# Homework Help: Drawing FBD for object inside elevator

1. Nov 2, 2004

### jhson114

I got alittle confused on something...
say an elevator is moving downward and its speed decreases.
If you are drawing a FBD of a block inside the elevator,
there are only two forces acting on the block, normal force from the elevator floor and weight force. But since elevator is decreasing in speed, netforce does not equal zero. Which vector, if any, should be longer than the other. Weight or Normal? or are they suppose to be equal?
I guess if you were drawing a FBD of the elevator itself, i'd be pretty easy, but i can seem to figure this one out. help please.

2. Nov 2, 2004

### cepheid

Staff Emeritus
This problems used to give me grief in high school! Memories...

Have you ever been in an elevator that was slowing to a stop just as it reached the desired floor? Of course? You no doubt observed that you felt heavier than usual when this happened. Why? Obviously the normal force on your feet increased. Again, why?

Think about it this way. As the elevator is descending, the net force on the elevator is zero...the elevator was given some initial downward velocity and then the upward force on the elevator cable was increased just enough to balance the weight of (car + passenger). The elevator descends evenly at a constant velocity. (of course, in real life, the elevator is not just hanging from a free cable, and the mechanism of descent is much better than a controlled fall! But let's model it this way for simplicity). In that case, in order to reach the desired floor, this descent must be retarded, otherwise the car will overshoot the destination. In our model this is accomplished by increasing the tension in the cable yet further. Rather than balancing the combined weight, the force in the cable is now greater than it. There is a net upward force on the car, and it slows down. Inside the car, the normal force on the person's feet is therefore greater.

3. Nov 2, 2004

### cepheid

Staff Emeritus
More detail:

The normal force is the force of the car pulling upwards on your feet. Car has mass M. You have mass m. Case 1: Descending at constant v. Applied force in cable is mg + Mg. What do you feel? You do not feel Mg because you feel the force of the *car* pulling upward on you. The car's weight is balanced by the Mg part of the applied force, so you only feel the additional force with which the car is pulled up i.e. mg. You feel your weight.

Case 2: Slowing the descent. F applied (to cable) = Mg + mg + Faddtional.

Again, you the car feels all of that, but it counteracts Mg, so you only feel mg + Faddtional, which is the newer, larger, normal force on your feet.

4. Nov 2, 2004

### jhson114

wow thanks. very detailed info. :)