Thanks alot for the explanation. For some reason I am still having issues wrapping my head around the fact that the Force of gravity and the Force of acceleration are opposite to each other. In your response you said up was positive, but you are using a positve value for g. I thought this would be negative? Of course you get the wrong answer this way and i avoided this using the real life example you pointed out. So I'm still trying to solve this with math.......sorry for the trouble as i know this is an easy one but still managing to irk me. I know the way you did was spot on b/c it also asnwers the reverse question if she was accelerating.
cheers
The force of gravity is pointed downwards meaning it's negative. This I have defined in my force equation. When dealing with gravity and forces,
g should always be plugged in as a positive value, but the force that it provides may be negative depending on your sign conventions. Take a look at the equation below.
F(normal) - F(gravity) = ma (notice the minus sign because the forces are in opposing directions, ignore the acceleration term for now).
or more simply
F(normal) - mg = ma
See the weight vector is opposite the normal vector, so a positive value for g will suffice. The sign for the acceleration simply depends on if she's slowing down or not, which she is. Thus her acceleration vector is pointed in the same direction as mg and should be negative (which it is because we found it to be -5m/s^2, so all we do is plug this value in and the rest takes care of itself). When you're doing problems that involve vectors, i.e. force problems, make sure you define a coordinate system and everything will work itself out in the end. If I was told that she was
decelerating in the elevator at say 7m/s^2, then that indicates the acceleration vector is pointing down, so I'd need to write my force equation like
F(normal) - F(gravity) = -ma (here there is a negative on the right side of the equation because I know that the acceleration vector is
down and I simply plug in a value of 7m/s^2 for a),
It's confusing I know, but drawing a free body diagram helps a lot when identifying positive and negative forces on a body.