An elevator (mass 4700 kg) is to be designed so that the maximum acceleration is 6.00×10^-2 g
What is the maximum force the motor should exert on the supporting cable?
What is the minimum force the motor should exert on the supporting cable?
forcemax=mass*amax=4700(.06+1)g
force min=mass*g that is the force to hold the elevator still.
I've tried both of those. Does not work.
ma = tension - weight so ma + weight = tension so 4700(.06)(9.8) + 4700(9.8)
To find the maximum and minimum forces the motor should exert on the supporting cable, we need to consider the maximum acceleration and the weight of the elevator.
First, let's calculate the weight of the elevator:
Weight = Mass x Acceleration due to gravity
Weight = 4700 kg x 9.8 m/s^2
Weight = 46,060 N (rounding to the nearest whole number)
Now, let's proceed with finding the maximum force:
Maximum Force = Weight + (Mass x Maximum Acceleration)
Maximum Force = 46,060 N + (4700 kg x (6.00x10^-2)g x 9.8 m/s^2)
Note: g represents the acceleration due to gravity - g = 9.8 m/s^2
Maximum Force = 46,060 N + (4700 kg x (6.00x10^-2) x 9.8 m/s^2)
Maximum Force = 46,060 N + 2782.8 N
Maximum Force ≈ 48,842.8 N (rounding to the nearest whole number)
Thus, the maximum force the motor should exert on the supporting cable is approximately 48,843 N.
For the minimum force, we assume that the elevator is not moving. In this case, the minimum force required from the motor should be equal to the weight of the elevator.
Therefore, the minimum force the motor should exert on the supporting cable is approximately 46,060 N.