# Horizontal momentum flux in the surface layer

Given the question below, how would the kinematic momentum flux be at the surface layer?

For letter a, why would it be that way (positive, negative, or 0)?

I’m not sure if I’ve done it correctly, but this is how I answered it based in your comment below.

1. First, I assumed that $$\overline{u’w’}$$ = $$\overline{w’u’}$$. So is this something like the mixing of horizontal momentum into areas of lower wind speed (as is the case in the surface layer)?

1. Which variable does $$\overline{u'w'}$$ affect most?
3. Imagine that you are on the line. If you go straight up (that is, $$w'>0$$) what would the difference be between the $$M$$ traced out and the graphed $$M$$? What is the sign of $$w' \times u'$$?
4. Do the same as number 3, except start at the top and go down ($$w'<0$$).
• When you said what the sign of w’ x u’ is, did u mean a cross product? Or something like, for step 3, since w>0 going up is (+) and u in this area would be (-), hence w’u’ < 0. Similarly, for step 4 going down, w’ < 0 and u’ > 0 giving w’u’ <0. So the average or net $\overline{w’u’}$ <0, meaning net momentum flux in this layer is downward? Thanks. – alberto_gar Oct 23 '20 at 2:32