Normal force for a bike on an incline So, in general I know how the find the normal force for an object on an incline, but this one is a bit harder, since the bike essentially has two normal forces like so:

where $L$ is the length of the wheel base and $h$ is the distance to the centre of gravity.
The idea is to find what the maximum angle of the slope is before gravity will overcome the friction between the tires and road, and supposedly, in this limiting case, $F=\mu N_{2}$. I'm not quite sure how to work out what $N_{2}$ is.
For reference here is the worked (and unexplained) solution:

Don't worry so much about the numerical answer right at the end, he's just subbed in the values, I'm more interested in the derivation.
I don't get where $\frac{h}{L}\sin(\theta)+ \frac{1}{2}\cos(\theta)$ came from and I also don't understand why were using torques? Basically I don't understand any of it.
 A: Why using torques? Because you have three unknowns, $\theta$, $F$ and $N_2$ and that requires three equations. You also have $N_1$ as unknown, but by using torques you can get rid of that! I would do first the torque part (the second half of the answer), then Newton's law and then the friction model formula (the first half).


*

*Find normal force $N_2$ by doing the torque balance around point A (now $N_1$ as well as friction $F$ doesn't matter):
$$
\sum \tau = 0 \quad\Leftrightarrow\quad
\tau_{N_2}-\tau_{w_x}-\tau_{w_y}= 0 \quad\Leftrightarrow\quad
N_2L-w_xh-w_y\frac{L}{2}= 0 \quad\Leftrightarrow\quad\\
N_2L-mg\sin(\theta)h-mg\cos(\theta)\frac{L}{2}= 0 \quad\Leftrightarrow\quad
N_2L=mg\left(\sin(\theta)h+\cos(\theta)\frac{L}{2}\right)
\quad\Leftrightarrow\quad\\
N_2=mg\left(\sin(\theta)\frac{h}{L}+\cos(\theta)\frac{1}{2}\right) 
$$

*Find friction $F$ with Newton's 1st law along the slope:
$$\sum F_x=0 \quad\Leftrightarrow\quad F-w_x=0 \quad\Leftrightarrow\quad F-mg\sin(\theta)=0 \quad\Leftrightarrow\quad F=mg\sin(\theta)$$

*And now find the critical angle $\theta$ from the friction model:
$$
F=\mu N_2 \quad\Leftrightarrow\quad\\
mg\sin(\theta)=\mu mg\left(\sin(\theta)\frac{h}{L}+\cos(\theta)\frac{1}{2}\right) 
\quad\Leftrightarrow\quad\\
\sin(\theta)=\mu \sin(\theta)\frac{h}{L}+ \cos(\theta)\frac{\mu}{2}
\quad\Leftrightarrow\quad\\
\sin(\theta)\left(1-\mu\frac{h}{L}\right)= \cos(\theta)\frac{\mu}{2}
\quad\Leftrightarrow\quad\\
\tan(\theta)\left(1-\mu \frac{h}{L}\right)=\frac{\mu}{2}
\quad\Leftrightarrow\quad\\
\tan(\theta)=\frac{\mu}{2\left(1-\mu \frac{h}{L}\right)}
\quad\Leftrightarrow\quad\\
\tan(\theta)=\frac{L\mu}{2\left(L-\mu h\right)}
\quad\Leftrightarrow\quad\\
\theta=\arctan\left(\frac{L\mu}{2\left(L-\mu h\right)}\right)
$$
$N_1$ is never introduced. That is why the torque balance is used. Newton's law in y and x directions could of course also be used, but might not be enough because they would introduce this fourth unknown $N_1$. Then you would need a fourth equation like the torque balance anyways.
