Exhibits, Shows and Displays

A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 8?

A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 839 N. (a) As the elevator moves up, the scale reading increases to 949 N, then decreases back to 839 N. Find the acceleration of the elevator. (b) As the elevator approaches the 74th floor, the scale reading drops to 782 N. What is the acceleration of the elevator?

Public Comments

1. we will apply newton's second law; the forces acting on the scale are the normal force (up), weight (down) and these combine to produce an ma term, in other words N-mg = ma the scale reading measures the normal force. if the reading at rest is 839N, the student's mass is 839N/9.8m/s/s = 85.6kg a) N-mg = ma if N=949, we have 949 = mg+ma 949=85.6g+85.6a a=1.28m/s/s in the positive direction b) if N=782N, then 782=mg+ma 782-85.6g=85.6a a=-0.66m/s/s (slowing down as it approaches the floor)