# How do you show the force of an elevator going up?

1. Oct 19, 2005

### dnt

for example, with an elevator not moving you have weight down and tension up.

but lets say it starts moving up...being on that elevator would make it FEEL heavier, correct? but the elevator is moving up so isnt the force going up? im trying to make a free body diagram to show a persons new weight on an elevator with say a = 1 m/s^2 going up (or any number).

i know the answer is weight = (mass)(acceleration) where a = 9.8 + 1 (due to elevator) but how do you show it and explain it?

thanks.

2. Oct 19, 2005

### Staff: Mentor

apparent weight vs real weight

I think you are asking how to find a person's apparent weight on an accelerating elevator. Don't confuse apparent weight (what it feels like, due to the force of the elevator floor pushing you up) with your real weight (the earth's gravitational pull, which doesn't depend on what the elevator is doing). The apparent weight is equal to the normal force that the elevator floor exerts on you. To find it, use Newton's 2nd law. First identify the forces acting on the person in the elevator (there are two).

3. Oct 19, 2005

### dnt

all i can think of are the persons weight pointing down and the force of the elevator (whihc points up). i cant figure out how that makes the apparent weight go up, even though i know it does.

4. Oct 19, 2005

### Staff: Mentor

Those are the only forces acting. Now use Newton's 2nd law to solve for the normal force which equals the apparent weight. (What's the net force on the person? Set that equal to ma.)

The apparent weight is the force that a supported object exerts on its support. (Its magnitude is equal to the normal force that the support exerts on it.) If the elevator accelerates upward, the floor must push against you with a force greater than your weight because it not only must balance your weight, but accelerate you as well. If the elevator cable were cut, the floor would stop supporting you at all (you and the elevator would be in free fall). In that case, your apparent weight would be zero--that's what being "weightless" means.

5. Oct 20, 2005

### dnt

so would it be

net force = weight + normal = ma

is that right?

6. Oct 20, 2005

### Staff: Mentor

That's right. Take care to use the correct directions for the forces and acceleration (which will determine their signs).