CP Elevator Design.? You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.60 times the passenger’s weight. The elevator accelerates upward with constant acceleration for a distance of 3.0 m and then starts to slow down. What is the maxi-mum speed of the elevator?

Solution 78P Step 1: The force exerted by the floor of the elevator on the person 1.60 × weight. Normal reaction N 1.60 × mg For simplicity we are going to replace the less than equal to symbol by the equal sign. So, N = 1.60 × mg . But we already know that, the normal reaction for a person in an upward accelerated elevator is, N mg = ma = force by the floor on the person. N = mg + F = mg + ma N = m(g + a) 1.60 × mg = m(g + a) 1.60g = g + a a = (1.60 1)g = 0.60 g = 0.60 × 9.8 = 5.88 m/s . So, the acceleration of the elevator we found is 5.88 m/s .