gives us a maximum force of approximately 5.08 x 10^4 N.

To calculate the minimum force, we use the same equation but with the minimum acceleration, which is -0.068g:

Minimum force = mass x minimum acceleration
= 4850 kg x (-0.068 x 9.8 m/s^2)

Calculating this gives us a minimum force of approximately 4.43 x 10^4 N.

Therefore, the maximum force the motor should exert on the supporting cable is approximately 5.08 x 10^4 N, and the minimum force is approximately 4.43 x 10^4 N.