Upward Normal Force Exerted by Floor on Elevator Passenger

[Heat]

Homework Statement

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

Homework Equations

F=ma

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.

 

[sapiental]

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]

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]

"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]

Your first problem is that w=mg not ma.

 

[sapiental]

Gravity is commonly measured in units of m s-2, (metres per second squared). hmmm