Physics-elevator problems?

Question

So these are confusing.... If you could clarify that'd be great:
On a downward trip, an elevator reaches it's max velocity of -15m/s, when it slows down quickly and stops within 20m.
Acceleration is 5.625m/s^2.
What does the scale read while decelerating (express it as a function. Ex. 1.2x of fn)

Answer 1

The scale measures normal force, ie the force that must be exerted by the surface to keep the mass in motion with the entire system. Let's assume the weight of an object is -mg, and therefore the force the scale must exert when the system is in translational equilibrium is mg. We can verify this with W + F_n = -mg + mg = 0. Thus the sum of the forces in the vertical direction is 0 and there is no net acceleration. Now, we can use the same principle in this problem the net force acting on the object is ma = 5.625 m. We set this equal to W + F_n = -mg + F_n = 5.625 m. solving for F_n, we get F_n = m(g + 5.625). Since the scale measures normal force, the weight the scale will read is m(9.8+5.625) N = 15.425m N where m is the mass of the object.

Answer 2

When you are riding in a downward moving elevator and the elevator’s speed begins to decrease, you will feel like your weight has increased. This happens, because the upward force which the floor of the elevator exerts on your feet is doing two jobs. The force is supporting your weight and causing your velocity to decrease. This is why the scale reading is equal to the sum of the person’s weight and the product of the person’s mass and upward acceleration. I hope this helps you understand how to solve this type problem.



Since the elevator’s velocity decreases to 0 m/s as it falls 20 m, the acceleration is negative. If the elevator is at rest or moving at a constant velocity, the scale reading is equal to the weight of the person. Since the elevator’s velocity is decreasing, the scale reading must be greater than the weight of the elevator.



Net force = m * g + m * a = m * 15.425 N



This is the upward force which the scale is exerting on the person who is standing on the scale. In this problem, the upward force which the scale is exerting is doing two jobs. The upward force is supporting the weight of the person and causing the person’s velocity to decrease from 15 m/s to 0 m/s as the elevator, scale, and person move 20 meters downward. That is why the scale reading is equal to the sum of the person’s weight and the product of the mass and acceleration.



Below is a website which has information about this type of problem.

http://hyperphysics.phy-astr.gsu.edu/hbase/elev.html#c1

Answer 3

Physics Elevator Problems

Answer 4

what scale are you talking about?