In Ch3 13P Step3: Where are those equations coming from? Question: An elevator (mass 4850kg) is to be designed so that themaximum acceleration is .0680g. What are the maximum andminimum forces the motor should exert on the supportingcable? I have no clue where to begin or why. There are things beingadded and subracted such as ft-mg=-ma. What exactly am Isolving for? Isn't ma=F and would'nt it be clearer/the same towrite F-mg=F. IS Ft= NORMAL FORCE? IS THIS EVEN A NORMALFORCE PROBLEM. Explain please and thank you!!! Test tom AM(6AM) In Ch3 13P Step3: Where are those equations coming from? Question: An elevator (mass 4850kg) is to be designed so that themaximum acceleration is .0680g. What are the maximum andminimum forces the motor should exert on the supportingcable? I have no clue where to begin or why. There are things beingadded and subracted such as ft-mg=-ma. What exactly am Isolving for? Isn't ma=F and would'nt it be clearer/the same towrite F-mg=F. IS Ft= NORMAL FORCE? IS THIS EVEN A NORMALFORCE PROBLEM. Explain please and thank you!!! Test tom AM(6AM)
Explanation / Answer
When the lift goes up The forces acting on it are T-mg=ma T=m(g+a) T=4850(9.8+0.0680g) T=50762.04N This is the maximum force. When the lift is going down the forces acting on it are, mg-T=ma T=m(g-a) T=4850(9.8-0.680g) T=15209.6N This is the minimum force of it. Hence we get by it.