forcemax=mass*amax=4700(.06+1)g
force min=mass*g that is the force to hold the elevator still.
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?
3 answers
I've tried both of those. Does not work.
ma = tension - weight so ma + weight = tension so 4700(.06)(9.8) + 4700(9.8)
1 answer