Question #246108

An elevator weighs 2500 𝑙𝑏; a cable, which can sustain a maximum tension of 6000 𝑙𝑏, supports it. If the elevator is going down with a velocity of 9 𝑓𝑡/𝑠, find the minimum safe distance it can travel before coming to a stop.


1
Expert's answer
2021-10-05T10:08:00-0400

Find the maximum acceleration that will create a 6000-lb-force on stop:


ma=mg+T,a=Tmg,d=v22a=v2m2Tmg=0.9 ft.ma=-mg+T,\\ a=\frac Tm-g,\\ d=\frac{v^2}{2a}=\frac{v^2m}{2T-mg}=0.9\text{ ft}.

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