Question

A small table is placed in an elevator. A light, frictionless pulley is attached to the end of the table.Two masses are connected by a string that passes over the pulley; m1 = 3.0 kg rests on the table, and m2 = 2.0 kg hangs off the side. Between m1 and the table there is no friction. While the elevator begins moving, the masses are measured by observers in the elevator to have an acceleration of a′.

(a) Find the acceleration of the elevator if a′ = 6.0 m/s2. Is the acceleration up or down?

(b) What value of a′ would indicate that the elevator is moving at constant speed.
(c) Describe what is happening to the elevator in the case that a′ = 0.

Answer 1

(a) The masss m1 is at rest on the table so the only force acting on it is the acceleration due to gravity. On the other hand the mass m2 is hanging off through the pulley so it will experience the tension force (due to string) in addition to gravitational force. As the elevator is moving the acceleration a' is observed by the observer. Applying Newton's second law along Y-axis (vertical direction)

Here, m is the total mass. We have chosen upward as the positive direction. Thus, we conclude from (1) that the scale reading T is greater than the weight mg if a (acceleration) is upward, so that a' is positive, and that the reading is less than the weight if a' is downward, so that a' is negative.

Suppose the acceleration is in the upward direction so a'=+6 m/s². Solving Eq. (1) for T

So the net acceleration of the elevator is

and the lift is moving in the upward direction so the direction of acceleration is upward.

(b) If the elevator is moving at a constant speed then its acceleration will be zero. Hence, a'=0m/s².

(c) For a'=0, the elevator will either be at rest or will move with the constant velocity.