What happened to the division of g in the Mach 3 force diagram problem?

[delve]

Here is an answer from a book of mine: From the force diagram we have N-mg=(\frac{mv^2}{R})e_r. The acceleration that the pilot feels is \frac{N}{m}=g+(\frac{mv^2}{R})e_r. I'm confused though; what happened to the g being divided by the last term as well?

 

[tiny-tim]

Hi delve!

Here is an answer from a book of mine: From the force diagram we have N-mg=(\frac{mv^2}{R})e_r. The acceleration that the pilot feels is \frac{N}{m}=g+(\frac{mv^2}{R})e_r. I'm confused though; what happened to the g being divided by the last term as well?

I'm not following you …

the equation is equivalent to N/m = g + (v²/R)e_r …

what do you mean by g being divided by something?

 

[delve]

Actually, you just helped me realize what I was confused about. Thank you!