# 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.

• If you write the 2nd Law of Newton for x components and for y components of the forces, then you will have a system of equations from which would get those relations. – rmhleo Jun 6 '16 at 12:31
• I get $mgcos(\theta)=N_{1}+N_{1}$ and $mgsin(\theta) = \mu N_{2}$. I still dont see where $\frac{h}{L}$ came from? – Lincoln77 Jun 6 '16 at 12:36
• Yes, I see. Not sure either why N1 and N2 are not taken as equal, since the center of mass is in the geometrical center. – rmhleo Jun 6 '16 at 13:04
• $N_1$ and $N_2$ do not have to be equal even if they are symmetrically placed around the centre of mass. If the bike was standing vertically on one wheel, that would still be symmetrical , but the bottom wheel would feel the whole normal force and the other wheel none. The trick in this question is to avoid the unknown $N_1$ - which is why they use the torque balance instead of just Newton's law several times. See the answer below. – Steeven Jun 6 '16 at 13:08
• Yes, the case you put is the only case in which this is possible, and involves standing in one wheel. If there are two points of support they definitely share the load, unless there is some weight distribution asymmetry. – rmhleo Jun 6 '16 at 13:15

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).

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

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

3. 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.

• This is it. I would only comment that the implicit assumption used of the friction being related only to N2 (there is only friction on the rear wheel) is not evident, and is in odds with establishing the center of mass in the geometrical center. – rmhleo Jun 6 '16 at 13:10
• @rmhleo True. I believe it is assumed that breaks are applied at the rear wheel while the front wheel is free to rotate (so no friction). Otherwise two frictions along the slope would be present, $F_1=\mu N_1$ and $F_2=\mu N_2$ . – Steeven Jun 6 '16 at 13:13
• hey thanks guys, this is really helpful :) though I still dont get how they got to tan at the end – Lincoln77 Jun 6 '16 at 13:25
• @Lincoln77 They got $\tan$ from the relation: $$\dfrac{\mu}2\cos\theta=\sin\theta\left(1-\mu\dfrac hL\right),$$ by dividing both sides by $\cos \theta$ and $1-\mu\dfrac hL$. – Workaholic Jun 6 '16 at 16:15
• ah isee, i tried that last night but i must have made a stupid mistake somewhere, thanks ! – Lincoln77 Jun 6 '16 at 22:06