# What's the normal force on a squishy ball on an inclined track?

https://billiards.colostate.edu/physics/Domenech_AJP_87%20article.pdf

...which analyzes rolling friction on a rolling ball, the author claims that the normal force on a ball rolling down an incline is given by:

Where $$R_b$$ is the regular radius of the ball and $$R_e$$ is the effective radius of the ball, which is less than the regular radius since the ball squishes under its own weight.

But I don't understand. Why does the fact that the ball "squishes" as it rolls change the magnitude of the normal force?

• @JoshuaRonis : If there are two normal forces with absolute values $N_1$, as in figure 3, and both of them are directed at an angle $\phi$ from the normal to the inclined plane, then $mg=2 N_1 \cos \phi$ and $\cos\phi=R_e/R_b$. Dec 29 '18 at 22:44
• @JoshuaRonis : Well, I probably made a mistake in my previous comment, I should have written $m g \cos\theta=2 N_1 \cos\phi$. Dec 29 '18 at 23:53
• @JoshuaRonis : $N=2N_1$. Dec 30 '18 at 0:47