Topic: Rigid Bodies
Concepts: Fictitious forces, Statics

---

Solution:

We go to the accelerating frame to reduce the setup to a statics problem, introducing the fictitious inertial force $-M\vec{a}$ on the bicycle’s CM. Choosing the CM as the pivot point, we can balance torques:

Since $f_1=\mu N_1$ and $f_2=\mu N_2$,

Balancing forces in the vertical direction, $N_1+N_2=Mg$ so

Since $N_2-N_1 \leq Mg$ (realized when $N_1=0, N_2=Mg$):

so the answer is A.

Topic: Rigid Bodies
Concepts: Fictitious forces, Statics

---

Solution:

We go to the accelerating frame to reduce the setup to a statics problem, introducing the fictitious inertial force $-M\vec{a}$ on the bicycle’s CM. Choosing the contact point of the front wheel with the ground as the pivot point, we can balance torques:

using the fact that the normal force $N_1 \geq 0$. Then

so the answer is B.

Topic: Rigid Bodies
Concepts: Fictitious forces, Statics

---

Solution:

We go to the accelerating frame to reduce the setup to a statics problem, introducing the fictitious inertial force $-M\vec{a}$ on the bicycle’s CM. Choosing the contact point of the front wheel with the ground as the pivot point, we can balance torques:

using the fact that the normal force $N_1 \geq 0$. Then

Note this is the same derivation and answer as the previous problem since we don’t assume anything about the coefficients of friction. The answer is E.