a 3200-lb elevator cab is supported by a cable in which the maximum safe tension is 4000 lb. (a)what is the greatest upward acceleration the elevator cab can have? (b) the greatest downward acceleration?

Answer 1

m=W/g
For upward motion
ma= T- mg
a=( T- mg)/m=(T-W)/m

For downward motion
ma=mg-T
a = (mg-T)/m=(W-T)/m