# Sign on a force

1. Jan 11, 2015

### joyce1029

You walk into an elevator, step on a scale, and push the "up" button. You recall that your normal weight is 625N. If the elevator has an acceleration of magnitude 2.50 m/s^2, what does the scale read?

I know how to solve this problem, but I don't know why I'm getting the wrong answer.

I first defined up as positive because it is the direction of acceleration. Then there are two forces: the normal and gravity.

m = w/g
m = (625/9.8) = 63.8 kg

N - mg = ma
N = ma + mg
N = m(a+g)
N = 63.8(2.5+(-9.8)) = -465.74N

Which is incorrect. This is correct if I make gravity positive, but I don't understand why it should be positive if I defined up as positive and gravity points down.

2. Jan 11, 2015

### Stephen Tashi

If you treat the magnitude of gravity as negative, then draw the force of gravity upward. The total force is the person is N + mg and the negative sign on the magnitudeof gravity tells you the direction of gravity is actually downward. If you draw the vector of gravity downward and compute N - mg then don't make the magnitude of gravity negative.

3. Jan 11, 2015

### joyce1029

So since I'm used to calculating the net force as the forces in the positive direction minus the forces in the negative direction (N-mg in this example), can I always keep everything positive(N,m,g)?

4. Jan 11, 2015

### Stephen Tashi

In theory, you can draw all forces in problem as if they were positive and give them negative magnitudes if they "really" point the the other way. In 2D, you could draw all vectors as if they wrere arrows in the first quadrant. The net force of two force vectors A and B is A+B. It is never really A-B. However, to aid human intuition, people disobey the conventions about the positive direction and they draw the arrows for vectors they way they think the vectors point. If you work that way then you must think of the magnitudes as all being positive and when you add two forces, you might need to compute the "sum" of the magnitudes as |A| - |B|.

Last edited: Jan 11, 2015
5. Jan 11, 2015

### my2cts

In the reference frame of the elevator there is a effective downward acceleration if the elevator is accelerated upward.
Condition is that the scale is set to zero in the elevator reference frame.

6. Jan 11, 2015

### joyce1029

What if I was calculating net force with the mass and acceleration? So say I have a 2kg object moving right and I set right to be positive. If the object is slowing down and I calculate the acceleration to be -5 m/s^2 and I'm told to solve for net force, would I take the absolute value of the acceleration. So F = ma = (2)*(|-5|) = 10N or would I keep acceleration to be negative resulting in -10N as the net force?

7. Jan 11, 2015

### Stephen Tashi

By your convention on the positive direction, the force is -10 N. However, some textbooks might give the answer as 10 N. They might put words in the problem that suggest you should use the left as the postive direction or make the positive direction ambiguous ( - words such as "Find the force that opposed the motion of the mass").

8. Jan 12, 2015

### dean barry

Just to confuse things.
Intuitively, the reading will be higher, i always feel heavier in a lift accelerating upwards.
If you draw the acceleration vectors (-9.8 and + 2.5) from a point on paper and measure the difference, this = 9.8 +2.5 = 12.3 m/s/s
then : 12.3 * 63.8 = 784.74 N