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

AI Thread Summary
The discussion revolves around the dynamics of a pilot experiencing acceleration in a force diagram. The equation presented indicates that the normal force (N) minus the gravitational force (mg) equals the centripetal force (mv²/R). The pilot's perceived acceleration is expressed as N/m = g + (mv²/R). A participant expresses confusion about the role of gravitational acceleration (g) in the equation, leading to a clarification that resolves the misunderstanding. The interaction highlights the importance of understanding the relationship between forces and acceleration in physics.
delve
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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?
 
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Hi delve! :smile:
delve said:
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)er …

what do you mean by g being divided by something? :confused:
 
Actually, you just helped me realize what I was confused about. Thank you!
 
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