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Heat
Oct24-07, 08:56 AM
1. The problem statement, all variables and given/known data

The upward normal force exerted by the floor is 620N on an elevator passenger who weighs 650N.


2. Relevant equations

F=ma

3. The attempt at a solution

I am reviewing all the Homework assigned so, I do got the answers. :)

Anyways, the way I set this problem up was F=ma.

First we need to know the mass, which we get by w=mg.

650 = m9.8
m = 66.33

now we plug into F=ma.

Fnet = 66.33a

I managed to get the acceleration of .455 , but my question here is , why is the net force 620-650 and not 650-620.

Do we always subtract from the smaller force first? My first instinct was since gravity was larger it would be 650-620.

What if it was this instead: "The upward normal force exerted by the floor is 700N on an elevator passenger who weighs 100N."?

Would it be normal force-w or will it still be w-n. ?

I just need a clarification. o:)

sapiental
Oct24-07, 09:09 AM
what exactly does the problem ask you to find? The normal force is less because the elevaotr is accelerating in what direction to make the passenger weight less?

Heat
Oct24-07, 09:18 AM
It tells me to find acceleration.

I did of -.455

I know because of it being negative it goes down.

but

what if I subtracted 650-620 = 30
then it would be 30/66.33 = .455

a would be positive.

sapiental
Oct24-07, 11:08 AM
"My first instinct was since gravity was larger it would be 650-620"

the normal force acts against the downward direction.

so u can think of it as 620N + (-650N) .



"Do we always subtract from the smaller force first? My first instinct was since gravity was larger it would be 650-620. "

no, newtons law is addition of forces so u need to keep your +/- sings with respect to the directions.

Tlocc
Oct24-07, 11:38 AM
Your first problem is that w=mg not ma.

sapiental
Oct24-07, 05:33 PM
Gravity is commonly measured in units of m s-2, (metres per second squared). hmmm