Why is Normal Force Negative in this situation?

[soccer123]

Homework Statement

The physics of circular motion sets an upper limit to the speed of human walking. (If you need to go faster, your gait changes from a walk to a run.) If you take a few steps and watch what's happening, you'll see that your body pivots in circular motion over your forward foot as you bring your rear foot forward for the next step. As you do so, the normal force of the ground on your foot decreases and your body tries to "lift off" from the ground.
(a)
. A person's center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass m at the top of a leg of length L.Find an expression for the person's maximum walking speed vmax.
Express your answer in terms of the variables L and appropriate constants.

Homework Equations

The Attempt at a Solution

Now I have the solution, which is reached in the following way:

mg - N = (mV^2)/r
N = 0 at Vmax. Also, let r = L.
mg = (mV^2)/L
Vmax = sqrt(Lg) = Answer. Now what I don't understand is why is it mg - N, rather than N - mg, seeing as how mg is pointing down and N is pointing up?

Thanks.

 

[CWatters]

You haven't actually defined up or down as positive in your answer.

However it appear you have chosen downwards as positive because you wrote +mg rather than -mg.

 

[soccer123]

You haven't actually defined up or down as positive in your answer.

However it appear you have chosen downwards as positive because you wrote +mg rather than -mg.

The problem is if you define negative as down, then the following will occur: N - mg = (mV^2)/L ----> Vmax = sqrt(-gL). This does not seem plausible given that you can't square root a negative number.

 

[vela]

You have to take into account the direction of the acceleration. Acceleration is a vector too.

 

[CWatters]

The problem is if you define negative as down, then the following will occur: N - mg = (mV^2)/L ----> Vmax = sqrt(-gL). This does not seem plausible given that you can't square root a negative number.

What about the sign of (mV^2)/L ? If you change the definition of +ve then the sign of the centripetal/centrifugal force must also change.

Personally I prefer to write the equations in the form...

A + B + C = 0

rather than..

A + B = C

That way you are forced to think about the sign of all the forces.