Asked by Arthur

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?
Answered by bobpursley
forcemax=mass*amax=4700(.06+1)g

force min=mass*g that is the force to hold the elevator still.
Answered by Arthur
I've tried both of those. Does not work.
Answered by nine
ma = tension - weight so ma + weight = tension so 4700(.06)(9.8) + 4700(9.8)
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